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I'm given equation that the joint pdf is $(2/3)(x + 2y)$ when $0 < x, y < 1$ and we want to find the probability that $X < 1/3 + Y$.

I understand how to do the actual math part, and that I have to do a double integral. But what are the inner and outer limits? I thought that the inner integral limits would be from $x - 1/3$ to $1$, but (if that part's even right) I can't figure out what the outer integral limits would be.

user642796
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Shell
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1 Answers1

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In the coordinate plane, draw the $1\times 1$ square on which the density function lives. Draw the line $x=\frac{1}{3}+y$, that is, $y=x-\frac{1}{3}$. We want the probability of being above the line.

The region above the line is a little messy, it is easier to find the probability of being below the line, that is, in a certain triangle that will be clear from your drawing.

Now it does not matter much whether we integrate first with respect to $x$ or with respect to $y$. Let's do it with respect to $y$. Then $y$ will go from $0$ to $x-\frac{1}{3}$. And after that, $x$ will go from its smallest possible value, namely $x=\frac{1}{3}$, to $x=1$.

After we have done the integration, if the result is $b$, the answer to the original problem is $1-b$.

Remark: If you want to find the probability directly, then you will have to break up the integral into $2$ integrals, the part where $0\le x\le \frac{1}{3}$ and the part where $\frac{1}{3}\le x\le 1$. You wrote down the correct limits on $y$ for the part where $x\ge \frac{1}{3}$. Then $x$ goes from $\frac{1}{3}$ to $1$.

But we will also have to deal with $0\le x\le \frac{1}{3}$. There, $y$ goes from $0$ to $1$.

I feel that drawing a careful picture is almost a necessity if one is to solve the problem with significant confidence of being right.

André Nicolas
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  • The region that directly answers your question is a $4$-sided figure, technically a special trapezoid. We can integrate over it, as pointed out in the Remark at the end of the answer. But we have to break up the integral into two parts, if we want to do it that way. – André Nicolas Apr 17 '14 at 02:21
  • Right, that makes sense and I definitely see that in the drawing looking at it. I get that the area of said trapezoid is 227/243... Is that correct...? – Shell Apr 17 '14 at 02:37
  • My arithmetic is not reliable, but I get that the probability (not area) is $\frac{199}{243}$. The area is simple, $\frac{5}{9}$, but irrelevant. – André Nicolas Apr 17 '14 at 02:51
  • Right... I meant probability, I got 5/9 as the area. – Shell Apr 17 '14 at 02:53
  • I keep getting 155/243, but I know that is incorrect... – Shell Apr 17 '14 at 02:55
  • It's either your 199/243, or I now have 227/243... – Shell Apr 17 '14 at 03:10
  • I did it again, and got $199/243$. So either same mistake, or it is right. First integral is $(2/3)(xy+y^2)$. Plug in, simplify. We get $(2/3)(2x^2-x+1/9)$. Now integrate from $1/3$ to $1$. We get $44/243$. Take $1$ minus this. – André Nicolas Apr 17 '14 at 03:19
  • You're right, I see my mistake... I didn't integrate the x to get the xy. I should get it this time. Thank you so much! – Shell Apr 17 '14 at 03:22
  • You are welcome. In case it was one source of the difficulty, I wrote the comment about $(2/3)(xy+y^2)$, and the substitution. – André Nicolas Apr 17 '14 at 04:06