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For $$ f(x,y) = \begin{cases}\frac{1}{y} & 0 < x < y < 1\\ 0 & \text{elsewhere} \end{cases} $$ find $P(X + Y > 1/2)$

First i should make it $1 - P(X + Y < 1/2)$

Then it will be a double integral but i dont know what to put for the bounds of the double integral.

I ended up concluding to this.

x < y < 1/2

if i subtract x on both sides

y < 1/2 - x so bounds for the y integral is x to 1/2 -x

and for x

0 < x < y
0 < x < 1/2 - x

if i add x.

0 < 2x < 1/2

and divide by 2

0 < x < 1/4.  so bounds for x is 0 to 1/4

Would this a correct way of doing it?

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  • $\begingroup$ what is the function f? probability distribution function? $\endgroup$ – MoonKnight Dec 13 '13 at 22:53
  • $\begingroup$ it is a Joint density function $\endgroup$ – user2036503 Dec 13 '13 at 22:58
  • $\begingroup$ Do you know how the area of $X+Y<\frac{1}{2}$ look like? $\endgroup$ – hhsaffar Dec 13 '13 at 23:02
  • $\begingroup$ Would it be a triangle? I actually dont know how to do the area can u give me a hint? $\endgroup$ – user2036503 Dec 13 '13 at 23:03
  • $\begingroup$ Your edit is correct. $\int_0^{1/4}dx\int_x^{1/2-x}dy$ is the correct integral boundary. $\endgroup$ – MoonKnight Dec 13 '13 at 23:41
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Draw a picture. Make sure that you see that the joint density function "lives" in the triangle with corners $(0,0)$, $(1,1)$, $(0,1)$, but is undefined at $(0,0)$.

Now draw the line with equation $x+y=\frac{1}{2}$. We want to find the probability that the pair $(X,Y)$ lands in the part of the triangle above that line.

We could find this probability directly, or else find the probability $(X,Y)$ lands below the line, and subtract the result from $1$.

We use the second approach, although it is really not much better.

So how shall we integrate? We can first integrate with respect to $y$, then with respect to $x$. This looks cleaner. Maybe you should try it.

But I will instead integrate first with respect to $x$. This has the disadvantage that we need ot break up things into two parts, $0\lt y\le \frac{1}{4}$ and $\frac{1}{4}\le y\le \frac{1}{2}$.

For the first integral, we want $$\int_{y=0}^{1/4}\left(\int_{x=0}^y \frac{1}{y}\,dx\right)\,dy.$$ The inner integral is beautifully easy, it is simply $1$! For the second integral, we want $$\int_{y=1/4}^{1/2}\left(\int_{x=0}^{1/2-y} \frac{1}{y}\,dx\right)\,dy.$$ Not too bad, the inner integral is $\frac{1}{2y}-1$.

Please recall that we will need to subtract the sum of the two integrals from $1$.

Other approaches will also work. Integrating first with respect to $y$ has the disadvantage that we get a $\ln$ immediately, which makes the seond integral somewhat harder.

Or else we can work directly with the region above the line $x+y=\frac{1}{2}$. Again, it is more efficient to integrate first with respect to $x$. We need two integral, $y=\frac{1}{4}$ to $\frac{1}{2}$, and $\frac{1}{2}$ to $1$.

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  • $\begingroup$ I really dont understand how you got the bounds but i appreciate such a long post. Can you explain how you got the bounds. Im still confused about that. $\endgroup$ – user2036503 Dec 13 '13 at 23:18
  • $\begingroup$ I drew a picture and looked at it. It is related to the picture drawn by MoonKnight, except that that one is not right, we want to be above $y=x$. But for my $1$ minus approach (the one I mainly wrote about) we are integrating over the pink region. I am integrating first with respect to $x$, so the story is different $y=0$ to $1/4$ and $y=1/4$ to $1/2$. For the first one, $x$ goes from $0$ to the line $y=x$, that is, to $y$. For the second one, we go from $x=0$ until we bump into the line $x+y=1/2$, so $x$ goes $0$ to $1/2-y$. $\endgroup$ – André Nicolas Dec 13 '13 at 23:25
  • $\begingroup$ I tried to algebraically find the bounds in my edit. Would what i wrote be correct? $\endgroup$ – user2036503 Dec 13 '13 at 23:28
  • $\begingroup$ More: If you are finding the probability we have $X+Y\le \frac{1}{2}$, your bounds are right. (I was also finding that probability, and said so.) I am not sure you know that's what the bounds give. My advice is to use the geometry primarily, and to use algebra only for minor numerical details. Your bounds are for integration first with respect to $y$, which is not optimal. $\endgroup$ – André Nicolas Dec 13 '13 at 23:34
  • $\begingroup$ I was told by my teacher thats what the bounds will be. $\endgroup$ – user2036503 Dec 13 '13 at 23:38
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The area satisfies $0<x<y<1$ is everywhere except the blue area. The area satisfies $x+y<1/2$ is the red area.

To find $P(x+y<1/2)$, you should integrate over the only red area.

enter image description here

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  • $\begingroup$ I did an edit , would that be a correct way of finding the bounds? $\endgroup$ – user2036503 Dec 13 '13 at 23:19
  • $\begingroup$ I gave you a +1 for the picture, but you should be above the blue line if $x<y$ $\endgroup$ – Eleven-Eleven Dec 13 '13 at 23:38
  • $\begingroup$ Thanks, Chris. I just edited my answer to correct that. $\endgroup$ – MoonKnight Dec 13 '13 at 23:42

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