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Let $X$ be a continuous random variable with a density $f_X(x)$ and let $A\subseteq \mathbb{R}$. Would it be true to claim that:

$E[X | X\in A] = \int_{x\in A} x \cdot f_X(x) \cdot dx$?

My understanding of dependent expectation is that:

$E[X|X\in A] = \int_{x\in\mathbb{R}} x \cdot f_{X|X\in A} (x) \cdot dx$

Are these expressions equivalent? If so, I'm getting a bit confused trying to show that, I'm sure it's likely only definition-based, but having a bit trouble.

I know that: $f_{X|X\in A} (x) = \frac{Something that says both happen}{P(X\in A)}$

Eric_
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    Almost. Your first expression is missing a normalization factor. You can see that factor in your definition of $f_{X\mid X\in A}$. – HackR Jun 08 '25 at 18:25
  • I think I see! You mean $E[X|X\in A] \cdot P(X\in A) = \int_{x\in A} f_X (x) \cdot dx$ ? – Eric_ Jun 09 '25 at 08:17
  • Again, almost. The rhs has to be the expectation, so you are missing an $x$ inside the integral. – HackR Jun 09 '25 at 08:33
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    Yes, that was a typo on my end :) I meant $E[X|X\in A] \cdot P(X\in A) = \int_{x\in A} x\cdot f_X (x) \cdot dx $, and I hope there are no more mistakes. – Eric_ Jun 09 '25 at 09:30

1 Answers1

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Assuming the integral in the numerator is finite and $\mathbb P(X \in A) \not =0$ (the integral in the denominator), you have the conditional expectation

$$\mathbb E[X \mid X\in A] = \dfrac{\int\limits_{x\in A} x \, f_X(x) \, dx}{\int\limits_{x\in A} f_X(x) \, dx}$$

which is equivalent to your second comment.

Henry
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