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I would like to find the cdf of $Z=X_1+X_2$, with $X_1\sim U(0,1) $, $X_2\sim U(0,2)$

I always prefer to find the cdf instead of the pdf with convolution, and this time I am having troubles with the last part of the cdf.

I split the cdf $F_Z(a)$ in 3 cases and I found:

$\frac{a^2}{2}-\frac{a^2}{4}\quad 0<a\le 1$

$\frac{a}{2}-\frac{1}{4}\, \quad 1<a\le 2$

$\frac{a}{2}-\frac{a^2}{4}+\frac{3}{4}\quad2<a\le 3$

I am omitting the part for $a\le0$ and $a>3$, but I don't know what to do with the last part of the cdf, I did this:

$\int_{a-2}^1F_{X_2}(a-x_1)dx_1$

I am fully convinced of the boundaries of integration of $X_1$, yet I can't get the correct answer for the last part of the cdf in that way. Any help would be greatly appreciated.

Drush
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  • Since the distributions are uniform, you could draw the rectangle $(0,1) \times (0,2)$ and consider the area below $x_1=x_2$? – Chappers Apr 27 '15 at 19:27
  • I have never understood how the graph can help with convolution, for instance, I can't visualize on the graph actually where is the sum x1+x2, I always used the analytical way and now I don't understand what isn't working because with that boundaries I ensured that $0\le a-x_1\le 2$ and $0\le x_1\le 1$ – Drush Apr 27 '15 at 19:31
  • The picture helps to set up the integral when we need to evaluate one, and in this case to bypass integration entirely. – André Nicolas Apr 27 '15 at 19:37
  • My point is that each point of $(0,1) \times (0,2) \cap { (x_1,x_2) : x_1+x_2 \leqslant a }$ is a possible value of $(X_1,X_2)$ that satisfies $X_1+X_2 \leqslant a$. Since all points are equally likely, you can just calculate the area of this and divide by the total area to find the probability. – Chappers Apr 27 '15 at 19:38
  • I may understand a bit better, but, can you split the cases with the graph? I need the distribution to evaluate the moments – Drush Apr 27 '15 at 19:41
  • You may most certainly divide the cases with the graph. On a sheet of graph paper, draw a rectangle with corners at $(0, 0), (1, 0), (0, 2)$, and $(1, 2)$. The horizontal axis represents $X_1$ and the vertical axis represents $X_2$. The line through $(0, a)$ and $(a, 0)$ represents $Z = X_1+X_2 = a$; the region to the lower left of that represents $Z < a$, and the area of that region represents the CDF. You will have to divide by the total area of the rectangle, which is $2$. You should find fairly quickly that you have the same cases, with the same expressions. – Brian Tung Apr 27 '15 at 20:27

1 Answers1

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I, too, feel it is easier to write the PDF or the CDF by inspection from the graph, but if you wish to stick with convolution for the CDF, your problem is that you are missing a term that represents the cases where $0 \leq X_1 \leq a-2$, and then $X_2$ can be anything (that is, it is certain that the sum is less than $a$, no matter the value of $X_2$). That is, the proper expression is

$$ F_Z(a) = \int_{x_1=a-2}^1 F_{X_2}(a-x_1) \, dx_1 \color{red}{ + \int_{x_1=0}^{a-2} \, dx_1} \qquad 2 \leq a \leq 3 $$

I think if you carry out that integration, you will get the proper answer of

$$ F_Z(a) = -\frac{a^2}{4} + \frac{3a}{2} - \frac{5}{4} \qquad 2 \leq a \leq 3 $$

Brian Tung
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