$\int_{\mathbb R+} P(Y-X>t)dt$ =$\int_{\mathbb R}P(X < y, Y >y)dy$

Question

i have $E[B]= \int_{\mathbb R+} P(Y-X>t)dt$ and i want to show this relation :
$\int_{\mathbb R+} P(Y-X>t)dt$=$\int_{\mathbb R}P(X < y, Y >y)dy$

I first began showing that for t ≥ 0, $ P(Y − X>t) = \int_{\mathbb R^2} 1_{y>x+t}f_{X|Y=y} (x) f_y(y) dxdy$

But i don't know how to continue

Even though there's already an accepted answer, I'd still like to link a closely related old post. — Lee David Chung Lin, Nov 23 '18 at 06:54

score 1 · Accepted Answer · answered Nov 21 '18 at 10:10

1

$\int_{\mathbb R} P(X<y,Y>y)dy=\int_{\mathbb R}\int I_{\{X<y<Y\}} dPdy=\int_{\mathbb R}\int I_{\{X<y<Y\}} dydP=\int (Y-X)^{+} dP$ and $\int (Y-X)^{+} dP=\int_0^{\infty} P(Y-X>t)dt$. I have used Fubini's Theorem and the fact that $Z=\int_0^{\infty} P(Z>t)\, dt$ for any non-negative random variable $Z$. Take $Z=(X-Y)^{+}$.

answered Nov 21 '18 at 10:10

Kavi Rama Murthy

311,013

thank you can you just specify what is the measure when you integrate with respect to dP – Farouk Deutsch Nov 21 '18 at 10:15
The random variables are defined on some probability space and $P$ is the probability measure on that space. – Kavi Rama Murthy Nov 21 '18 at 10:19
but this is two random variable that's the thing who stuck me what is the boundary of the second integral and what is the the space ? – Farouk Deutsch Nov 21 '18 at 10:25
Are you familiar with probability spaces and random variables on them? You are using density functions in your argument but not every random variable has a density. My integrals are over the sample space $\Omega$. – Kavi Rama Murthy Nov 21 '18 at 10:29
thank you finally i get it – Farouk Deutsch Nov 21 '18 at 10:35

$\int_{\mathbb R+} P(Y-X>t)dt$ =$\int_{\mathbb R}P(X < y, Y >y)dy$

1 Answers1