Proof: $E[X\mathbb{I}_{\{Y\leq y\}}] = E[X|Y\leq y]P(Y\leq y)$ Correct?

Question

I've seen the equation:

$$ E[X\mathbb{I}_{\{Y\leq y\}}] = E[X|Y\leq y]P(Y\leq y) \Leftrightarrow \frac{E[X\mathbb{I}_{\{Y\leq y\}}]}{P(Y\leq y)} = E[X|Y\leq y]$$

but never found a proof (I'm sure there is one), so here is my attempt:

$$ E[X|Y\leq y] := \int x f_{X|Y}(x|y) dx $$

\begin{align*} E[X\mathbb{I}_{\{Y\leq y\}}] &= \int_\mathcal{S_X} \int_\mathcal{S_Y} x\mathbb{I}_{\{Y\leq y\}}f_{X, Y}(x,y) dxdy \\ &= \int_\mathcal{S_X} \int_\mathcal{S_Y}x\mathbb{I}_{\{Y\leq y\}}f_{X|Y}(x|y)f_Y(y) dxdy \\ &= \int_\mathcal{S_X} \bigg(\int_\mathcal{S_Y} \mathbb{I}_{\{Y\leq y\}}f_Y(y)dy\bigg)xf_{X|Y}(x|y) dx \\ &= P(Y\leq y) \int_\mathcal{S_X} xf_{X|Y}(x|y) dx \\ &= P(Y\leq y) E[X|Y\leq y] \\ \end{align*}

Is my argument correct?

Answer 1

Let me give you an effort to justify what I asserted in a comment to your question:

the equality that you are trying to prove is a definition hence needs no proof.

Let's start with a probability space $(\Omega,\mathcal A,P)$ and let $A$ be an event with $P(A)>0$. If $X$ is some integrable rv defined on the space then its expectation is of course determined by $P$.

Now let us define a new probability measure $P(\cdot|A)$ on the space by:$$P(B|A):=\frac{P(B\cap A)}{P(A)}$$or equivalently:$$P(B|A)P(A)=P(B\cap A)$$ It is straightforward to prove (it is only here where we can use that word properly) that the integral of $X$ wrt this probability measure equals:$$\frac{\mathbb EX1_A}{P(A)}$$ Denoting this integral as $\mathbb E[X|A]$ we arrive at:$$\mathbb E[X|A]P(A)=\mathbb EX1_A$$

Answer 2

In general, the mean value of an integrable function $f:\Gamma \to Y$ in a finite measure space $(\Gamma ,\mathcal{F},\mu)$ is defined as $\frac{\int_{\Omega }f\,d \mu }{\int_{\Omega }\,d \mu }$. If $\mu$ is a probability measure (i.e. a finite measure such that $\mu(\Omega )=1$) then the mean value of $f$ reduces to just $\int_{\Omega }f\,d \mu$, and this value is denoted by $\operatorname{E}[f]$.

Now, the notation $\operatorname{E}[X|A]$, for some measurable set $A\subset \Omega $ who measure is not zero, is defined as the mean value of $X$ restricted to $A$, that is

$$ \operatorname{E}[X|A]:=\frac{\int_{A}X\,d P}{\int_{A}\,d P}=\frac{\int_{\Omega }\mathbf{1}_{A}X\,d P}{P[A]}=\frac{\operatorname{E}[X\mathbf{1}_{A}]}{P[A]} $$

If $A=\{\omega \in \Omega : Y(\omega )\in B\}$ (for some borel set $B\subset \mathbb{R}$ and random variable $Y$) then the above becomes

$$ \operatorname{E}[X|Y\in B]=\frac{\operatorname{E}[X\mathbf{1}_{Y^{-1}(B)}]}{P[Y\in B]} $$

So there is no possible proof for your identity, as this identity is a consequence of the definition of the notation $\operatorname{E}[\,\cdot\, |\,\cdot\, ]$.

Note: using the pushforward measure $P_Y:=P\circ Y^{-1}$ then we have the identity $\int_{\Omega }X\mathbf{1}_{Y^{-1}(B)}\,d P=\int_{B}h\,d P_Y$, where $h$ is the Radon-Nikodym derivative of the measure $\mu(B):=\int_{\Omega }X\mathbf{1}_{Y^{-1}(B)}\,d P$ respect to $P_Y$ (note that $\mu\ll P_Y$ so there is some function $h$ such that $h\,d P_Y=\,d \mu$). This function $h$ is usually noted as $\operatorname{E}[X|Y=t]$ respect to $\,d P_Y(t)$ and named as the "conditional expectation of $X$ respect to $Y$".