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This is a problem on Freund's Mathematical statistics book on page 101

If the independent random variables $X$ and $Y$ have the marginal densities

$f(x) = \begin{cases} 1/2 \space \text{for} \space 0<x<2 \\ 0 \space \text{elsewhere} \end{cases}$

$\pi(y) = \begin{cases} 1/3 \space \text{for} \space 0<y<3 \\ 0 \space \text{elsewhere} \end{cases}$

Find: $\space$ (a)the joint probability density of $X$ and $Y$ (b) the value of $P(X^2 + Y^2 > 1)$

I got the first answer since it's independent $f(x,y) = \begin{cases} 1/6 \space \text{for} \space 0<x<2, 0<y<3 \\ 0 \space \text{elsewhere} \end{cases}$

How do I approach the second one, like I am not too sure of the limits of the integrals

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  • $\begingroup$ You need to integrate over the region of the joint density where this statement holds. Think of this as the continuous analog of "adding up all the probabilities." $\endgroup$ – dsaxton Dec 1 '18 at 17:39
  • $\begingroup$ I get that, but I'm having trouble with the limits will it be $\int_{1}^{2} \int_{\sqrt(1-x)}^{3} 1/6 dy dx$ ? $\endgroup$ – Sumukh Sai Dec 1 '18 at 17:44
  • $\begingroup$ Draw a graph, find the area such that $X^2 + Y^2 > 1$. Integral the joint pdf on the area, you get the probability. $\endgroup$ – user158565 Dec 1 '18 at 17:47
  • $\begingroup$ Are you sure your limits in $\int_{1}^{2} \int_{\sqrt(1-x)}^{3}$ are correct? The answer is 1-$\pi/24$ $\endgroup$ – user158565 Dec 1 '18 at 17:51
  • $\begingroup$ No, that's where I'm having trouble. Don't know if my limits are correct $\endgroup$ – Sumukh Sai Dec 1 '18 at 17:54
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$X^2 + Y^2 > 1$ = area A

$\Pr(A) = 1 - \Pr(-A)$

$\Pr (-A) = \int_{0}^{1} \int_{0}^{\sqrt{1-y^2}} \frac 16 dy dx $

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  • $\begingroup$ Oh good grief! What you call the region "$-A$" (ugh!) has area $\pi/4$ and so the probability that $X^2+Y^2$ exceeds $1$ is just $1-\frac{\pi}{24}$, no mss, no fuss, no complicated integrals etc. $\endgroup$ – Dilip Sarwate Dec 1 '18 at 19:55

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