The expected value as an integral

Asked Apr 29 '17 at 18:55

Active Apr 29 '17 at 18:55

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Let $X \geq 0$ be continuous random variable with the CDF $F(x) := \displaystyle\int_0^x f(x) \; \mathrm dx$ where $f(x)$ is the PDF.

I want to express the (finite) expected value $E[X] := \displaystyle\int_0^\infty x f(x) \; \mathrm dx < \infty$ as an integral that includes the failure rate function $r(t) = \frac{f(x)}{1 - F(x)}$ but not the PDF, CDF or survival function $G(x) = 1 - F(x)$.

asked Apr 29 '17 at 18:55

Markus Klyver

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Nothing fancy here: $\displaystyle E[X] = \int_0^{\infty} S(t)dt = \int_0^{\infty} \exp\left{-\int_0^t r(u)du\right}dt$ – BGM Apr 29 '17 at 19:07
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@BGM Could you justify your steps? – Markus Klyver Apr 29 '17 at 20:38
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@MarkusKlyver Hope it helps: https://math.stackexchange.com/questions/172841/integral-of-cdf-equals-expected-value?rq=1 – Ramil Apr 29 '17 at 22:15

The expected value as an integral

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