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Let $X$ be a positive countinuous random variable. How do I show that $\mathbb{E}[X\mathbb{1}_{\lbrace X\le a \rbrace}]=\mathbb{P}(X\le a)$? It seems that I have missed something important.
$\mathbb{E}[X\mathbb{1}_{\lbrace X\le a \rbrace}]=\int\limits_{-\infty}^{\infty}x\cdot f_{X\mathbb{1}_{\lbrace X\le a \rbrace}}(x)dx$ and i want to justify it equals to $\int\limits_{-\infty}^{a}x\cdot f_X(x)dx$.

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    $\begingroup$ No, this equation is false. It must be $$E[1_{\lbrace X\le a \rbrace}] = P(X\le a)$$ $\endgroup$
    – NN2
    Mar 18, 2021 at 23:37
  • $\begingroup$ @NN2 then what does it equal? can I say something about P(X≤a) ? the question arise from this comment, where I don't get the transition of (2): math.stackexchange.com/a/4065150/696493 $\endgroup$
    – Bernard
    Mar 18, 2021 at 23:47
  • $\begingroup$ Do you mean $$\mathbb E[X1_{\lbrace X\le a \rbrace}] = a\mathbb{P}(X\le a)$$ maybe? $\endgroup$
    – Clement C.
    Mar 18, 2021 at 23:52
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    $\begingroup$ @Benny You can leave a comment under my answer if you need any further clarification :) $\endgroup$ Mar 19, 2021 at 0:14
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    $\begingroup$ @peacefulbreeze thanks for the support. I think I got it finally :) $\endgroup$
    – Bernard
    Mar 19, 2021 at 0:33

4 Answers 4

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The equation is false because $E(X\mathbb{I}_{\{X \le a \}}) \ne P(X \le a )$. In fact, we have rather $P(X \le a ) = E(\mathbb{I}_{\{X \le a \}})$.

For your second question: why we have $E[X \mathbb{1}_{\{X>a\}}]\le \sqrt{E[X^2]}\sqrt{P\{X>a\}}$ ?

In fact, we apply the Cauchy-Schwarz inequality here, for $A,B$ two random variable $$E(AB) \le \sqrt{E(A^2)}\sqrt{E(B^2)} \tag{1}$$ Just replacing $A = X$ and $B = \mathbb{I}_{\{X \le a \}} $ in $(1)$ and noticing 2 things below: $$B^2 = B = \mathbb{I}_{\{X \le a \}} $$ and $$E(\mathbb{I}_{\{X \le a \}})= P(X \le a )$$

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    $\begingroup$ that was very helpful. sometimes the trivial things are not trivial at all to others. thank you very much :) $\endgroup$
    – Bernard
    Mar 18, 2021 at 23:58
  • $\begingroup$ @Benny you're welcome $\endgroup$
    – NN2
    Mar 18, 2021 at 23:59
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The equation you provided is false. The correct one is $$ \mathbb{E}[\mathbb{1}_{X\leq a}] = \mathbb{P}(X ≤ a) .$$

In fact, $$ \mathbb{E}[\mathbb{1}_{X ≤ a}] = \int_{-\infty}^{\infty} \mathbb{1}_{X ≤ a} f_X(x) dx = \int_{-\infty}^a f_X(x) dx = \mathbb{P}(A) . $$

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Thie is very very false. Take $X$ to be an exponetial r.v. with parameter $1$. Then LHS is $\int_0^{a} xe^{-x}dx$ and RHS is $\int_0^{a} e^{-x}dx$. You can compute these integrals and see that these two are not equal for any $a \in (0,\infty)$.

$EX1_{X\leq a}=\int_0^{a}xf_X(x)dx$ is the correct formula.

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  • $\begingroup$ then what does it equal? can I say something about $P(X\le a)$ ? the question arise from this comment, where I don't get the transition of (2): math.stackexchange.com/a/4065150/696493 $\endgroup$
    – Bernard
    Mar 18, 2021 at 23:45
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That is not quite correct. It is that : $\mathsf E(\mathbf 1_{X\leqslant a})=\mathsf P(X\leqslant a)$


What $\mathsf E(X\,\mathbf 1_{X\leqslant a})$ equals is $\int\limits_0^a\mathsf P(y\lt X\leqslant a)\,\mathrm d y$

The key is that $X$ is a positive continuous random variable; meaning that what ever the support for $f$ may be, it does not intercept the negative reals. So for any positive $a$...

$$\begin{align}\mathsf E(X\,\mathbf 1_{X\leqslant a})&=\int_0^a x~f(x)\,\mathrm d x\\&=\int_0^a \left(\int_0^x \mathrm d y\right)~f(x)~\mathrm d x\\&=\int_0^a\int_y^a f(x)\,\mathrm d x\,\mathrm d y\\&=\int_0^a\mathsf P(y\lt X\leqslant a)~\mathrm d y\end{align}$$

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