let $X$ be a standard Gaussian random variable. Show that $(X,X)$ is not absolutely continuous.

Question

i'm trying to understand a proof of the following statement:

let $X$ be a standard Gaussian random variable. Show that $(X,X)$ is not absolutely continuous.

I'll write down the proof in such a way that i can show you where i have troubles.

Proof:

We assume that $V:=(X,X)$ is absolutely continuous, meaning that there exists a map $f:\Bbb{R}^2\to \Bbb{R}^2$ which is Lebesgue measurable and such that $\int_{\Bbb{R}^2} f d\lambda^2=1$, where $\lambda^2$ denotes the Lebesgue measure on $\Bbb{R}^2$.

Then for all Borel measurable maps $h:\Bbb{R}^2\to\Bbb{R}^2$, we have $$\Bbb{E}[h(X,X)]=\Bbb{E}[V] \overset{by def}{=} \int_{\Omega}h(V)d\Bbb{P}\overset{(1)}{=}\int_{\Bbb{R}^2}h(x,y)d\Bbb{P}_V(x,y) = \int_{\Bbb{R}^2}h(x,y)f(x,y)d\lambda^2(x,y).$$

On the other hand we have $$\Bbb{E}[h(X,X)]\overset{bydef}{=}\int_{\Omega}h(X,X)d\Bbb{P}\overset{(2)}{=}\int_{\Bbb{R}}h(x,x)d\Bbb{P}_X(x)=\frac{1}{\sqrt{2\pi}}\int_{\Bbb{R}}h(x,x)e^{-\frac{x^2}{2}}d\lambda(x)$$

Now choose $h:\Bbb{1}_A(x,y)$, where $A:=\{(x,y)\in \Bbb{R}^2:x=y\}$. By the first relation we have: $$\int_{\Bbb{R}^2}h(x,y)f(x,y)d\lambda^2(x,y)=0$$ Since the measure of a line is $0$ with respect to $\lambda^2$. By the second relation we have $$\frac{1}{\sqrt{2\pi}}\int_{\Bbb{R}}h(x,x)e^{-\frac{x^2}{2}}=1$$

Since we reached a contradiction we can deduce that $(X,X)$ is not absolutely continuous.

What i don't understand are the equalities $(1),(2)$ What is the "trick" behind the definition of $V:=(X,X)$. Why does this definition lead to the two different equalities $(1),(2)?$ Thank you for any help

Answer 1

Both integral representations are applications of the following: If $Y$ is a random variable with values in $\mathbb{R}^n$, then $$ {\rm E}[u(Y)]:=\int_{\Omega} u(Y)\,\mathrm dP=\int_{\mathbb{R}^n}u(y)\,P_Y(\mathrm dy),\tag{1} $$ for measurable, bounded $u$. Here $P_Y$ is the distribution of $Y$ which is the measure on $\mathbb{R}^n$ given by $$ P_Y(A)=P(Y^{-1}(A)), $$ for Borel sets $A\subseteq\mathbb{R}^n$.

The first equality follows from $(1)$ with $Y=V$ and $u=h$. The second equality follows from $(1)$ with $Y=X$ and $u=h\circ g$, where $g:\mathbb{R}\to\mathbb{R}^2$ is given by $g(x)=(x,x)$.

Answer 2

Define $\triangle=\left\{ \left(x,x\right)\mid x\in\mathbb{R}^{2}\right\} $ here.

Suppose that $f\left(x,y\right)$ serves as PDF here.

Then we get the contradiction: $$1=P\left\{ \left(X,X\right)\in\mathbb{R}^{2}\right\} =P\left\{ \left(X,X\right)\in\triangle\right\} =\int_{\triangle}fd\lambda=0$$ The last equality as a consequence of $\lambda\left(\triangle\right)=0$.

This shows that there is no PDF, i.e. that the distribution of $(X,X)$ is not absolutely continuous.

The fact that $X$ is standard Gaussian plays no part in this.

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addendum

In my view it is handsome to write $\mu\left(f\right)$ for integral $\int fd\mu$.

In fact $\mu$ is no longer looked at as a function on measurable sets but as a function (with nice properties) on integrable functions. The original measure of set $A$ can be found back as $\mu\left(1_A\right)$.

Defining $\delta:\mathbb{R}\rightarrow\mathbb{R}^{2}$ by $x\mapsto\left(x,x\right)$ we come to $V=\delta\circ X:\Omega\rightarrow\mathbb{R}^{2}$.

Here: $$P_{V}\left(h\right)=P\left(h\circ V\right)=P\left(h\circ\delta\circ X\right)=P_{X}\left(h\circ\delta\right)$$ $P$ denotes the original probability measure on $\Omega$ and $P_{V}$ and $P_{X}$ are induced measures on $\mathbb{R}^{2}$ and $\mathbb{R}$ respectively.