Suppose $A$ and $B$ are random variables and the only restriction on the probability density function $f_{A,B}(a,b)$ is $P(B>A) = 1$. Also suppose $p(x) = P(A \leq x \leq B)$ where $x$ is $fixed$ and is not considered a random variable. How do you show the expected length of [A,B] interval is
$\int_{-\infty}^{\infty}p(x)dx$ ?
So by definition of expectation we know $E[length([A,B]] = $
$= E[b-a] $
Target$= \int_{-\infty}^{\infty}(B-A) f_{A,B}(a,b)dA dB $ ---> SHOW = $\int_{-\infty}^{\infty}p(x)dx$
So now we know our target. Starting from :
$\int_{-\infty}^{\infty}p(x)dx = $
$= \int_{-\infty}^{\infty} P(A \leq x \leq B)dx$ (by definition of $p(x)$)
$= \int_{-\infty}^{\infty} P( (A \leq x) \cap (B \geq x)dx$ (by definition of intersection)
This means A ranges from $(-\infty,x]$ and B ranges from $[x,\infty)$ and can be re-written as 2 integrals using the joint PDF $f_{A,B}$
$= \int_{-\infty}^{\infty} dx \int_{-\infty}^{x} dA \int_{x}^{\infty} f_{A,B}dB$
This does not look like our target, but noticing that $\int_{a}^{b}dx = b-a$ I try to change the order of integration of $x$ and attempt to move it from outer-most to inner-most integral, pushing A, and B outward.
$= \int_{-\infty}^{\infty} da \int_{a}^{\infty} f_{A,B} dB \int_{a}^{b} dx$
$= \int_{-\infty}^{\infty} da \int_{a}^{\infty} f_{A,B} dB (b-a)$
$= \int_{-\infty}^{\infty}(b-a) da \int_{a}^{\infty} f_{A,B} dB $
So now I am much closer to my target, except I have an integral $\int_{a}^{\infty}$ that needs to somehow be canceled out. I know this has something to do with $P(B \geq A)=1$ but I'm not sure how to make this last connection.
This is Tonelli: start from the identity between random variables $$\mathsf{length}[A,B]=\int_A^B\mathrm dx=\int_{-\infty}^\infty\mathbf 1_{A\leqslant x\leqslant B}\,\mathrm dx.$$ Integrating both sides with respect to $P$ and exchanging the order of the integrals with respect to $P$ on $\Omega$ and with respect to $\mathsf{Leb}$ on $\mathbb R$ yields $$E(\mathsf{length}[A,B])=\int_{-\infty}^\infty E(\mathbf 1_{A\leqslant x\leqslant B})\,\mathrm dx=\int_{-\infty}^\infty P(A\leqslant x\leqslant B)\,\mathrm dx=\int_{-\infty}^\infty p(x)\,\mathrm dx.$$