Expectation of Random variable times indicator function of zero probability event is equal zero

Question

I was reading here and I bumped into $$\mathbb{E}[X1_E] = 0$$ where $X$ is an integrable Random Variable, $E$ is a zero-probability event and $1_E$ is the indicator function for $E$. The link doesn't give a proof, and I tried making one of my own but with no success.

My Try

All I know is:

• $P(E) = 0$ or better $P(X \in E) = P(\{w \in \Omega: X(w) \in E\}) = 0$ where $w$ is a sample point of the sample space $\Omega$.
• $1_E = 1$ if $w \in E$ and $1_E = 0$ if $w \notin E$.
• $\mathbb{E}[1_E] = P(E)$

However, we don't know whether $X1_E$ is a discrete RV, an absolutely continuous RV or neither of the two. Thus we don't know whether we have to calculate the expectation with a finite sum, with an infinite sum, with an integral or with a Riemann-Stieltjes Integral (or a Lebesgue).

As the link suggests, we know that $X1_E$ is going to be $0$ whenever $w \notin E$, by definition of the indicator function, and we also know that $X1_E$ is going to be $X$ when $w\in E$. However, contrary to what the link says, we really don't know that the various terms will be multiplies by the probability of event $E$. That's because we don't know whether the two random variables $I_E$ and $X$ are independent or not.

Answer 1

Edited because there was a mistake in the answer

$E[XI_A]$ can be written as:

\begin{equation} E[XI_A] = \int_{X} x I_A(x) dX \end{equation}

Not that we can split the integral over a partition of the support into $A$ and $A^c$:

\begin{equation} \int_{X} x I_A(x) dX = \int_{A} x I_A(x) dX + \int_{A^c} x I_A(x) dX \end{equation}

The second term goes away because $I_A(x) = 0 \hspace{2mm} \forall x \in A^c$. It should seem obvious here that because $X$ gives no probability to $A$ that the remaining term is $0$, but to rigorously prove it. Specifying $x^* = \sup_{x\in A} ||x||$, we have that:

\begin{equation} \begin{split} \int_{X} x I_A(x) dX &= \int_{A} x I_A(x) dX\\ & \Rightarrow ||\int_{A} x I_A(x) dX|| \leq \int_A||x||I_A(x)dX \leq \int_A||x^*||I_A(x)dX \\ & =||x^*||\int_A I_A(x)dX = ||x^*||P[x \in A]\\ & = ||x^*|| 0 = 0 \end{split} \end{equation}

Because the $0$ vector is the unique element with norm $0$, this gives us that:

\begin{equation} \int_{X} x I_A(x) dX = 0 \Rightarrow E[XI_A] =0 \end{equation}

Second Edit

Some care needs to be taken when stating that \begin{equation} \int_A||x^*||I_A(x)dX = ||x^*||\int_AI_A(x)dX = ||x^*||P[x \in A] = ||x^*||0 = 0 \end{equation} This is because $||x^*||$ could be infinite or outside of $A$ if $A$ is not compact. In these situations $x^*$ is a boundary point of $A$ and we can construct a sequence $x_n \in \textrm{Interior}(A)$ such that $x_n \rightarrow x^*$ gives us $\lim_{n\rightarrow \infty} ||x_n|| =||x^*||$.

Then we have that, using the monotone convergence theorem: \begin{equation} \begin{split} \int_A||x^*||I_A(x)dX &= \int_A\lim_{n\rightarrow \infty} ||x_n||I_A(x)dX \\ &= \lim_{n\rightarrow \infty} \int_A||x_n||I_A(x)dX \\ &= \lim_{n\rightarrow \infty} ||x_n|| \int_A I_A(x)dX \\ &= \lim_{n\rightarrow \infty} ||x_n|| P[X \in A] \\ &= \lim_{n\rightarrow \infty} ||x_n|| 0 \\ &= \lim_{n\rightarrow \infty} 0\\ &= 0\\ \end{split} \end{equation}

Answer 2

Let $\langle\Omega,\mathcal A, P\rangle$ be a probability space.

Let $\mathcal R$ denote the collection of non-negative random variables on $\langle\Omega,\mathcal A, P\rangle$ that have finite image.

If $\{y_1,\dots,y_n\}\subseteq[0,\infty)$ denotes the image of some $Y\in\mathcal R$ and the $y_i$ are distinct then the sets $\{Y=y_i\}$ form a partition of $\Omega$ and by definition:$$\mathsf EY=\sum_{i=1}^ny_i\mathsf P(Y=y_i)\tag1$$

Observe that $\mathsf EY\in[0,\infty)$ for every $Y\in\mathcal R$.

If $Z:\Omega\to\mathbb R$ is a non-negative random variable on $\langle\Omega,\mathcal A, P\rangle$ then by definition:$$\mathsf EZ=\sup\{\mathsf EY\mid Y\in\mathcal R\wedge Y\leq Z\}\tag2$$

Observe that $\mathsf EZ\in[0,\infty]$ and that this definition expands the definition given in $(1)$.

If $Z:\Omega\to\mathbb R$ is a random variable on $\langle\Omega,\mathcal A, P\rangle$ then it induces non-negative random variables $Z_+:=\max(Z,0)$ and $Z_-:=\max(-Z,0)$. Observe that $Z=Z_+-Z_-$ and $|Z|=Z_++Z_-$.

$\mathsf EZ$ is defined if at least one $\mathsf EZ_+$ and $\mathsf EZ_-$ is finite and this by:$$\mathsf EZ=\mathsf EZ_+-\mathsf EZ_-\tag3$$

This under the convention that $\infty-a=\infty$ and $a-\infty=-\infty$ for any $a\in\mathbb R$. This definition expands the definition given in $(2)$.

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Now have a look at $Z=X1_E$ where $X$ is a random variable on $\langle\Omega,\mathcal A, P\rangle$ and $E\in\mathcal A$ with $P(E)=0$.

In order to find $\mathsf EX1_E$ we must find $\mathsf E(X1_E)_+$ and $\mathsf E(X1_E)_-$.

If $Y\in\mathcal R$ with $Y\leq\mathsf (X1_E)_+$ having image $\{y_1,\dots,y_n\}$ then for every non-negative $y_i$ we have $\{Y=y_i\}\subseteq E$ and consequently $P(Y=y_i)=0$.

Then with $(1)$ we find $\mathsf EY=0$ and $(2)$ justifies the conclusion that $\mathsf E(X1_E)_+=0$.

Likewise we find $\mathsf E(X1_E)_-=0$.

Then our final conclusion based on $(3)$ is: $$\mathsf EX1_E=\mathsf E(X1_E)_+-\mathsf E(X1_E)_-=0-0=0$$