Prove an equality using Fubini's Theorem

Asked Mar 01 '22 at 20:02

Active Mar 01 '22 at 20:02

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Let $f(x)$ be integrable and non-negative. Let $\int_{-\infty} ^\infty f(x)dx = 1$ and let $F(x)=\int_{-\infty} ^xf(y)dy$. Also, assume $xf(x)$ is integrable. Use Fubini's Theorem to show that:

$\int_{0}^\infty (1-F(x))dx = \int_0 ^\infty xf(x) dx$.

Well, my initial idea was to substitute the definition of $F(x)$ into the LHS. But that "1" there stands in the way . I tried to split the integral to include the 1 in it somehow, but it also does not work. What is the trick here?

asked Mar 01 '22 at 20:02

Desintegration

1

It may help that $1-F(x)=\int_x^{\infty}f(y)dy$. – Martin R Mar 01 '22 at 20:06
yes this is common proof of why tail sum formula gives expectation in probability, apply what Martin said and then swap the integrals – John D Mar 01 '22 at 20:08
Or this: https://math.stackexchange.com/a/2335395/42969 – Martin R Mar 01 '22 at 20:14

Prove an equality using Fubini's Theorem

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