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How to prove E(I(X ∈ A)) = P(X ∈ A)?

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    $I$ is the indicator function ? – 2017-01-22
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    I is an indicator function – 2017-01-22
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    I think that the LHS is meant to be $\mathbb E1_A(X)$. If $I$ on its own is a function on its own allready then in your question it has a rather strange argument: the event "$X\in A$". – 2017-01-22
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    As mentioned, this question is too context-dependent to be usefully answered. – 2017-01-22
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    @drhab Actually, $\mathbf 1_A(X)$ and $\mathbf 1_{X\in A}$ and $\mathbf 1(X\in A)$ are synonyms. – 2017-01-22
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    @Did The third is new to me. – 2017-01-22

3 Answers 3

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Depends on your knowledge of measure theory. In a discrete space, you can do : $${\rm E}({\mathbf 1}(X\in A)) = \sum_{\omega\in\Omega} {\mathbf 1}(\omega\in A)p_\omega = \sum_{\omega\in A} p_\omega = P(A)$$ In a continuous space, you can do : $${\rm E}({\mathbf 1}(X\in A)) = \int_\Omega {\mathbf 1}(\omega\in A){\rm d}P(\omega) = \int_A {\rm d}P(\omega) = P(A)$$

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    Merci, Nicolas. I believe that the probability that an X is in A is P(ω) so that I can sum up P(ω) of each X that is A. Thanks, again. – 2017-01-22
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    What exactly is $\mathbf1$ here? A function? If so then what is its domain? – 2017-01-22
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    @drhab : yes, it's a function, and you can see ${\mathbf 1}(X\in A)$ as a random variable : ${\mathbf 1}(X\in A)(\omega)=1$ if $X(\omega)\in A$, else $0$. – 2017-01-22
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Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space and $X:\Omega\to \mathbb{R}$ be a random variable. For every Borel set, we have $$X^{-1}(A)=\{\omega\in \Omega\ : X(\omega)\in A\}\in\mathcal{F}$$ Sometimes, we use $X\in A$ instead of $X^{-1}(A)$. We have \begin{align*}\mathbb{E}[I_{X\in A}]&=\mathbb{E}[I_{X^{-1}(A)}]=\int_{\Omega}I_{X^{-1}(A)} d\mathbb{P}=\int_{\Omega\,\cap \,X^{-1}(A)} d\mathbb{P}=\int_{ X^{-1}(A)} d\mathbb{P}=\mathbb{P}(X^{-1}(A))\\&=\mathbb{P}(X\in A)\end{align*}

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    What is F? Why do we use X-inverse of A? – 2017-01-22
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    A sigma algebra. See it: https://en.wikipedia.org/wiki/Sigma-algebra – 2017-01-22
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    Yes, I get it now. Sukran/thanks, Behrouz. – 2017-01-22
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    Please, good luck. – 2017-01-22
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I am not familiar with the notation used in your question and preassume in this answer that you are asked to prove that:$$\mathbb E\mathbb1_A(X)=P(X\in A)$$ where $A$ denotes a measurable set and $X$ denotes a random variable.

Then $1_A(X)$ is also a random variable, and observe that it only takes values in $\{0,1\}$.

On set $\{X\in A\}:=\{\omega\in\Omega\mid X(\omega)\in A\}$ it takes value $1$ and on its component it takes value $0$.

From this we conclude that:$$\mathbb E\mathbb1_A(X)=1\times P(\{X\in A\})+0\times P(\{X\notin A\})=P(X\in A)$$ Here $P(X\in A)$ is an abbreviation of $P(\{X\in A\})$.