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Let the continuous random variables $X$ and $Y$ have the joint probability density function given by $f(x) = 3/2x$ for $0<x<2$, $0<y<1$, $x<2y$. Find Pr $(x<1.5|y>0.5)$.

This was how I approached the problem. But I know that Pr $(x<1.5 and y>0.5)$/$Pr(y>0.5)$. My problem is how to solve the numerator? I have tried several ways but keep getting $0$. Is there a better way to approach this?

J.R.
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  • You do not show what you have done to find the probability. So no one can say if there is a better approach. – Karl Nov 08 '14 at 18:10

1 Answers1

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Outline: Draw a picture. The joint density "lives" in the triangle $T$ with corners $(0,0)$. $(2,1)$, and $(0,1)$.

For the probability in the numerator, we want to integrate the joint density over the part $P$ of the triangle $T$ which is to the left of the line $x=1.5$ and above the line $y=0.5$. Draw these two lines to make the geometry of $P$ clear.

The issue is that the region $P$ does not have a very nice shape. But if you break it up suitably into two parts, the integrals will not be difficult. I suggest letting $P_1$ be the part of $P$ to the left of the line $x=1$, and $P_2$ the part between $x=1$ and $x=1.5$.

André Nicolas
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  • you have been very helpful. I did as suggested above and had two integrals. For the first part P1, I integrated $x$ from 0 to 1 and $y$ from 1/2 to 1. and for P2, I integrated $x$ from 1 to 3/2 and then $y$ from $x/2$ to 1. Is that a logical thing to do? – J.R. Nov 08 '14 at 18:38
  • And eventually got 11/32 for the numerator and 7/8 for the denominator? Is that how the problem needs to be approached? – J.R. Nov 08 '14 at 18:46
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    I calculated and got $11/32$ (the $7/8$ is almost instant). For the numerator, one can integrate first with respect to $y$ (which was my choice, hence $P_1$ and $P_2$), or with respect to $x$. Am puzzled by your "and then" wording because the and then part is done first. However, we probably did the same thing, certainly agree on the result. – André Nicolas Nov 08 '14 at 19:48
  • The "and then" wording meant having two separate integrals and summing them up. But I am tempted to believe both attempt should be right like you mentioned. – J.R. Nov 08 '14 at 20:22
  • I tried finding $P(x<1.5|y=0.5)$ but got 2.25. That doesnt appear right since the probability cannot be greater than one. What am I doing wrong? – J.R. Nov 08 '14 at 20:46
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    You got the right numbers, for the conditional probability divide, we get $11/28$. As to order of integration, we can use either if we get the limits right, but the second (outer) integral cannot be to a variable limit. – André Nicolas Nov 08 '14 at 21:45
  • For the probability given $y=0.5$, we need to find the conditional density function. That is a different type of problem. – André Nicolas Nov 08 '14 at 21:47
  • The conditional density function, I got $f(x|y)= x/(2y^2)$. I then integrated $x$ from $0$ to $1.5$ of the function above after substituting $y=0.5$. There must be something wrong somewhere. What could the problem be? – J.R. Nov 08 '14 at 22:26
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    Perhaps you will want to ask a separate question. But if you don't, note that the conditional density of $x$ given $y=0.5$ must be $0$ for $x\gt 1$, since given $y=0.5$, we must have $x\lt 1$. – André Nicolas Nov 08 '14 at 22:33
  • I finally solved it. I had $P(x<1.5|y=0.5)=1$. Your input definitely helped. – J.R. Nov 08 '14 at 23:45
  • Let us continue this discussion in chat. – J.R. Nov 09 '14 at 00:55