So I have two independent random variables $X$ and $Y$, $Y$ ~ $U[0, 3]$ and the density function of $X$ is as follows:

$f(x) = 1/3$ if $0\le x \le 1$

$f(x) = 2/3$ if $1\le x \le 2$

$f(x) = 0$ otherwise

I calculated the joint density of this function by multiplying them, since they are independent. And now I'm trying to find $P(Y\gt X)$

Thanks in advance!

  • $\begingroup$ if you think the answer is useful @bernas, can you please accept and/or upvote it? $\endgroup$ – gunes Feb 12 at 10:13

The support is the rectangle $\mathcal{R}=[0,3]\times[0,2]$, and for $P(X>Y)$, you'll integrate the area under $Y=X$ line: $$\int_{\mathcal R} f_{X,Y}(x,y)dydx=\int_{0}^1\int_0^x \frac{1}{9} dy dx+\int_1^2\int_0^x\frac{2}{9} dydx$$

  • $\begingroup$ Actually I made a mistake in the post, what I want is P(Y>X), what changes would have to be made? $\endgroup$ – bernas Feb 11 at 19:38
  • 3
    $\begingroup$ come on! subtract it from 1 $\endgroup$ – gunes Feb 11 at 19:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.