Probability XIII: Gaussian Vectors

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---
template: inter-slide

# Gaussian Vectors

### 2021-12-03

#### [Probabilités Master I MIDS](http://stephane-v-boucheron.fr/courses/probability/)

#### [Stéphane Boucheron](http://stephane-v-boucheron.fr/)

---
template: inter-slide

## <svg aria-hidden="true" role="img" viewBox="0 0 576 512" style="height:1em;width:1.12em;vertical-align:-0.125em;margin-left:auto;margin-right:auto;font-size:inherit;fill:white;overflow:visible;position:relative;"><path d="M0 117.66v346.32c0 11.32 11.43 19.06 21.94 14.86L160 416V32L20.12 87.95A32.006 32.006 0 0 0 0 117.66zM192 416l192 64V96L192 32v384zM554.06 33.16L416 96v384l139.88-55.95A31.996 31.996 0 0 0 576 394.34V48.02c0-11.32-11.43-19.06-21.94-14.86z"/></svg>

### Univariate Gaussian Distributions

### Gaussian Vectors

### Gaussian Spaces and Independence

### Convergence of Gaussian Vectors

### Gaussian Conditionning

### Norms of Gaussian vectors

### Gaussian Concentration

---
template: inter-slide

## Univariate Gaussian distribution

---

### The standard Gaussian distribution

- Density

`$$\phi(x) = \frac{\mathrm{e}^{- \frac{x^2}{2} }}{\sqrt{2 \pi}}$$`

- .ttc[cumulative distribution function]

`$$\Phi(x) = \int_{-\infty}^x \phi(t) \mathrm{d}t$$`

- .ttc[survival function]

`$$\overline{\Phi}(x) = 1- \Phi(x) = \int_x^{\infty} \phi(t) \mathrm{d}t$$`

`$\mathcal{N} (0, 1)$` (expectation `$0$`, variance `$1$`) denotes
the standard Gaussian probability distribution, that is the probability distribution
defined by density `$\phi$`

---

Any affine transform of a standard Gaussian random variable is distributed according to a univariate
Gaussian distribution

If `$X \sim \mathcal{N} (0, 1)$` then

- `$\sigma X + \mu \sim \mathcal{N} \left( \mu, \sigma^2 \right)$`

- with density: `$\frac{1}{\sigma}\phi\left(\frac{\cdot- \mu}{\sigma}\right)$`

- with CDF:  `$\Phi\left(\frac{\cdot - \mu}{\sigma}\right)$`

---

---

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---

### Stein's identity

The standard Gaussian distribution is characterized by the next identity.

.bg-light-gray.b--light-gray.ba.bw1.br3.shadow-5.ph4.mt5[

Let `$X \sim \mathcal{N}(0,1)$`,

let `$g$` be an absolutely continuous function with derivative `$g'$` such that
`$\mathbb{E}[ |X g(X)|]<\infty$`,

then

`$g'(X)$` is integrable and

`$$\mathbb{E}[g'(X)] = \mathbb{E}[Xg(X)] \, .$$`

]

---

### Proof of Stein's identity

The proof relies on integration by parts (IPP).

First note that replacing `$g$` by `$g - g(0)$` changes neither `$g'$`,
nor  `$\mathbb{E}[Xg(X)]$`.

We may assume that `$g(0)=0$`.

`$$\begin{array}{rl} \mathbb{E}[Xg(X)] & = \int_{\mathbb{R}} xg(x) \phi(x) \mathrm{d}x \\& = \int_0^\infty xg(x) \phi(x) \mathrm{d}x + \int_{-\infty}^0 xg(x) \phi(x) \mathrm{d}x \\& = \int_0^\infty x \int_0^\infty g'(y) \mathbb{I}_{y\leq x}\mathrm{d}y \phi(x) \mathrm{d}x -\int^0_{-\infty} x \int^0_{-\infty} g'(y) \mathbb{I}_{y\geq x}\mathrm{d}y \phi(x) \mathrm{d}x\\& = \int_0^\infty g'(y) \int_0^\infty  \mathbb{I}_{y\leq x} x\phi(x)\mathrm{d}x  \mathrm{d}y -\int_{-\infty}^0 g'(y) \int^0_{-\infty}  x \phi(x)\mathbb{I}_{y\geq x}\mathrm{d}x \mathrm{d}y  \\& = \int_0^\infty g'(y) \int_y^\infty x\phi(x)\mathrm{d}x  \mathrm{d}y -
\int_{-\infty}^0 g'(y) \int^y_{-\infty}  x \phi(x)\mathrm{d}x \mathrm{d}y  \\& = \int_0^\infty g'(y) \phi(y) \mathrm{d}y -
\int_{-\infty}^0 - g'(y) \phi(y)\mathrm{d}y \\& = \int_{-\infty}^\infty g'(y) \phi(y) \mathrm{d}y\end{array}$$`

The last inequality is justified by Tonelli-Fubini's Theorem. Then, we rely on `$\phi'(x)=-x \phi(x)$`.

---

The characteristic function is a very efficient tool when handling Gaussian distributions.

.bg-light-gray.b--light-gray.ba.bw1.br3.shadow-5.ph4.mt5[

The characteristic function of  `$\mathcal{N}(\mu,\sigma^2)$`  is

`$$\widehat{\Phi}(t) = \mathbb{E}\left[\mathrm{e}^{\imath t X}\right] = \mathrm{e}^{\imath t \mu - \frac{t^2 \sigma^2}{2}}$$`
]

---

### Proof

It is enough to check the proposition for `$\mathcal{N}(0,1)$`. As  `$\phi$` is even,

`$$\begin{array}{rcl}\widehat{\Phi}(t) &= & \int_{-\infty}^{\infty} \mathrm{e}^{\imath t x} \frac{\mathrm{e}^{- \frac{x^2}{2}}}{\sqrt{2 \pi}} \mathrm{d} x \\& = & \int_{-\infty}^{\infty} \cos(tx) \frac{\mathrm{e}^{- \frac{x^2}{2}}}{\sqrt{2 \pi}} \mathrm{d} x\end{array}$$`

Derivation with respect to `$t$`, interchanging derivation and expectation  (why can we do that?)

`$$\begin{array}{rcl}\widehat{\Phi}'(t)  & = &    \int_{-\infty}^{\infty} -x \sin(tx) \frac{\mathrm{e}^{- \frac{x^2}{2}}}{\sqrt{2 \pi}} \mathrm{d} x\end{array}$$`

---

### Proof (continued)

Now relying on Stein's Identity with `$g(x)=-\sin(tx)$` and `$g'(x)=-t\cos(tx)$`

`$$\begin{array}{rcl}\widehat{\Phi}'(t)  & = & - t \int_{-\infty}^{\infty} \cos(tx) \phi(x) \mathrm{d} x \\ & = & -t \widehat{\Phi}(t)\end{array}$$`

We immediately get `$\widehat{\Phi}(0)=1$`, and solving the differential equation leads to

`$$\log \widehat{\Phi}(t) =  - \frac{t^2}{2}$$`

---

The fact that the characteristic function completely defines the probability distribution
provides us with a converse of Stein's Lemma.

.bg-light-gray.b--light-gray.ba.bw1.br3.shadow-5.ph4.mt5[

### Stein's Lemma (bis)

Let `$X$` be a real-valued  random variable  on some probability space.

If, for any differentialle function `$g$` such that  `$g'$` and `$x \mapsto xg(x)$` are integrable,
the following holds

`$$\mathbb{E}[g'(X)] = \mathbb{E}[X g(X)]$$`

then the distribution of `$X$` is standard Gaussian.

]

---

### Proof

Consider the real `$\widehat{F}$` and the imaginary part  `$\widehat{G}$` of  the characteristic
function of the distribution of `$X$`.

The identity entails that

`$$\widehat{F}'(t) = -t \widehat{F}(t) \quad \text{and} \quad \widehat{G}'(t) = -t \widehat{G}(t)$$`

with  `$\widehat{F}(0)=1$` and  `$\widehat{G}(0)=0$`

Solving the two differential equations leads to  `$\widehat{F}(t) = \mathrm{e}^{-t^2/2}$` and `$\widehat{G}(t)=0$`

We just checked that the characteristic function of the distribution of `$X$` is the characteristic function of  `$\mathcal{N}(0,1)$`

---

### The sum of two independent Gaussian random variables is a Gaussian random variable

.bg-light-gray.b--light-gray.ba.bw1.br3.shadow-5.ph4.mt5[

### Proposition

If  `$X$` and  `$Y$`  are two independent random variables distributed according to
`$\mathcal{N} (\mu, \sigma^2)$` and  `$\mathcal{N} (\mu', \sigma^{\prime 2})$`

then

`$$X + Y \sim \mathcal{N} \left(\mu + \mu', \sigma^2 + \sigma^{\prime 2}\right)$$`

]

---

### Moment generating function

`$$s \mapsto \mathbb{E} \left[ \mathrm{e}^{s X} \right] = \text{e}^{\frac{s^2}{2}}$$`

---

.bg-light-gray.b--light-gray.ba.bw1.br3.shadow-5.ph4.mt5[

### Proposition: Mill's ratios

For `$x \geq 0,$`

`$$\frac{\phi(x)}{x} \left( 1 - \frac{1}{x^2} \right)  \leq \overline{\Phi} (x) \leq \min \left( \mathrm{e}^{-\frac{x^2}{2}}, \frac{\phi(x)}{x} \right)$$`

]

---

### Proof

The proof boils down to repeated integration by parts.

`$$\begin{array}{rcl} \overline{\Phi}(x) & = & \int_x^{\infty}  \frac{1}{ \sqrt{2 \pi}} \mathrm{e}^{- \frac{u^2}{2}} \mathrm{d} u\\ & = & \left[ -  \frac{1}{ \sqrt{2 \pi} u} \mathrm{e}^{- \frac{u^2}{2}} \right]^{\infty}_x - \int_x^{\infty}  \frac{1}{ \sqrt{2 \pi}} \frac{1}{u^2} \mathrm{e}^{- \frac{u^2}{2}} \mathrm{d} u\end{array}$$`

As the second term is non-positive,

`$$\overline{\Phi}(x)\leq  \left[ -  \frac{1}{ \sqrt{2 \pi} u} \mathrm{e}^{- \frac{u^2}{2}} \right]^{\infty}_x =  \frac{\phi(x)}{x}$$`

This is the first part of the right-hand inequality, the other part
comes from Markov's inequality.

---

For the left-hand inequality, we have to upper bound

`$$\int_x^{\infty}  \frac{1}{ \sqrt{2 \pi}}  \frac{1}{u^2} \mathrm{e}^{-
\frac{u^2}{2}} \mathrm{d} u$$`

`$$\begin{array}{rcl} \int_x^{\infty}  \frac{1}{ \sqrt{2 \pi}}  \frac{1}{u^2} \mathrm{e}^{- \frac{u^2}{2}} \mathrm{d} u & = & \left[ \frac{- 1}{ \sqrt{2 \pi}} \frac{1}{u^3} \mathrm{e}^{- \frac{u^2}{2}} \right]_x^{\infty} - \int_x^{\infty}  \frac{1}{ \sqrt{2 \pi}}  \frac{3}{u^4} \mathrm{e}^{-\frac{u^2}{2}} \mathrm{d} u\\ & \leq  & \frac{1}{ \sqrt{2 \pi}}  \frac{1}{x^3} \mathrm{e}^{- \frac{x^2}{2}}\end{array}$$`

---

.bg-light-gray.b--light-gray.ba.bw1.br3.shadow-5.ph4.mt5[

### Proposition, Gaussian Moments

For a standard Gaussian random variable,

`$$\mathbb{E} \left[ X^k \right] = \begin{cases} 0 &  \text{ if } k \text{ is odd}\\ \frac{k!}{2^{k / 2} (k / 2) !} = \frac{\Gamma (k + 1)}{2^{k / 2} \Gamma (k / 2 + 1)} & \text{ if }  k \text{ is even.}\end{cases}$$`

]

---

### Proof

Thanks to distributional symmetry,
`$\mathbb{E} \left[ X^k \right]=0$`  for all odd `$k$`.

We handle even powers using integration by parts:

`$$\begin{array}{rcl} \mathbb{E} \left[ X^{k+2} \right] & = & (k+1) \mathbb{E} \left[ X^{k} \right]\end{array}$$`

Induction on  `$k$`  leads to,

`$$\begin{array}{rcl} \mathbb{E} \left[ X^{2k} \right] &= & \prod_{j=1}^k (2j-1) =  \frac{(2k) !}{2^k k! }\end{array}$$`

---

Note that  `$(2k)!/(2^k k!)$` is
also the number of partitions of  `$\{1, \ldots, 2k\}$`
into subsets of cardinality `$2$`.

---

The  _skewness_ is null,

the _kurtosis_ (ratio of fourth centred moment over
squared variance) equals `$3$`:

`$$\mathbb{E}[X^4] =  3 \times  \mathbb{E}[X^2]^2$$`

---
template: inter-slide

## Gaussian vectors

---

A Gaussian vector is a collection of univariate Gaussian random variables that
satisfies a very stringent property:

.bg-light-gray.b--light-gray.ba.bw1.br3.shadow-5.ph4.mt5[

### Definition, Gaussian vector

A random vector `$X = (X_1, \ldots, X_n)^T$` is a _Gaussian vector_

iff

for any real vector  `$\lambda = (\lambda_1, \lambda_2, \ldots, \lambda_n)^T$`, the distribution of
the univariate random variable

`$$\langle \lambda, X\rangle = \sum_{i = 1}^n \ \lambda_i X_i$$`

is Gaussian.

]

---

The random vector `$(X, \epsilon X)$` with `$X \sim \mathcal{N}(0.1)$`, independent of `$\epsilon$` which is worth `$\pm 1$` with probability `$1/2$`, is not a Gaussian vector although both `$X$` and `$\epsilon X$` are univariate Gaussian random variables.

---

.bg-light-gray.b--light-gray.ba.bw1.br3.shadow-5.ph4.mt5[

### Proposition

If  `$X_1, \ldots, X_n$` is a sequence of independent Gaussian random variables,

then

`$$(X_1, \ldots, X_n)^t = \begin{pmatrix}X_1 \\ \vdots \\ X_n\end{pmatrix}$$`

is a Gaussian vector.

]

---

In the sequel,  a _standard Gaussian vector_ is a random vector with independent coordinates
with each coordinate distributed according to `$\mathcal{N}(0,1)$`.

---

We will see how to construct general Gaussian vectors.

Before this, let us check that the joint distribution of a Gaussian random vector
is completely characterized by its covariance matrix and its expectation vector.

---

Recall that the  _covariance_ of random vector `$X= (X_1, \ldots, X_n)^T$` is the matrix
`$K$` with dimension `$n \times n$` with coefficients

`$$K [i, j] = \operatorname{Cov} (X_i, X_j) = \mathbb{E} [X_i X_j] - \mathbb{E} [X_i] \mathbb{E} [X_j] .$$`

Without loss of generality, we may assume that random vector `$X$`  is centered

For every  `$\lambda = (\lambda_1, \ldots, \lambda_n)^T \in \mathbb{R}^n$`, we have:

`$$\operatorname{var}(\langle \lambda, X \rangle) =  \lambda^t K \lambda = \text{trace} (K \lambda \lambda^t)\,$$`

---

Indeed,

`$$\begin{array}{rcl} \operatorname{var}(\langle \lambda, X \rangle) & = & \mathbb{E} \left[ \left( \sum_{i=1}^n \lambda_i X_i\right)^2\right]  \\ & = & \sum_{i,j=1}^n \mathbb{E} \left[\lambda_i \lambda_j X_i X_j \right] \\ & = & \sum_{i,j=1}^n \lambda_i \lambda_j K[i,j] \\ & = & \lambda^t K \lambda\end{array}$$`

The characteristic function of a Gaussian vector  `$X$` with expectation vector  `$\mu$`  and covariance `$K$`
satisfies

`$$\mathbb{E} \mathrm{e}^{\imath \langle \lambda, X \rangle } =  \mathrm{e}^{\imath \langle \lambda, \mu \rangle - \frac{\lambda^t K \lambda}{2}}$$`

---

A linear transform of a Gaussian vector is a Gaussian vector.

.bg-light-gray.b--light-gray.ba.bw1.br3.shadow-5.ph4.mt5[

### Proposition

If  `${Y} = (Y_1, \ldots, Y_n)^T$` is a Gaussian vector with
covariance `$K$` and `$A$` a real matrix with dimensions `$p \times n$`, then `$A \times Y$` is Gaussian vector
with expectation `$A \times \mathbb{E}Y$` and covariance matrix

`$$A K A^T$$`

]

---

### Proof

Without loss of generality, we assume  `$Y$`  is centred.

For any  `$\lambda \in \mathbb{R}^p$`,

`$$\langle \lambda , A Y \rangle = \langle A^T \lambda, Y \rangle$$`

thus `$A \times Y$`  is  Gaussian with variance

`$$\lambda^T A K A^T \lambda$$`

The covariance of `$A \times Y$` is determined by this observation.

---

To manufacture Gaussian vectors with general covariance matrices, we rely
on an important notion from matrix analysis.

.bg-light-gray.b--light-gray.ba.bw1.br3.shadow-5.ph4.mt5[

### Definition: Semi-definite positive matrices

A symmetric matrix  `$M$` with dimensions `$k \times k$`  is  Definite Positive
(respectively Semi-Definite Positive) iff, for any non-null vector `$v \in \mathbb{R}^k$`,

`$$v^T M v > 0 \qquad (\text{resp.} \qquad v^T M v \geq 0)$$`

We denote by  `$\textsf{dp}(k)$` (resp. `$\textsf{sdp}(k)$`), the cones
of Definite Positive (resp. Semi-Definite Positive) matrices.

]

---

.bg-light-gray.b--light-gray.ba.bw1.br3.shadow-5.ph4.mt5[

### Proposition

If `$K$` is the covariance matrix of a random vector, `$K$`
is symmetric, Semi-Definite Positive.

]

---

### Proof

If  `$X$`  is a  `$\mathbb{R}^k$`-valued random vector, with covariance `$K$`,
for any vector  `$\lambda \in \mathbb{R}^n$`,

`$$\lambda^T K \lambda = \sum_{i,j\leq k} K_{i,j} \lambda_i \lambda_j = \operatorname{cov}(\langle \lambda, X \rangle, \langle \lambda, X \rangle)$$`

soit  `$\lambda^T K \lambda =  \operatorname{var}(\langle \lambda, X \rangle)$`. The variance of a univariate random variable is always non-negative.

---

The next  observation is the key to the construction to general Gaussian vectors.

.bg-light-gray.b--light-gray.ba.bw1.br3.shadow-5.ph4.mt5[

### Proposition: Cholesky's factorization

If `$A$` is a  Semi-definite  Positive symmetric matrix then there exists
(at least) a real matrix `$B$` such that `$A = B^T B$`.

]

---

We do not check this proposition. This is a basic Theorem from  matrix analysis.

It can be established from the _spectral decomposition theorem_ for symmetric matrices.

It can also be established by a simple constructive approach:

A positive definite  matrix `$K$` admits a _Cholesky decomposition_, in other words, there exists a triangular matrix lower than `$L$` such that
`$$K = L \times L^T$$`

---

The next proposition is a corollary of the general formula for image densities.

.bg-light-gray.b--light-gray.ba.bw1.br3.shadow-5.ph4.mt5[

### Proposition

If  `$A$` is a symmetric positive definite matrix  ( `$A \in \textsf{dp}(n)$` ),

then

the distribution `$\mathcal{N}(0, A)$` of the centred Gaussian vector with covariance matrix `$A$`
is absolutely continuous with respect to Lebesgue's measure on `$\mathbb{R}^n$`, with density

`$$\frac{1}{({2 \pi})^{n/2}  \operatorname{det}(A)^{1/2}} \exp\left( - \frac{x^t A^{-1} x}{2} \right)$$`

]

---

### Proof

The density formula is trivially correct for standard Gaussian vectors.

For the general case, it is enough to invoke the image density formula to the image of the standard Gaussian vector
by the bijective linear transformation defined by the Cholesky factorization of `$A$`.

The determinant of the Cholesky factor is the square root of the determinant of `$A$`.

---

Is the distribution of a Gaussian vector `$X$` with _singular_ covariance matrix absolutely continuous
with respect to Lebesgue measure?

---
template: inter-slide

## Gaussian spaces and independence

---

.bg-light-gray.b--light-gray.ba.bw1.br3.shadow-5.ph4.mt5[

### Definition: Gaussian space

If `$X= (X_1, \ldots, X_n)^T$` is a centered Gaussian vector with covariance matrix `$K$` ,
the set

`$$\Big\{ \sum_{i = 1}^n \lambda_i X_i = \langle \lambda, X\rangle ; \lambda \in \mathbb{R}^n\Big\}$$`

is the Gaussian space generated by  `$X = (X_1, \ldots, X_n)^T$`

]

---

The Gaussian space is a real vector space.

If `$(\Omega, \mathcal{F},P)$` denotes the  probability space, `$X$` lives on,
the Gaussian space is a subspace of  `$L^2_{\mathbb{R}}(\Omega, \mathcal{F},P)$`.

It inherits the inner product structure from `$L^2_{\mathbb{R}}(\Omega, \mathcal{F},P)$`.

This inner-product is completely defined by the covariance matrix  `$K$`.

---

`$$\begin{array}{rcl} \left\langle \sum_{i = 1}^n \lambda_i X_i, \sum_{i = 1}^n \lambda'_i X_i \right\rangle & \equiv & \mathbb{E}_P \left[ \left( \sum_{i = 1}^n \lambda_i X_i \right) \left( \sum_{i = 1}^n \lambda'_i X_i \right) \right]\\ & = & \sum^n_{i, i' = 1} \lambda_i \lambda_{i'}'  K [i, i']\\ & = & (\lambda_1, \ldots, \lambda_n) K \left(\begin{array}{c} \lambda'_1\\
\vdots\\ \lambda'_n \end{array}\right) \\ & = & \text{trace} \left( K \left(\begin{array}{c} \lambda_1\\
\vdots\\ \lambda_n \end{array}\right) \left(\begin{array}{ccc} \lambda'_1 & \dots & \lambda'_n \end{array}\right) \right)\\ & = & \left\langle K, \left(\begin{array}{c} \lambda_1\\
\vdots\\ \lambda_n \end{array}\right) \left(\begin{array}{ccc} \lambda'_1 & \dots & \lambda'_n \end{array}\right)\right\rangle_{\text{HS}}\end{array}$$`

---

Different Gaussian vectors may generate the same Gaussian space. Explain how and why.

---

Gaussian spaces enjoy remarkable properties.

Independence of random variables belonging to the same Gaussian space may be checked very easily.

.bg-light-gray.b--light-gray.ba.bw1.br3.shadow-5.ph4.mt5[

### Proposition: independence in Gaussian space

Two random variables  `$Z$` and `$Y$`, belonging to the same Gaussian space, are independent

iff

they are orthogonal (or decorrelated), that is

iff

`$$\operatorname{Cov}_P [Y ,Z] = \mathbb{E}_P [Y Z] = 0 .$$`

]

Without loss of generality, we assume covariance matrix `$K$` is positive definite.

---

### Proof

Independence always implies orthogonality.

Without loss of generality, we assume that the Gaussian space is
generated by a standard Gaussian vector,  let
`$Z = \sum_{i = 1}^n \lambda_i X_i$` and  `$Y = \sum_{i = 1}^n \lambda'_i X_i$`.

If  `$Z$` and `$Y$` are orthogonal (or non-correlated)

`$$\mathbb{E} [ZY] =  \sum_{i = 1}^n \lambda_i \lambda_{i}' = 0$$`

To show that  `$Z$` and  `$Y$` are independent, it is enough to check that
for all `$\mu$` and  `$\mu'$` in `$\mathbb{R}$`

`$$\mathbb{E} \left[ \mathrm{e}^{\imath \mu Z} \mathrm{e}^{\imath \mu' Y} \right] = \mathbb{E} \left[ \mathrm{e}^{\imath \mu Z}  \right] \times \mathbb{E} \left[ \mathrm{e}^{\imath \mu' Y} \right]$$`

---

### Proof (continued)

.small[

`$$\begin{array}{rcl}
\mathbb{E} \left[ \mathrm{e}^{\imath \mu Z} \mathrm{e}^{\imath \mu' Y} \right] &
= & \mathbb{E} \left[ \mathrm{e}^{\imath \mu \sum_i \lambda_i    X_i}
\mathrm{e}^{\imath \mu' \sum_i \lambda'_i    X_i} \right]\\
& = & \mathbb{E} \left[ \prod_{i = 1}^n    \mathrm{e}^{\imath (\mu
\lambda_i + \mu' \lambda'_i)    X_i} \right] \qquad  (X_i \text{ are  independent} \ldots)\\
& = & \prod_{i = 1}^n  \mathbb{E} \left[ \mathrm{e}^{\imath (\mu \lambda_i +
\mu' \lambda'_i)    X_i} \right]\\
& = & \prod_{i = 1}^n \mathrm{e}^{- (\mu \lambda_i + \mu' \lambda'_i)
^2 / 2}\\
& = & \exp \left( - \frac{1}{2} \sum_{i = 1}^n \mu^2 \lambda_i^2 + 2 \mu
\mu' \lambda_i \lambda'_i + \mu'^2 \lambda'^2_i    \right)\qquad  (\text{orthogonality})\\
& = & \exp \left( - \frac{1}{2} \sum_{i = 1}^n \mu^2 \lambda_i^2 + \mu'^2
\lambda'^2_i    \right)\\
&  & \ldots\\
& = & \mathbb{E} \left[ \mathrm{e}^{\imath \mu Z}  \right] \times
\mathbb{E} \left[ \mathrm{e}^{\imath \mu^\prime Y} \right]\end{array}$$`

]

---

.bg-light-gray.b--light-gray.ba.bw1.br3.shadow-5.ph4.mt5[

### Corollary

If `$E$` and `$E^\prime$` are two linear sub-spaces of the Gaussian space
generated by the Gaussian vector with  independent coordinates
`$X_1, \ldots, X_n$`,

the (Gaussian) random  variables belonging  to subspace  `$E$`
and the random (Gaussian) variables belonging to the `$E^\prime$` space are independent
if and only these two subspaces are orthogonal.

`$$\left(\forall (X, Y) \in E \times E', \quad X \perp Y \right) ⇔ \left(\left(\forall (X, Y) \in E \times E', \quad X \perp\!\!\!\perp Y \right)\right)$$`

]

---
template: inter-slide

## Convergence of Gaussian vectors

---

.bg-light-gray.b--light-gray.ba.bw1.br3.shadow-5.ph4.mt5[

### Lévy continuity theorem <svg aria-hidden="true" role="img" viewBox="0 0 512 512" style="height:1em;width:1em;vertical-align:-0.125em;margin-left:auto;margin-right:auto;font-size:inherit;fill:currentColor;overflow:visible;position:relative;"><path d="M201.5 174.8l55.7 55.8c3.1 3.1 3.1 8.2 0 11.3l-11.3 11.3c-3.1 3.1-8.2 3.1-11.3 0l-55.7-55.8-45.3 45.3 55.8 55.8c3.1 3.1 3.1 8.2 0 11.3l-11.3 11.3c-3.1 3.1-8.2 3.1-11.3 0L111 265.2l-26.4 26.4c-17.3 17.3-25.6 41.1-23 65.4l7.1 63.6L2.3 487c-3.1 3.1-3.1 8.2 0 11.3l11.3 11.3c3.1 3.1 8.2 3.1 11.3 0l66.3-66.3 63.6 7.1c23.9 2.6 47.9-5.4 65.4-23l181.9-181.9-135.7-135.7-64.9 65zm308.2-93.3L430.5 2.3c-3.1-3.1-8.2-3.1-11.3 0l-11.3 11.3c-3.1 3.1-3.1 8.2 0 11.3l28.3 28.3-45.3 45.3-56.6-56.6-17-17c-3.1-3.1-8.2-3.1-11.3 0l-33.9 33.9c-3.1 3.1-3.1 8.2 0 11.3l17 17L424.8 223l17 17c3.1 3.1 8.2 3.1 11.3 0l33.9-34c3.1-3.1 3.1-8.2 0-11.3l-73.5-73.5 45.3-45.3 28.3 28.3c3.1 3.1 8.2 3.1 11.3 0l11.3-11.3c3.1-3.2 3.1-8.2 0-11.4z"/></svg>

A sequence of probability distributions `$(P_n)_{n \in \mathbb{N}}$` sur `$\mathbb{R}^k$`
converges weakly towards a probability distribution

iff

there exists a function  `$f$` over  `$\mathbb{R}^k$`, continuous at `$\vec{0}$`, such that for all `$\vec{s} \in \mathbb{R}^k$`:

`$$\mathbb{E}_{P_n} \left[ \mathrm{e}^{\imath \langle \vec{s}, \vec{X} \rangle} \right] \rightarrow f(\vec{s})$$`

Then, function `$f$` is   the characteristic function of some probability distribution `$P$`.

]

The continuity condition at  `$0$`  is necessary: the characteristic function of a probability distribution is always continuous at `$0$`.

Continuity at `$0$` warrants the _tightness_ of the sequence of probability distributions.

---

.bg-light-gray.b--light-gray.ba.bw1.br3.shadow-5.ph4.mt5[

### Proposition

If a sequence of `$k$`-dimensional Gaussian vectors `$(X_n)$` is defined by a `$\mathbb{R}^k$`-valued sequence  `$(\vec{\mu}_n)_n$` and a `$\textsf{SDP}(k)$`-valued sequence  `$(K_n)_n$` and

`$$\begin{array}{rcl}\lim_n \vec{\mu}_n & = & \mu \in \mathbb{R}^k\\ \lim_n K_n & = & K \in \textsf{SDP}(k)\end{array}$$`

then

the sequence  `$(X_n)_n$`  converges in distribution towards `$\mathcal{N}\left(\vec{\mu}, K\right)$` (if `$K = 0$`,
the limit distribution is `$\delta_\mu$`).

]

---
template: inter-slide

## Gaussian conditioning

---

Let `$(X_1,\ldots,X_n)^T$`  be a Gaussian vector with distribution `$\mathcal{N}(\mu, K)$` where   `$K \in \textsf{DP}(n)$`.

The  covariance matrix  `$K$`  is  partitioned into blocks

`$$K = \left[\begin{array}{cc} A & B^t \\ B & W \end{array}\right]$$`

where  `$A \in \textsf{DP}(k)$`, `$1 \leq k < n$`, and `$W \in \textsf{DP}(n-k)$`.

We are interested in the conditional expectation of  `$(X_1, \ldots, X_k)^T$`  with repsect to `$\sigma(X_{k+1},\ldots,X_n)$`
and in the conditional distribution of `$(X_1, \ldots, X_k)^T$`  with respect to  `$\sigma(X_{k+1},\ldots,X_n)$`.

The Schur complement of `$A$`  in `$K$`  is defined as

`$$W - B A^{-1} B^T\, .$$`

This definition makes sense for symmetric matrices when `$A$` is non-singular.

If `$K \in \textsf{DP}(n)$`  then the Schur complement of  `$A$` in  `$K$` also belongs to `$\textsf{DP}(n-k)$`

---

In the statement of  the next theorems,  `$A^{-1/2}$` denotes the Cholesky factor
of  `$A^{-1}$`:   `$A^{-1} = A^{-1/2} \times (A^{-1/2})^T$`.

.bg-light-gray.b--light-gray.ba.bw1.br3.shadow-5.ph4.mt5[

### Theorem

The conditional expectation

`$(X_{k+1}, \ldots, X_n)^t$`  with respect to `$(X_{1},\ldots,X_k)^t$` is an affine transformation of `$(X_{1},\ldots,X_{k})^t$`:

`$$\mathbb{E}\left[ \begin{pmatrix} X_{k+1} \\ \vdots \\ X_{n}\end{pmatrix} \mid \begin{matrix}
X_{1} \\ \vdots \\ X_k \end{matrix}\right] = \begin{pmatrix} \mu_{k+1} \\ \vdots \\ \mu_n \end{pmatrix}   + \left(B A^{-1}  \right) \times  \left( \begin{pmatrix} X_{1} \\ \vdots \\ X_{k} \end{pmatrix}  - \begin{pmatrix} \mu_{1} \\ \vdots \\ \mu_k\end{pmatrix}\right)$$`

]

---

.bg-light-gray.b--light-gray.ba.bw1.br3.shadow-5.ph4.mt5[

### Theorem

The conditional distribution of `$(X_{k+1}, \ldots, X_n)^T$` with respect
to `$\sigma(X_{1},\ldots,X_k)$` is a Gaussian distribution with

- expectation: the conditional expectation `$(X_{k+1}, \ldots, X_n)^T$` with respect to `$\sigma(X_{1},\ldots,X_k)$`
- covariance: the Schur complement of the covariance of `$(X_{1},\ldots,X_k)^T$`
in the covariance matrix of `$(X_1, \ldots, X_n)^T$`.

]

---

We will first study the conditional density, and,  with a minimum amount of calculation, establish that it is Gaussian. Conditional expectation will be calculated as expectation under conditional distribution.

---

To characterize conditional density, we rely on a distributional representation argument (any Gaussian vector is distributed as the image of a standard Gaussian vector by an affine transformation) and a matrix analysis result that is at the core of the Cholesky factorization of positive semi-definite matrices.

`$(X_1, \ldots, X_n)^T$` is distributed as the image of standard Gaussian vector
by a block triangular matrix

Then we use standard properties of conditional distributions
in order to prove both Theorems

---

.bg-light-gray.b--light-gray.ba.bw1.br3.shadow-5.ph4.mt5[

### Proposition

Let `$K$` be a symmetric definite positive matrix with dimensions
`$n \times n$`

`$$K = \left[ \begin{array}{cc} A & B^t \\ B & W\end{array} \right]$$`

where  `$A$` has dimensions  `$k \times k$`, `$1 \leq k < n$`.

Then,  the Schur-complement of `$A$` with respect to  `$K$`

`$$W - B A^{-1} B^t$$`

is positive definite ...

]

---

.bg-light-gray.b--light-gray.ba.bw1.br3.shadow-5.ph4.mt5[

### Proposition (continued)

Sub-matrices `$A$` and   `$W - B A^{-1} B^t$` both have a Cholesky decomposition

`$$A = L_1 L_1^t \qquad  W - B A^{-1} B^t = L_2 L_2^t$$`

where   `$L_1, L_2$`  are lower triangular.

The factorization of `$K$` reads like:

`$$K = \left[ \begin{array}{cc} L_1 & 0 \\ B (L_1^t)^{-1}  & L_2 \end{array} \right] \times \left[\begin{array}{cc} L_1^t & L_1^{-1} B^t \\
0 & L_2^t\end{array}\right]$$`

]

---

### Proof

Without loss of generality, we check the statement on centered vectors.
The Cholesky factorization of `$K$` allows us to write

`$$\begin{pmatrix} X_1 \\ \vdots \\
X_n \end{pmatrix} \sim \left[ \begin{array}{cc}
L_1 & 0 \\ B (L_1^t)^{-1}  & L_2 \end{array} \right] \times \begin{pmatrix} Y_1 \\ \vdots \\ Y_n \end{pmatrix}$$`

where  `$( Y_1, \ldots, Y_n)^t$` is a centered standard Gaussian vector.

In the sequel,  we assume  `$(X_1, \ldots,X_n)^T$`  and
`$(Y_1,\ldots,Y_n)^T$` live on the same probability space.

As `$L_1$` is  invertible, the `$\sigma$`-algebras generated by  `$(X_1, \ldots,X_k)^T$` and
`$(Y_1, \ldots,Y_k)^T$` are equal. We agree on  `$\mathcal{G}=\sigma(X_1, \ldots,X_k)$`.
The conditional expectations and conditional distributions also coincide .

---

`$$\begin{array}{rcl}\mathbb{E} \left[ \begin{pmatrix} X_{k+1} \\ \vdots \\ X_n
\end{pmatrix} \mid \mathcal{G} \right] &= &\mathbb{E} \left[ B (L_1^t)^{-1} \begin{pmatrix} Y_{1} \\ \vdots \\ Y_k \end{pmatrix} \mid \mathcal{G} \right] + \mathbb{E} \left[ L_2 \begin{pmatrix} Y_{k+1} \\ \vdots \\ Y_n \end{pmatrix} \mid \mathcal{G} \right]  \\ & = & B (L_1^t)^{-1} L_1^{-1}\begin{pmatrix} X_{1} \\ \vdots \\ X_k\end{pmatrix} =  B A^{-1} \begin{pmatrix}X_{1} \\\vdots \\ X_k\end{pmatrix} \, ,
\end{array}$$`

as `$(Y_{k+1}, \ldots,Y_n)^t$` is centered and independent from `$\mathcal{G}$`.

---

Note that residuals

`$$\begin{pmatrix} X_{k+1} \\ \vdots \\ X_n \end{pmatrix} -\mathbb{E} \left[ \begin{pmatrix} X_{k+1} \\\vdots \\ X_n\end{pmatrix} \mid \mathcal{G} \right] = L_2 \begin{pmatrix} Y_{k+1} \\ \vdots \\ Y_n \end{pmatrix}$$`

are independent from  `$\mathcal{G}$`. This is a Gaussian property. <svg aria-hidden="true" role="img" viewBox="0 0 496 512" style="height:1em;width:0.97em;vertical-align:-0.125em;margin-left:auto;margin-right:auto;font-size:inherit;fill:currentColor;overflow:visible;position:relative;"><path d="M248 8C111 8 0 119 0 256s111 248 248 248 248-111 248-248S385 8 248 8zm80 168c17.7 0 32 14.3 32 32s-14.3 32-32 32-32-14.3-32-32 14.3-32 32-32zm-160 0c17.7 0 32 14.3 32 32s-14.3 32-32 32-32-14.3-32-32 14.3-32 32-32zm194.8 170.2C334.3 380.4 292.5 400 248 400s-86.3-19.6-114.8-53.8c-13.6-16.3 11-36.7 24.6-20.5 22.4 26.9 55.2 42.2 90.2 42.2s67.8-15.4 90.2-42.2c13.4-16.2 38.1 4.2 24.6 20.5z"/></svg>

The conditional distribution of  `$(X_{k+1},\ldots, X_n)^T$` with respect to `$(X_1,\ldots, X_k)^T$`
coincides with the conditional distribution of

`$$B (L_1^t)^{-1}   \times \begin{pmatrix} Y_1\\ \vdots \\ Y_k \end{pmatrix} + L_2 \times \begin{pmatrix}
Y_{k+1}\\ \vdots \\ Y_n \end{pmatrix}$$`

conditionally on   `$(Y_1,\ldots, Y_k)^T$`.

---

As  `$(Y_1,\ldots, Y_k)^t = L_1^{-1}(X_1,\ldots,X_k)^T$`, the conditional distribution we are looking for is Gaussian
with expectation

`$$B A^{-1}   \times \begin{pmatrix} X_1\\ \vdots \\ X_k \end{pmatrix}$$`

(the conditional expectation) and variance
`$L_2 \times L_2^t = W - B A^{-1} B^t$`.

---

If  `$(X,Y)^T$`  is a  centered Gaussian vector with
- `$\operatorname{var}(X)=\sigma_x^2$`,
- `$\operatorname{var}(Y)=\sigma^2_y$` and
- `$\operatorname{cov}(X,Y)= \rho \sigma_x \sigma_y$`,

the conditional distribution of  `$Y$` with respect to `$X$` is

`$$\mathcal{N}\left( \rho \sigma_y/\sigma_x X, \sigma^2_y (1- \rho^2)  \right)$$`

The quantity `$\rho$` is called  the _linear correlation coefficient_ between  `$X$` and  `$Y$`.

By the Cauchy-Schwarz Inequality, `$\rho \in [-1,1]$`.

---

These two theorems are usually addressed in the order in which they are stated.

Conditional expectation is characterized by adopting the `$L^2$` (predictive) viewpoint:

> the conditional expectation of the random vector `$Y$` knowing `$X$` is defined as the best `$X$`-measurable predictor of the vector `$Y$`  with respect to  quadratic error (the random vector `$Z$`, `$X$`-measurable that minimizes `$\mathbb{E} \left[ \| Y- Z\|^2 \right]$`).

---

In order to characterize conditional expectation,
we first compute the optimal affine predictor of
`$(X_{k+1},\ldots,X_n)^T$`  based on `$(X_{1},\ldots,X_k)^T$`.

This optimal affine predictor is

`$$\begin{pmatrix} \mu_{k+1} \\ \vdots \\ \mu_n \end{pmatrix}   + \left(B A^{-1}  \right) \times  \left( \begin{pmatrix}
X_{1} \\ \vdots \\ X_{k} \end{pmatrix}  - \begin{pmatrix} \mu_{1} \\ \vdots \\ \mu_k \end{pmatrix}\right)$$`

---

The optimal affine predictor is a Gaussian vector.

One can check that the residual vector

`$$\begin{pmatrix} X_{k+1}\\ \vdots \\ X_n \end{pmatrix} - \left\{ \begin{pmatrix} \mu_{k+1} \\ \vdots \\ \mu_n \end{pmatrix}   + \left(B A^{-1}\right) \times  \left( \begin{pmatrix}
X_{1} \\ \vdots \\ X_{k}
\end{pmatrix}  - \begin{pmatrix}
\mu_{1} \\ \vdots \\ \mu_k
\end{pmatrix}\right) \right\}$$`

is also Gaussian and orthogonal to the affine predictor. The residual vector  is independent
from the  affine predictor.

---

This is enough to establish that the affine predictor is the orthogonal projection of
`$(X_{k+1}, \ldots, X_n)^T$` on the closed linear subspace of  square-integrable
`$(X_{1},\ldots,X_k)^T$`-measurable random vectors.

This proves that the affine predictor is the conditional expectation.

---

We dealt with a special case of linear conditioning.

To figure out general linear conditioning, consider  `$X \sim \mathcal{N}(0, {K})$`
(we assume centering to alleviate notation and computations, translating does not change the relevant `$\sigma$`-algebras and thus conditioning), where `${K} \in \textsf{DP}(n)$`, and a linear transformation  defined by matrix
`$H$` with dimensions `$m \times n$`. `$H$` is assumed to have rank `$m$`. Agree on `$Y= {H} X$`.
Considering the Gaussian vector `$[ X^T : Y^T]$` with covariance matrix

`$$\left[ \begin{array}{cc} {K} & {K} {H}^t \\ {H}{K} & {H} {K} {H}^t \end{array} \right]$$`

and adapting the previous computations (the covariance matrix is not positive definite any more),
we may check that the conditional distribution of `$X$` with respect to `$Y$` is Gaussian with expectation
$$  K H^T (HKH^T)^{-1}$$ and variance $$  K - K H^t (HKH^T)^{-1} H K \, .$$

The linearity of conditional expectation is a property of Gaussian vectors and linear conditioning.
If you condition with respect to the norm  `$\| X\|_2$`, the conditional distribution is not Gaussian anymore.

---
template: inter-slide

## Gamma distributions

---

Investigating the norm of Gaussian vectors will prompt us
to introduce  `$\chi^2$` distributions, a sub-family of Gamma distributions.

.bg-light-gray.b--light-gray.ba.bw1.br3.shadow-5.ph4.mt5[

### Definition: Gamma Distributions

A Gamma distribution with parameters `$(p, \lambda)$`}
( `$\lambda \in \mathbb{R}_+$` and  `$p \in \mathbb{R}_+$` ),
is a distribution on `$(\mathbb{R}_+, \mathcal{B}(\mathbb{R}_+))$` with density

`$$g_{p, \lambda} (x) = \frac{\lambda^p}{\Gamma (p)} \mathbf{1}_{x \geq 0} x^{p - 1} e^{- \lambda x}$$`

where  `$\Gamma (p) =\int_0^{\infty} t^{p - 1} e^{- t} \mathrm{d} t$`

- `$p$` is called the _shape_ parameter,
- `$\lambda$` is called the _rate_  or _intensity_  parameter,
- `$1/\lambda$` is called the _scale_ parameter

]

---

If  `$X \sim  \text{Gamma}(p,1)$` then `$\sigma X \sim \text{Gamma}(p,1/\sigma)$` for  `$\sigma>0$`

The Euler  `$\Gamma ()$` function  interpolates the factorial.

For every  positive real `$p$`, `$\Gamma (p + 1) = p \Gamma(p)$`

If  `$p$` is integer,  `$\Gamma (p + 1) = p!$`

---

.bg-light-gray.b--light-gray.ba.bw1.br3.shadow-5.ph4.mt5[

### Proposition

If  `$X \sim \mathrm{Gamma}(p, \lambda)$`
`$\mathbb{E}X = \frac{p}{\lambda}$` and `$\operatorname{var}(X) = \frac{p}{\lambda^2}$`.

]

---

The sum of two independent Gamma-distributed random variables is Gamma distributed
if they have the same intensity (or scale) parameter.

.bg-light-gray.b--light-gray.ba.bw1.br3.shadow-5.ph4.mt5[

### Proposition

If `$X ⟂\!\!\!⟂ Y$` are independent Gamma-distributed random variables
with the same intensity parameter `$\lambda$`:  `$X \sim \mathrm{Gamma}(p, \lambda), Y\sim \mathrm{Gamma}(q, \lambda)$`

then

`$$X + Y \sim \mathrm{Gamma}(p+q, \lambda)$$`

]

---

### Proof

The density of the distribution of `$X+Y$` is the convolution of the densities
 `$g_{p, \lambda}$` et `$g_{q, \lambda}$`.
`$$\begin{array}{rcl}
g_{p, \lambda} \ast g_{q, \lambda} (x) & = & \int_{\mathbb{R_{}}}
g_{p, \lambda} (z) g_{_{q, \lambda}} (x - z) \mathrm{d} z\\
& = & \int_0^x    g_{p, \lambda} (z) g_{_{q,
\lambda}} (x - z) \mathrm{d} z\\
& = & \int_0^x  \frac{\lambda^p}{\Gamma (p)} z^{p - 1} \mathrm{e}^{- \lambda
z}     \frac{\lambda^q}{\Gamma (q)} (x - z)^{q - 1} \mathrm{e}^{-
\lambda (x - z)} \mathrm{d} z\\
& = & \frac{\lambda^{p + q}}{\Gamma (p) \Gamma (q)} \mathrm{e}^{- \lambda x}
\int_0^x z^{p - 1} (x - z)^{q - 1}    \mathrm{d} z\\
&  & \operatorname{changement} \operatorname{de} \operatorname{variable} z = x u\\
& = & \frac{\lambda^{p + q}}{\Gamma (p) \Gamma (q)} \mathrm{e}^{- \lambda x}
x^{p + q - 1} \int_0^{1} u^{p-1} (1 - u)^{q - 1}
\mathrm{d} u\\
& = & g_{p + q, \lambda} (x) \frac{\Gamma(p+q)}{\Gamma(p)\Gamma(q)} \int_0^{1} u^{p-1} (1 - u)^{q - 1}
\mathrm{d} u\end{array}$$`

---

We may pocket the next observation:

`$$B(p,q):=   \int_0^{1} u^{p-1} (1 - u)^{q - 1}\mathrm{d} u$$`

satisfies  `$B(p,q) = \frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}.$`

---

Gamma distributions  with parameters  `$(k / 2, 1 / 2)$` for `$k \in \mathbb{N}$` deserve to be named

.bg-light-gray.b--light-gray.ba.bw1.br3.shadow-5.ph4.mt5[

### Proposition: Chi-square distributions

The  `$\chi^2$` distribution with `$k$` degrees of freedom, denoted by `$\chi^2_k$`  has density

`$$\mathbb{I}_{x>0} \frac{x^{ \frac{1}{2} (k - 2)}}{2^{k / 2} \Gamma (k /2)} \mathrm{e}^{- \frac{x}{2}}$$`

]

---

.bg-light-gray.b--light-gray.ba.bw1.br3.shadow-5.ph4.mt5[

### Corollary

The sum of  `$k$` independent squared standard Gaussian random variables is distributed according to the chi-square
distributions with `$k$` degrees of freedom  `$\chi^2_k$`.

]

---

### Proof

It  suffices to establish the proposition `$k = 1.$`

Let `$X \sim \mathcal{N}(0,1)$`, for `$t\geq 0$`,

`$$\begin{array}{rcl} \mathbb{P} \left\{ X^2 \leq t\right\} & = & \Phi(\sqrt{t}) - \Phi(-\sqrt{t}) \\ & = & 2 \Phi(\sqrt{t})  - 1\end{array}$$`

Now, differentiating with respect to `$t$`, applying
the chain rule provides us with a formula for the density:

`$$2 \frac{1}{2\sqrt{t}} \phi(\sqrt{t}) = \frac{1}{\sqrt{2\pi t}} \mathrm{e}^{-\frac{t}{2}} = \left(\frac{1}{2}\right)^{1/2} \frac{t^{-1/2}}{\Gamma(1/2)} \mathrm{e}^{-\frac{t}{2}}$$`

---
template: inter-slide

## Norms of centered Gaussian Vectors

---

The distribution of the squared Euclidean norm of
a centered Gaussian vector only depends on the spectrum of its covariance
matrix.

.bg-light-gray.b--light-gray.ba.bw1.br3.shadow-5.ph4.mt5[

### Theorem

Let   `${X}:= (X_1, X_2, \ldots, X_n)^{^T} \sim \mathcal{N}\left(0, A\right)$`
with  `$A = L L^T$` ($L$ lower triangular).

If  `$M \in \mathrm{SDP}(n)$`,

then

`$${X}^T M {X} \sim \sum_{i = 1}^n \lambda_i Z_i$$`

where `$(\lambda_i)_{i \in \{1, \ldots, n\}}$` denote the eigenvalues of `$L^T \times M\times L$`
and  where  `$Z_i$` are independent  `$\chi^2_1$`-distributed random variables.

]

---

This is a corollary of an important property of standard Gaussian vectors: _rotational invariance_.

The standard Gaussian distribution is invariant under orthogonal transform

A matrix `$O$`  is  orthogonal iff   `$OO^T=\text{Id}$`

---

### Proof

Matrix `$A$` may be factorized as

`$$A = LL^t$$`

and  `${X}$` is  distributed like `$L {Y}$` where `${Y}$` is standard Gaussian.

The quadratic form `${X}^T M {X}$` is thus distributed like `${Y}^T {L}^T M {L}  {Y}$`.

There exist an orthogonal transform `$O$` such that

`$$L^T M L = O^t \operatorname{diag} (\lambda_i) O$$`

Random vector `$O {Y}$` is distributed like `$\mathcal{N} (0, I_n)$`.

---
template: inter-slide

## Norm of Non-Centered Gaussian Vectors

---

The distribution of the squared norm of a Gaussian vector with covariance matrix `$\sigma^2 \operatorname{Id}$`
depends on the norm of the  expectation but does  not depend on its direction. In addition, this distribution stochastically can be compared with  the distribution of the  squared norm of a centred Gaussian vector with the same covariance.

.bg-light-gray.b--light-gray.ba.bw1.br3.shadow-5.ph4.mt5[

### Definition: Stochastic Ordering

In a probability space endowed with distribution `$\mathbb{P}$`, a real random  variable   `$X$`  is
_stochastically smaller_ than  random variable `$Y$`, if

`$$\mathbb{P} \{ X \leq Y \} =  1$$`

The distribution of `$Y$`  is said to stochastically dominate the distribution of  `$X$`

]

---

If  `$X$`  is stochastichally less than `$Y$`  and if  `$F$`  and  `$G$`  denote the cumulative distribution functions of `$X$` and `$Y$`, then for all  `$x \in  \mathbb{R}$`, `$F(x)\geq G(x)$`. Quantile functions `$F^\leftarrow, G^\leftarrow$`  satisfy  `$F^\leftarrow(p) \leq G^\leftarrow(p)$` for `$p \in (0,1)$`.

---

Conversely.

.bg-light-gray.b--light-gray.ba.bw1.br3.shadow-5.ph4.mt5[

### Proposition

If  `$F$` and `$G$`  are two cumulative distribution functions that satisfy  `$\forall x \in \mathbb{R}$` `$F(x)\geq G(x)$`

then

there exists a probability space equipped with a probability distribution `$\mathbb{P}$` and
two random variables `$X$` and `$Y$`  with cumulative distribution functions `$F, G$`  that satisfy:

`$$\mathbb{P}\{ X \leq Y\} = 1$$`

]

---

The proof proceeds by a  _quantile coupling_ argument.

### Proof

It is enough to endow  `$([0,1], \mathcal{B}([0,1])$`  with the uniform distribution.

Let `$X (\omega)=  F^{\leftarrow}(\omega)$`,  `$Y(\omega) = G^\leftarrow(\omega)$`.

Then the distribution of  `$X$`  (resp. `$Y$`) has cumulative distribution function `$F$` (resp. `$G$`) and the
following holds:

`$$\mathbb{P} \{ X \leq Y\}  = \mathbb{P} \{ F^{\leftarrow}(U) \leq  G^{\leftarrow}(U)\} = 1$$`

---

.bg-light-gray.b--light-gray.ba.bw1.br3.shadow-5.ph4.mt5[

### Theorem

If `$X \sim \mathcal{N}\left( 0, \sigma^2 \operatorname{Id}\right)$` and
`$Y \sim \mathcal{N}\left( \theta, \sigma^2 \operatorname{Id}\right)$` with `$\theta \in \mathbb{R}^d$`
then

`$$\left\Vert Y \right\Vert^2 \sim  \left( (Z_1 + \|\theta\|_2)^2 + \sum_{i=1}^d Z_i^2 \right)$$`

where  `$Z_i$`  are  i.i.d. according to `$\mathcal{N}(0,\sigma^2)$`.

For every  `$x \geq 0$`,

`$$\mathbb{P} \left\{ \| Y \|\leq x\right\} \leq \mathbb{P} \left\{ \| X \| \leq x \right\}$$`

]

---

### Proof

The Gaussian vector `$Y$` is  distributed like `$\theta + X$`.   There exists an orthogonal transform
`$O$` such that

`$$O \theta = \begin{pmatrix} \| \theta\|_2 \\ 0 \\ \vdots \\ 0\end{pmatrix}$$`

Vectors  `$OY$` and `$OX$` respectively have the same norms as  `$X$` and `$Y$`.

The squared norm of  `$Y$`
is distributed as the squared norm of `$OY$`, that is like  `$(Z_1+ \|\theta\|_2)^2 +\sum_{i=2}^d Z_i^2$`.
This proves the first part of the theorem.

To establish the second part of the theorem, it suffices to check that for every `$x\geq 0$`,

`$$\mathbb{P} \left\{ (Z_1+ \|\theta\|_2)^2  \leq x \right\} \leq \mathbb{P} \left\{ X_1^2 \leq x \right\}$$`

that is

`$$\mathbb{P} \left\{ |Z_1+ \|\theta\|_2|  \leq \sqrt{x} \right\} \leq \mathbb{P} \left\{ |X_1| \leq \sqrt{x} \right\}$$`

`$$\Phi(\sqrt{x}- \|\theta\|_2) - \Phi(-\sqrt{x}-\|\theta\|_2)  \leq \Phi(\sqrt{x}) -  \Phi(-\sqrt{x})$$`

For  `$y>0$`, the function mapping `$[0,\infty)$` to `$\mathbb{R}$`, defined by `$a \mapsto \Phi(y-a) - \Phi(-y-a)$`
is non-increasing with respect to `$a$`:
it derivative with respect to `$a$` equals  `$-\phi(y-a)+\phi(-y-a)=\phi(y+a)-\phi(y-a)\leq 0$`.
The conclusion follows

---

The last step of  the proof reads as

`$$\mathbb{P} \left\{ X \in \theta + C \right\} \leq \mathbb{P} \left\{ X \in C\right\}$$`

where  `$X \sim \mathcal{N}(0,\operatorname{Id}_1)$`, `$\theta \in \mathbb{R}$` and `$C = [-\sqrt{x},\sqrt{x}]$`.

This inequality holds in dimension `$d\geq 1$` if   `$C$` is compact, convex, symmetric.

This (subtle) result is called Anderson s Lemma.

---
template: inter-slide

## Cochran Theorem and  consequences

---

.bg-light-gray.b--light-gray.ba.bw1.br3.shadow-5.ph4.mt5[

### Theorem (Cochran)

Let `$X \sim \mathcal{N}(0, \text{I}_n)$`  and
`$\mathbb{R}^n = \oplus_{j=1}^k E_j$` where  `$E_j$`  are pairwise orthogonal linear subspaces of
`$\mathbb{R}^n$`.

Denote by  `$\pi_{E_j}$`  the orthogonal projection on `$E_j$`.

The collection of Gaussian vectors  `$\left( \pi_{E_j} X\right)_{j \leq k}$`
is  independent and for each `$j$`

`$$\| \pi_{E_j} X\|_2^2 \sim \chi^2_{\text{dim}(E_j)}$$`

]

---

### Proof

The  covariance matrix of  `$\pi_{E_j} X$`  is `$\pi_{E_j} \pi_{E_j}^t = \pi_{E_j}$`.

The eigenvalues of `$\pi_{E_j}$`  are `$1$` with  multiplicity `$\text{dim}(E_j)$` and  `$0$`.

The statement about the distribution of `$\| \pi_{E_j} X\|_2^2$` is
a corollary of results on norms of centered Gaussian vectors

---

To prove stochastic independence, let us consider `$\mathcal{I}, \mathcal{J} \subset \{1,\ldots,k\}$`
with `$\mathcal{I} \cap \mathcal{J} = \emptyset.$`

It is enough to check that
for all `$(\alpha)_{j \in \mathcal{I}}, (\beta_j)_{j \in \mathcal{J}}$`, the characteristic functions of

`$$\left(\sum_{j\in \mathcal{I}} \langle \alpha_j, \pi_{E_j} X \rangle, \sum_{j\in \mathcal{J}} \langle \beta_j, \pi_{E_j} X \rangle\right)$$`

can be factorized. It suffices to check that the
two Gaussians are orthogonal.

`$$\begin{array}{rcl}
{ \mathbb{E} \left[ \left(\sum_{j\in \mathcal{I}} \langle \alpha_j, \pi_{E_j} X \rangle \right) \times \left(\sum_{j'\in \mathcal{J}} \langle \beta_{j'}, \pi_{E_{j'}} X \rangle\right)\right]}
& = & \sum_{j \in \mathcal{I}, j' \in \mathcal{J}}
\alpha_j^t \pi_{E_j} \pi_{E_{j'}} \beta_{j'}
=  0 \, .
\end{array}$$`

---

The next result is a cornerstone of statistical inference in Gaussian models.

It is a corollary of Cochran s Theorem.

.bg-light-gray.b--light-gray.ba.bw1.br3.shadow-5.ph4.mt5[

### Theorem (Student)

If `$(X_1, \ldots, X_n) \sim_{\text{i.i.d.}} \mathcal{N} (\mu, \sigma^2)$`,

let if  `$\overline{X}_n = \sum^n_{i = 1}  X_i / n$` and  `$V= \sum^{n}_{i = 1} (X_i - \overline{X}_n)^2$`,

then

i. `$\overline{X}_n$` is distributed according to `$\mathcal{N} (\mu, \sigma^2/n)$`,
i. `$V$` is  independent from `$\overline{X}_n$`
i. `$V/\sigma^2$` is distributed according to `$\chi_{n - 1}^2$`.

]

---

### Proof

Without loss of generality, we may assume that `$\mu=0$` et `$\sigma=1$`.

`$$\begin{pmatrix}\overline{X}_n \\\vdots\\\overline{X}_n \\ \end{pmatrix} =  \frac{1}{n} \begin{pmatrix} 1 \\ \vdots\\ 1 \\ \end{pmatrix} \times \begin{pmatrix} 1 & \ldots & 1 \end{pmatrix} X$$`

the vector `$(\overline{X}_n, \ldots , \overline{X}_n)^t$`  is the orthogonal projection
of the standard Gaussian vector `$X$` on the
line generated by  `$(1, \ldots, 1)^t$`.

Vector  `$(X_1- \overline{X}_n, \ldots , X_n -\overline{X}_n)^t$` is the orthogonal projection fo Gaussian vector `$X$`
on the hyperplane which is orthogonal to `$(1, \ldots, 1)^t$`.

---

### Proof (continued)

According to the Cochran Theorem, random vectors  `$(\overline{X}_n, \ldots , \overline{X}_n)^t$`,
and   `$(X_1- \overline{X}_n, \ldots , X_n -\overline{X}_n)^t$` are independent.

The distribution of  `$\overline{X}_n$` is trivially Gaussian.

The distribution of `$V$` is characterized using the Cochran Theorem.

---

.bg-light-gray.b--light-gray.ba.bw1.br3.shadow-5.ph4.mt5[

### Definition (Student t-distribution)

If `$X \sim \mathcal{N}(0,1)$`, `$Y \sim \chi_p^2$` and if `$X$`  and  `$Y$` are independent,

then

`$Z =  X/ \sqrt{Y/p}$` is distributed according to a  (centered) Student  distribution with
`$p$`  degrees of freedom

]

---
template: inter-slide

## Gaussian concentration

---

The very definition of Gaussian vectors characterizes the distribution of any affine function of a
standard Gaussian vector.

If the linear part of the affine function is defined by a vector `$\lambda$`, we know that
the variance will be `$\|\lambda\|^2_2$`.

What happens if we are interested in fairly regular functions of a standard Gaussian vector?

---

For example if we consider `$L$`-lipschitzian functions?

These are generalizations of affine functions.

We cannot therefore expect a general bound on the variance of the `$L$`-Lipschitzian functions of a standard Gaussian vector better than `$L^2$` (in the linear case the Lipschitz constant is the Euclidean norm of `$\lambda$`).

It is remarkable that the bound provided for linear functions extends to Lipschitzian functions.

It is even more remarkable that this bound does not involve the dimension of the ambient space.

---

.bg-light-gray.b--light-gray.ba.bw1.br3.shadow-5.ph4.mt5[

### Theorem

Let  `$X \sim \mathcal{N}(0 , \text{Id}_d)$`.

If  `$f$` is  differentiable on `$\mathbb{R}^d$`,
`$$\operatorname{var}(f(X)) \leq \mathbb{E} \| \nabla f \|^2   \qquad \text{(Poincaré  Inequality)}$$`

If  `$f$`  is  `$L$`-Lipschitz on  `$\mathbb{R}^d$`,

`$$\operatorname{var}(f(X)) \leq L^2$$`

`$$\log \mathbb{E} \mathrm{e}^{\lambda(f(X)-\mathbb{E}f)} \leq \frac{\lambda^2 L^2}{2}\qquad \forall \lambda >0$$`

`$$\mathbb{P} \left\{ f(X) - \mathbb{E} f(X) \geq t \right\} \leq \mathrm{e}^{-\frac{t^2}{2 L^2}}\qquad \forall t>0$$`

]

---

The proof relies on

.bg-light-gray.b--light-gray.ba.bw1.br3.shadow-5.ph4.mt5[

### Covariance identity

Let  `$X,Y$`  be two independent `$\mathbb{R}^d$`-valued standard Gaussian vectors, let `$f,g$`  be two
differentiable functions from `$\mathbb{R}^d$` to `$\mathbb{R}$`.

`$$\operatorname{cov}(f(X),g(X)) = \int_0^1 \mathbb{E}\left\langle \nabla f(X) , \nabla g\left(\alpha X +\sqrt{1- \alpha^2} Y \right) \right\rangle \mathrm{d} \alpha$$`

]

---

We start by checking this proposition  on  functions `$x \mapsto  \mathrm{e}^{\imath \langle \lambda, x\rangle}, x \in \mathbb{R}^d$`.

---

### Proof

Let us first check the Poincaré  Inequality.

We choose `$f=g$`. Starting from the covariance identity, thanks to  the Cauchy-Schwarz Inequality:

`$$\begin{array}{rcl} \operatorname{var}(f(X) )     &= & \operatorname{cov}(f(X),f(X)) \\ & = & \int_0^1  \mathbb{E}\left\langle \nabla f(X) , \nabla f\left(\alpha X +\sqrt{1- \alpha^2} Y \right) \right\rangle \mathrm{d} \alpha \\ & \leq & \int_0^1 \left( \mathbb{E}\|  \nabla f(X) \|^2\right)^{1/2} \times \left(\mathbb{E} \|\nabla f\left(\alpha X +\sqrt{1- \alpha^2} Y\right)\|^2 \right)^{1/2} \mathrm{d} \alpha \end{array}$$`

The desired results follows by noticing that  `$X$` and  `$\alpha X + \sqrt{1- \alpha^2}Y$`  are both
 `$\mathcal{N}(0,\text{Id})$`-distributed.

---

### Proof (continued)

To obtain the exponential inequality, choose  `$f$` differentiable and  1-Lipschitz, and `$g = \exp(\lambda f)$` pour `$\lambda\geq 0$`.

Without loss of generality, assume  `$\mathbb{E}f(X)=0$`.

The covariance identity and the chain rule imply

`$$\begin{array}{rcl}\operatorname{cov}\left(f(X),\mathrm{e}^{\lambda f(X)}\right)  & = & \lambda \int_0^1  \mathbb{E}\left[\left\langle \nabla f(X) , \nabla f\left(\alpha X +\sqrt{1- \alpha^2} Y \right) \right\rangle \mathrm{e}^{\lambda f\left(\alpha X +\sqrt{1- \alpha^2} Y \right)}\right] \mathrm{d} \alpha \\ & \leq & \lambda L^2 \int_0^1  \mathbb{E}\left[ \mathrm{e}^{\lambda f\left(\alpha X +\sqrt{1- \alpha^2} Y \right)}\right] \mathrm{d} \alpha \\ & = & \lambda L^2 \mathbb{E}\left[ \mathrm{e}^{\lambda f\left(X\right)}\right]\end{array}$$`

---

### Proof (continued)

Define  `$F(\lambda):= \mathbb{E}\left[ \mathrm{e}^{\lambda f\left(X\right)}\right]$`

Note that we have just established a differential inequality for  `$F$`, checking
`$\operatorname{cov}( f , \mathrm{e}^{\lambda f})= F'(\lambda)$` since `$f$` is centred:

`$$F'( \lambda) \leq \lambda L^2 F(\lambda)$$`

Solving this differential inequality under  `$F(0)=1$`, for `$\lambda\geq 0$`

`$$F( \lambda) \leq \mathrm{e}^{\frac{\lambda^2L^2}{2}}$$`

The same approach works for `$\lambda<0$`.

It is enough to invoke the  Markov exponential inequality and to optimize with respect to  `$\lambda=t/L^2$`.

---

Concentration inequalities describe (among other things) the behavior of the norm of high-dimensional Gaussian vectors

.bg-light-gray.b--light-gray.ba.bw1.br3.shadow-5.ph4.mt5[

### Corollary

If  `$X$` is a standard `$d$`-dimensional Gaussian vector, then

`$$\operatorname{var}(\|X\|_2) \leq 1$$`

and

`$$\sqrt{d-1} \leq \mathbb{E} \|X\|_2 \leq \sqrt{d}$$`

]

---

### Proof

The Euclidean norm is  `$1$`-Lipschitz (triangle inequality)

The first inequality follows fron the Poincaré Inequality.

The upper bound on expectation follows from the Jensen Inequality

The lower bound on expectation follows from

`$$\Big(\mathbb{E} \|X\|_2\Big)^2 = \mathbb{E} \|X\|_2^2 - \operatorname{var}(\|X\|_2)= d -\operatorname{var}(\|X\|_2)$$`

and from the variance upper bound.

---

Show

`$$\operatorname{Var}(Z) \leq \max_{i \leq d } K_{i,i}:= \max_{i \leq d} \operatorname{Var} (X_i)$$`

Show

`$$\sqrt{2n-1} \leq   \mathbb{E}[\|X-Y\|] \leq \sqrt{2 n}$$`

and

`$$\mathbb{P}   \left\{ \|X-Y\| - \mathbb{E}[\|X-Y\|] \geq t \right\} \leq \mathrm{e}^{-t^2}$$`

---
exclude: true

## Bibliographic remarks

Gaussian literature is very abundant, see for example [@janson1997gaussian].
Much of this literature is relevant to statistics.

The Stein lemmas that characterize the Gaussian standard are the starting point of Stein's (Charles) method to demonstrate the central limit theorem (and many other results). This relatively recent development is described in [@2011arXiv1109.1880R].

Matrix analysis and algorithmics play an important role in Gaussian analysis and statistics.
The books [@HorJoh90], and if we wish to go further [@Bha97], provide an introduction to the concepts and techniques of matrix factorization and elements of perturbation theory.

There is a multi-dimensional version of the laws of `$\chi^2$` that
appear when determining the law of variance
empirical. These are the laws of Wishart. They were the subject of
intensive studies in random matrix theory, see for example [@AnGuZe10]

Gaussian concentration plays an important role in non-parametric statistics and is a source of inspiration in statistical learning. M. Ledoux's book [@ledoux:2001] provides an elegant perspective on this issue.

---

template: inter-slide

background-image: url('./img/pexels-cottonbro-3171837.jpg')
background-size: 112%

# The End