2
$\begingroup$

I have a joint PDF of the following form:

$f_{X,Y}(x,y) = 6e^{-(2x+3y)}$ for $x\geq 0$, and $y\geq 0$; and $f_{X,Y}(x,y) = 0$ otherwise.

My goal is to find the probability that $X > Y$. Since I have no clue where to start I've peaked at the solution but am still confused as to why it is set up the way it is:

\begin{align*} P[X\geq Y] &= \int_{0}^\infty \int_{0}^x 6e^{-(2x+3y)} \,dydx\\ &=\int_{0}^\infty 2e^{-2x}\left(-e^{-3y}\bigg|_{y=0}^{y=x}\right)\,dx\\ &=\int_0^\infty \left[2e^{-2x}-2e^{-5x}\right]\,dx \\ &=\frac{3}{5} \end{align*}

Can anybody please explain to me why the first line is set with the inside integral 0 to x? I'm also confused as to how the second line is derived from the first. I tried plugging in the inside integral into wolfram but got a totally different answer.

$\endgroup$
  • 2
    $\begingroup$ $Y$ is smaller or equal to $X$ in case of $P(X\geq Y)$. That´s why $y$ goes from $0$ to $x$ $\endgroup$ – callculus May 29 '17 at 22:09
  • $\begingroup$ Furthermore to @callculus comment - If you draw out an $x$-$y$ domain with the relationship of $x$ to $y$ then you can quickly see the valid domains. This also allows one to find simpler transformations to make the integral simpler in some cases. $\endgroup$ – Chinny84 May 29 '17 at 22:19
  • $\begingroup$ Thanks @callculus Could you explain how the second line is derived from the first by any chance? I don't get how the second line is structured. $\endgroup$ – MarksCode May 29 '17 at 22:27
  • $\begingroup$ Which line are you talking about? within the actually derivation or the from the first line with the problem to the first line of the derivation? . $\endgroup$ – Chinny84 May 29 '17 at 22:33
  • 1
    $\begingroup$ Please use MathJax and type out all images. Formatting tips here. $\endgroup$ – Em. May 29 '17 at 22:37
2
$\begingroup$

To calculate the integral w.r.t $y$ the variable $x$ can be treated like a constant. So it is a good idea to split.

$6e^{-(2x+3y)}=6e^{-2x}\cdot e^{-3y}$

$\int_0^{\infty}6e^{-2x} \left(\int_0^x e^{-3y} \, dy \right)\, dx$

The antiderivative of $e^{-3y} $is $-\frac13 e^{-3y}$

The factor $ \frac13$ and the factor 6 can be summarized to $2$

Therefore it remains to insert the limits $0$ and $x$ into $e^{-3y}$

$\endgroup$

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.