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I am having some difficulty with this probability that I need to compute, and I could use some help from other eyes to see where I am messing up:

Joint PDF: $x+y$ for $0 < x < 1$, $0 < y < 1$.

I need to find the $P(x+y\leq 0.5)$.

For the double integration, I have the following bounds:

  • Outer bound is respectively to $x$ and is from $0$ to $0.5$.

  • Inner bound is respective to $y$ and is from $0$ to $0.5 - x$.

Here is the initial set up:

$\int_0^.5 \int_0^{.5-x}(x +y) dydx$.

My steps of integration:

1: $\int_0^.5 \left(xy + \frac{y^2}{2}\right)\Big|_0^{.5-x}dx$.

2: $\int_0^.5$ $\left(\frac{x}{2} -x^2+\frac{(0.5-x)^2}{2}\right) dx $.

3: $(\frac{x^2}{4} - \frac{x^3}{3} + \frac{(0.5-x)^3}{6})\Big|_0^{0.5}$.

I am getting a negative number based on this final equation, and the answer should be $1/24$. Can someone help me figure out where I went wrong?

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Your made the sign error in step 2 (as indicated by KaviRamaMurthy).

As an alternative, you can simplify in step 2 before integrating: $$\int_0^{0.5}\left(\frac{x}{2} -x^2+\frac{(0.5-x)^2}{2}\right) dx= \int_0^{0.5}\left(\require{cancel}\cancel{\frac{x}{2}} -x^2+\frac{1}{8}-\cancel{\frac x2}+\frac{x^2}{2}\right) dx=\\ \int_0^{0.5}\left(-\frac{x^2}{2}+\frac18\right) dx= \left(-\frac{x^3}{6}+\frac x8\right)|_0^{0.5}= -\frac{1}{48}+\frac{1}{16}= \frac 1{24}.$$

Answering your comment, yes, for the upper limit $0.5$ the term will be $0$, but for the lower limit $0$, the term will be $\frac1{48}$: $$(\frac{x^2}{4} - \frac{x^3}{3} \color{red}{-} \frac{(0.5-x)^3}{6})\Big|_0^{0.5}=\left(\frac{1}{16}-\frac1{24}-0\right)-\left(0-0-\frac1{48}\right)=\frac1{24}.$$

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  • $\begingroup$ Farruhota, thank you for the awesome feedback. I really appreciate your help. $\endgroup$ – MitterHai Dec 5 '18 at 11:42
  • $\begingroup$ You are welcome. Good luck. $\endgroup$ – farruhota Dec 5 '18 at 11:43
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Anti-derivative of $\frac {(0.5-x)^{2}} 2$ is $-\frac {(0.5-x)^{3}} 6$. You are missing the minus sign.

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  • $\begingroup$ Hi Kavi, thank you for taking on this problem, I figured out how the anti-derivative should have a minus sign, but when you evaluate it at 0.5, doesn't this part of the equation just go to 0? $\endgroup$ – MitterHai Dec 5 '18 at 10:38
  • $\begingroup$ @MitterHai When you evaluate $-(0.5 - x)^3/6$ at $0$, you get a non-zero value. This fixes the problem. $\endgroup$ – littleO Dec 5 '18 at 11:22
  • $\begingroup$ What a face palm moment. Thanks Pro..fe.. I mean littleO! $\endgroup$ – MitterHai Dec 5 '18 at 11:40

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