In a book I met a formula for math. expectation of a random variable $\xi$ with distribution function $F(x)$:


I wonder how do I prove it?

My attempt follows:


Integrating in parts, I obtain


Passing to the limit, I get


So in order to prove the initial statement, I need to prove that for arbitrary distribution function $F$




however I have no idea how to prove it and moreover I doubt that it's true.

  • $\begingroup$ I guess the last 2 assumptions should be true if the expected value is finite. $\endgroup$ – kludg Mar 24 at 17:34

If the expectation is finite then both limits $$\lim\limits_{a\to-\infty}\int_{-\infty}^a x\,dF(x) \text{ and } \lim\limits_{b\to\infty}\int_b^{\infty} x\,dF(x)$$ are zero. Then $$ 0=\lim\limits_{a\to-\infty}\int_{-\infty}^a x\,dF(x) \leq \lim\limits_{a\to-\infty}a \int_{-\infty}^a dF(x) =\lim\limits_{a\to-\infty} aF(a)\leq 0. $$ And $$ 0=\lim\limits_{b\to\infty}\int_b^{\infty} x\,dF(x) \geq \lim\limits_{b\to\infty}b \int_b^{\infty} dF(x) =\lim\limits_{b\to\infty} b(1-F(b))\geq 0. $$


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