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I'm looking for a rigorous proof of an identity I came across many times (in the context of Gÿongy's lemma). Suppose $X$ and $Y$ are two r.v. We know that $\mathbb{E}\left[X|Y\right]$ is $\sigma(Y)$-measurable so there exist a function $\varphi$ such that $\mathbb{E}\left[X|Y\right]=\varphi(Y)$. We define $\mathbb{E}\left[X|Y=y\right] := \varphi(y)$.

The identity I'm after is then the following: $$ \mathbb{E}\left[X|Y=y\right] = \frac{\mathbb{E}\left[X\delta(Y-y)\right]}{\mathbb{E}\left[\delta(Y-y)\right]} $$

Where $\delta$ is the Dirac delta distribution.

I intuitively think this may be traced back to $$ p_\left[X|Y\right] = \frac{p_\left[X,Y\right]}{p_\left[Y\right]} $$ where $p_\left[X|Y\right]$ is the probability distribution of $\mathbb{E}\left[X|Y\right]$, $p_\left[X,Y\right]$ the joint probability distribution of $(X,Y)$, and $p_\left[Y\right]$ the marginal distribution of $Y$.

Any help would be appreciated.

Aguelmame
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1 Answers1

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If we interpret $\delta(Y-y)$ as $1_{\{Y=y\}}$ (the random variable which is $1$ on the event $\{Y=y\}$ and $0$ otherwise), then the formula is true as written for any discrete random variable $Y$.

Suppose $X$ and $Y$ are real-valued random variables with a joint density $f(x,y)$ on $\mathbb{R}^2$. Let us interpret $E[g(X)\delta(Y-y)]$ as $$ \int_{\mathbb{R}} g(x)f(x,y)\,dx. $$ Then $E[\delta(Y-y)]=\int_{\mathbb{R}}f(x,y)\,dx$. If we define $$ \varphi(y) = \frac{E[X\delta(Y-y)]}{E[\delta(Y-y)]} $$ whenever $E[\delta(Y-y)]>0$, and $\varphi(y)=0$ otherwise, then $E[X\mid Y]=\varphi(Y)$.

These results are Propositions 6.39 and 6.41 in these notes. In the latter case, we can use the Lebesgue differentiation theorem to prove that for Lebesgue almost every $y\in\mathbb{R}$, if $E[\delta(Y-y)]>0$, then $$ E[X \mid Y \in (y - \varepsilon, y + \varepsilon)] = \frac{E[X1_{\{|Y-y|<\varepsilon\}}]}{E[1_{\{|Y-y|<\varepsilon\}}]} \to \frac{E[X\delta(Y-y)]}{E[\delta(Y-y)]} $$ as $\varepsilon\to0$.

Jason Swanson
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