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Suppose that $X_1$ and $X_2$ are i.i.d random variables and that each of them has a uniform distribution on the interval [0,1]. Find the probability density function $Y = X_1 + X_2$

I understand that a uniform distribution is given as $f(x_1,x_2) = 1$ for $0 < x_1 < 1$ and $0 < x_2 < 1$.

Given that: $Pr[Y \le y]=Pr[X_1+X_2 \le y] = Pr[X_1 \le y - X_2]$

Where do I go from here?

I know that it must be within the interval $0 < y-x_1<1$.

While using the equation for the sum of two random variables:

$g(y) = \int_{-\infty}^{\infty}f(y-z, z)dz$

I know that $f(y-z, z) = 1$ because it is on a uniform distribution. When computing the boundaries, do I go from $0 $ to $y-x$, or $0$ to $y$ and why?

Having a quick look at the solution to this exercise, It seems to be of two parts. The first is what I am looking for, and the second is computing the integral when $1 < y <2$, though I cannot understand where these boundaries are taken from?

Update: Found this youtube link that explains this clearly also

Stackcans
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  • The reason for a split is that if $y=x_1+x_2 < 1$ then $0 \le x_1 \le y$ while if $y =x_1+x_2>1$ then $y-1 \le x_1 \le 1$ and that leads to two different limits on the integral – Henry Aug 16 '21 at 12:51
  • @Henry This may be a silly question, but why is $y = x_1+x_2>1$? – Stackcans Aug 16 '21 at 12:55
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    $Y$ is the sum of two values between $0$ and $1$ so $Y$ must be between $0$ and $2$, in which case it can be $<1$ or $>1$ or $=1$. You will find it easier to find the solution if you treat the $0 \le Y<1$ and $1 < Y\le 2 $ cases separately - you can put the $Y=1$ case into either – Henry Aug 16 '21 at 12:57
  • @Henry Thank you for the explanation. Then when we re-arrange for $x_1$, we get $y-x_2 \le x_1 \le 1$ because x falls between 0 & 1, and get the secondary boundary you mentioned or is this achieved with a different reason? – Stackcans Aug 16 '21 at 13:00
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    Draw the region of support of $(X_1,X_2)$ on the $(x_1,x_2)$ plane. Draw on a line for $y=x_1+x_2=c$ for the cases where $0 – Glen_b Aug 16 '21 at 15:02
  • If you use R, the following code may help with visualization: `x1 = runif(10^6,0,1); x2 = runif(10^6,0,1); y = x1+x2; hist(y, prob=T, br=30)`. Also this link to [Irwin-Hall](https://en.wikipedia.org/wiki/Irwin%E2%80%93Hall_distribution) distributions. – BruceET Aug 16 '21 at 16:58
  • @BruceET Thank you for the visualisation! It has helped me better understand the boundaries for X. However, I seem to be mistaken in thought on noting the start and endpoint of the segment as mentioned by Glen_b in helping me determine the boundaries for $\int_0^y$ and $\int_{y-1}^1$. Could you kindly support this with an illustration? – Stackcans Aug 16 '21 at 17:56
  • Joint dist'n of $(X,Y)$ is uniform on square with vertices at $(0,0), (1,1).$ To find $P(Y = X+Y \le a),$ for $a < 1,$ you need to integrate over triangle bounded by x-axis, y-axis and line $x+y=a.$ For $Y > 1$ you need to exclude a triangle at upper right. Draw a sketch. Treat $P(Y < a)$ as two cases, depending on $a < 1$ or $a > 1.$ Differentiate CDF to get triangular PDF with apex at $y=1.$ – BruceET Aug 16 '21 at 20:24

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