$X$ and $Y$ independent and uniform on $[0,1]$. Find PDF of $X+Y$.

Question

The idea here is to use convolution. The books does the following:

Let $z=X+Y$. I have that $$f_Z(x)=\int_{-\infty}^{\infty}f_Y(x-u)f_X(u) \ du.\tag 1$$

First we note that the range of $Z$ is $[0,2],$ so we need to pick our $x$ from that interval. Now $f_X(u)=1$ if $u$ is between $0$ and $1$ and $0$ otherwise. For the other factor in the integrand, note that $f_Y(x-u)=1$ if its argument $x-u$ is between $0$ and $1$, that is $\color{green}{0\le x-u \le 1}$, and $0$ otherwise. Since $x$ is fixed and $u$ is the variable, we rewrite this as $\color{red}{x-1\le u \le 1}.$ Hence for any $x\in[0,2]$ the two inequalities

$$0\le u\le 1 \quad \text{and} \quad \color{brown}{x-1\le u\le x}\tag 2$$

must be satisfied and if they are, the integrand equals $1$. We need to distinguish between 2 cases:

Case 1: $x\in[0,1].$ In this case, the two conditions in $(2)$ are satisfied if $0\le u \le x.$ Hence, for $0\le x \le 1$

$$f_Z(x)=\int_0^xdu=x, \quad 0\le x\le 1.$$

Case 2: $x\in[1,2]$. Here, the conditions are satisfied if $x-1\le u \le 1$, ad we get

$$f_Z(x)=\int_{x-1}^1du=2-x, \quad 1\le x \le 2.$$

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Questions:

1. I don't understand how they go from the green, to red to brown inequality. Is there a typo in the book?
2. I still don't understand the need to split this into two cases and how they came up with the bounds.

Answer 1

The inequalities come from simple equivalent transformations. E.g., $x-u\leq 1$, add $u-1$ to both sides, and then you obtain $x-1\leq u$.

You need to split into two cases because the relevant interval, where $f$ is nonzero, is $[0,1]$. The value of $f(x-u)$ and $f(u)$ depend on whether or not $x-u$ or $u$ is in or out of the interval $[0,1]$.

Answer 2

An intuitive view for the second question:

Consider the $x$-$y$ plane. The point $(x,y)$ is uniformly distributed over a square. For a fixed $u$, the equation $u=x+y$ specifies a $45^\circ$ line across the plane. The PDF of $u$ is thus proportional to the intersection length of the line with the square, which obviously is in two pieces, and piecewise linear.

Answer 3

For identifying the case distinction of the ranges of the density of $X+Y$, a picture of the unit square would be most helpful to find the distribution function, as has been reiterated in several posts of this question here.

The algebra is purely mechanical.

First of all, the red part should read $x-1\le u\le x$, so that $(2)$ is clear.

But for a single range of $u$, $(2)$ is equivalent to saying $\max(x-1,0)\le u\le\min(x,1)$.

That means your final integration is of the form

\begin{align} f_Z(x)&=\int_{\max(x-1,0)}^{\min(x,1)}\,du\,\mathbf1_{0<x<2} \\ \\&=[\min(x,1)-\max(x-1,0)]\,\mathbf1_{0<x<2} \end{align}

Naturally, you would have to break the range at $x=1$ and the result follows that $$f_Z(x)=x\,\mathbf1_{0<x<1}+(2-x)\,\mathbf1_{1<x<2}$$