# Joint PDF P[X+Y<=0.5]

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?

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 $$\frac1{48}$$: $$(\frac{x^2}{4} - \frac{x^3}{3} \color{red}{-} \frac{(0.5-x)^3}{6})\Big|_0^{0.5}=\left(\frac{1}{16}-\frac1{24}-0\right)-\left(0-0-\frac1{48}\right)=\frac1{24}.$$

• Farruhota, thank you for the awesome feedback. I really appreciate your help. – MitterHai Dec 5 '18 at 11:42
• You are welcome. Good luck. – farruhota Dec 5 '18 at 11:43

Anti-derivative of $$\frac {(0.5-x)^{2}} 2$$ is $$-\frac {(0.5-x)^{3}} 6$$. You are missing the minus sign.

• 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? – MitterHai Dec 5 '18 at 10:38
• @MitterHai When you evaluate $-(0.5 - x)^3/6$ at $0$, you get a non-zero value. This fixes the problem. – littleO Dec 5 '18 at 11:22
• What a face palm moment. Thanks Pro..fe.. I mean littleO! – MitterHai Dec 5 '18 at 11:40