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For any distribution function and any $a \geq 0$, $\int_{-\infty}^{\infty} (F(x+a)-F(x))dx = a$.

In this case, "distribution function" means a right continuous function F with $F(-\infty) = 0$, $F(\infty)= 1$.

This problem comes after a section in the book on mathematical expectation, so I suspect I should use some property of expectation to prove it. Expectation appears to be something that applies to random variables, so I think I should be somehow transforming the problem into something in terms of expectations of random variables, but I'm not sure how. I know that given a distribution function, I can get a probability measure, but I'm not sure how to go from there to a random variable and actually get a relation between the expectation of the random variable and this integral. I'm sure whether this is the right way to go about this problem, but if it is, I'm missing a connection somewhere.

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If $X$ is a random variable with distribution function $F$, then $$ F(x+a)-F(x)=P(x<X\leq x+a)={\rm E}[\mathbf{1}_{x<X\leq x+a}]. $$ Since everything is non-negative we have by Tonelli that $$ \int_{-\infty}^\infty (F(x+a)-F(x))\,\mathrm dx={\rm E}\Big[\int_{-\infty}^\infty \mathbf{1}_{x<X\leq x+a}\,\mathrm dx\Big]. $$ Now evaluate the inner integral and conclude.

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  • $\begingroup$ I'm sorry to bother you, but could you please elaborate on how to evaluate the inner integral? $\endgroup$ – Ant Oct 4 '14 at 18:40
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    $\begingroup$ @Ant: Note that $\mathbf{1}_{x<X\leq x+a}=\mathbf{1}_{X-a\leq x<X}$ and hence the integral is $\int_{X-a}^X\mathrm dx=a$. $\endgroup$ – Stefan Hansen Oct 4 '14 at 18:47
  • $\begingroup$ Great! Thank you very much :D $\endgroup$ – Ant Oct 4 '14 at 18:51
  • $\begingroup$ This looks so obvious now that you've said it. I'm just having some weird kind of cognitive difficulty moving between ideas from measure theory and ideas from probability. $\endgroup$ – Xindaris Oct 4 '14 at 18:56

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