William Curry

2021-12-14

Let's say Z=X+Y, where X and Y are independent uniform random variables with range [0,1]. Then the PDF is

$f\left(z\right)=\{\begin{array}{ll}z& for0z1\\ 2-z& xfor1\le z2\\ 0& otherwise.\end{array}$

How was this PDF obtained?

sonSnubsreose6v

Beginner2021-12-15Added 21 answers

If we want to use a convolution, let $f}_{X$ be the full density function of X, and let $f}_{Y$ be the full density function of Y. Let Z=X+Y. Then

$fz\left(z\right)={\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}{f}_{X}\left(x\right){f}_{Y}(z-x)dx$ .

Now let us apply this general formula to our particular case. We will have${f}_{Z}\left(z\right)=0f{\textstyle \phantom{\rule{1em}{0ex}}}\text{or}{\textstyle \phantom{\rule{1em}{0ex}}}z<0,$ and also for $z\ge 2$ . Now we deal with the interval from 0 to 2. It is useful to break this down into two cases (i) $0<z\le 1{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\left(ii\right)1<z<2$ .

(i) The product${f}_{X}\left(x\right){f}_{Y}(z-x)$ is 1 in some places, and 0 elsewhere. We want to make sure we avoid calling it 1 when it is 0. In order to have ${f}_{Y}(z-x)=1,we\ne edz-x\ge 0,t\hat{i}s,x\le z$ . So for (i), we will be integrating from x = 0 to x = z. And easily

${\int}_{0}^{z}1dx=z$

Thus${f}_{Z}\left(z\right)=zf{\textstyle \phantom{\rule{1em}{0ex}}}\text{or}{\textstyle \phantom{\rule{1em}{0ex}}}0<z\le 1$

(ii) Suppose that 1<z<2. In order to have fY(z−x) to be 1, we need$z-x\le 1,t\hat{i}s,we\ne edx\ge z-1$ . So for (ii) we integrate from z−1 to 1. And easily

${\int}_{1}^{z-1}1dx=2-z$

Thus${f}_{Z}\left(z\right)=2-zf{\textstyle \phantom{\rule{1em}{0ex}}}\text{or}{\textstyle \phantom{\rule{1em}{0ex}}}1<z<2$ .

Another way: (Sketch) We can go after the cdf${F}_{Z}\left(z\right)ofZ$ , and then differentiate. So we need to find $Pr(Z\le z)$ .

For a few fixed z values, draw the lines with equation x+y=z on an x-y axis plot. Draw the square S with corners (0,0), (1,0), (1,1), and (0,1).

Then$Pr(Z\le z)$ is the area of the part S that is "below" the line x+y=z.
That area can be calculated using basic geometry. For example, when z
is 2, the whole square area is under the line so Pr=1. There is a switch
in basic shape at z=1.

Now let us apply this general formula to our particular case. We will have

(i) The product

Thus

(ii) Suppose that 1<z<2. In order to have fY(z−x) to be 1, we need

Thus

Another way: (Sketch) We can go after the cdf

For a few fixed z values, draw the lines with equation x+y=z on an x-y axis plot. Draw the square S with corners (0,0), (1,0), (1,1), and (0,1).

Then

ol3i4c5s4hr

Beginner2021-12-16Added 48 answers

By the hint of jay-sun, consider this idea, if and only if $fX(z-y)=1when0\le z-y\le 1$ . So we get

$z-1\le y\le z$

however,$z\in [0,2]$ , the range of y may not be in the range of [0,1] in order to get ${f}_{X}(z-y)=1$ , and the value 1 is a good splitting point. Because $z-1\in [-1,1]$ .

Consider$\left(i\right){\textstyle \phantom{\rule{1em}{0ex}}}\text{if}{\textstyle \phantom{\rule{1em}{0ex}}}z-1\le 0then-1\le z-1\le 0t\hat{i}sz\in [0,1]$ , we get the range of $y\in [0,z]\mathrm{sin}cez\in [0,1]$ . And we get ${\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}{f}_{X}(z-y)dy={\int}_{0}^{z}1dy=z{\textstyle \phantom{\rule{1em}{0ex}}}\text{if}{\textstyle \phantom{\rule{1em}{0ex}}}z\in [0,1]$ .

Consider$\left(ii\right){\textstyle \phantom{\rule{1em}{0ex}}}\text{if}{\textstyle \phantom{\rule{1em}{0ex}}}z-1\ge 0t\hat{i}sz\in [1,2]$ , so we get the range of y $\in [z-1,1],{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}{\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}{f}_{X}(z-y)dy={\int}_{z-1}^{1}1dy=2-z{\textstyle \phantom{\rule{1em}{0ex}}}\text{if}{\textstyle \phantom{\rule{1em}{0ex}}}z\in [1,2]$

To sum up, consider to clip the range in order to get${f}_{X}(z-y)=1$ .

however,

Consider

Consider

To sum up, consider to clip the range in order to get

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