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Let $X\text{~Uniform} [0,1]$ and $Y\text{~Uniform}[0,2]$. Find the distribution of their sum, $Z = X + Y$, using the convolution method.

I understand that I have to break this into cases for $Z=[0,1],[1,2],[2,3]$, but I'm having difficulties determining the bounds on each integration.

asosnovsky
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ittiekat
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  • If you write the density function of $Y$ as if it were defined on all real numbers - like $0$ if $t<0$, $\frac12$ if $0\le t\le 2$, and $0$ if $t>2$ (and similarly for the density function of $X$) - then, when you write down the convolution integral, the integrand should naturally tell you where you need to split the integral up and what form the integrand takes there. – Greg Martin Feb 24 '15 at 23:32
  • Duplicate of: http://math.stackexchange.com/questions/1126317/sum-of-uniform-random-variables-u0-1-and-u0-a – wolfies Feb 25 '15 at 13:40

2 Answers2

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You need to consider: the support of the integrated variable $X$, the support of the convoluted variable $Z-X$, and the support of the summed random variable $Z$

Use indicator functions to determine the integration support intervals of $x$ with respect to $z\;$. $$\begin{align} f_X(x) & = {\bf 1}_{0\leq x \leq 1} \\[2ex] f_Y(y) & = \tfrac 1 2 \;{\bf 1}_{0\leq y\leq 2} \\[2ex] f_{X+Y}(z) & = \int_{-\infty}^{\infty} f_X(x)f_Y(z-x) \operatorname d x \; {\bf 1}_{0\leq z\leq 3} \\[1ex] & = \int_{-\infty}^{\infty}\tfrac 1 2\; {\bf 1}_{0\leq x \leq 1}\;{\bf 1}_{0\leq z-x \leq 2}\;{\bf 1}_{0\leq z \leq 3}\; \operatorname d x \\[1ex] & = \tfrac 1 2\;\int_{-\infty}^{\infty} {\bf 1}_{0\leq x \leq 1}\;{\bf 1}_{z-2\leq x \leq z}\;{\bf 1}_{0\leq z \leq 3}\; \operatorname d x \\[1ex] & = \tfrac 1 2\;\int_{-\infty}^{\infty} {\bf 1}_{\max(0,z-2)\leq x \leq \min(1,z)}\;{\bf 1}_{0\leq z \leq 3}\; \operatorname d x \\[1ex] & = \tfrac 1 2\;\int_{-\infty}^{\infty} {\bf 1}_{0\leq x \leq z}\;{\bf 1}_{0\leq z \leq 1} + {\bf 1}_{0\leq x \leq 1}\;{\bf 1}_{1 < z \leq 2} + {\bf 1}_{z-2\leq x \leq 1}\;{\bf 1}_{2 < z \leq 3}\; \operatorname d x \\[1ex] & = \tfrac 1 2\;\left(\int_{0}^{z} {\bf 1}_{0\leq z \leq 1}\operatorname d x + \int_0^1\;{\bf 1}_{1 < z \leq 2}\operatorname d x + \int_{z-2}^1\;{\bf 1}_{2 < z \leq 3}\; \operatorname d x\right) \end{align}$$

A more compact way to arrive at this is to consider that you need to integrate over the supports: $x\in [0;1] , (z-x)\in [0;2], z\in [0;3]$

Which is: $x\in [0;1], x\in [z-2;z], z\in[0;3] \\[2ex] \langle x,z\rangle \in [\min(0, z-2); \max(1, z)]\times[0;3] \\[2ex] \langle x,z\rangle \in [\min(0, z-2); \max(1, z)]\times[0;1]\;\cup\;[\min(0, z-2); \max(1, z)]\times(1;2]\;\cup\;[\min(0, z-2); \max(1, z)]\times(2;3] \\[2ex] \langle x,z\rangle \in [0; z]\times[0;1]\;\cup\;[0; 1]\times(1;2]\;\cup\;[z-2; 1]\times(2;3] \\$

This gives the integration ranges of $x$ for values of $z$ within each of three intervals.

Graham Kemp
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Try and visualize your bounds here.

We are looking at a random variable $Z$ that is the sum of $X+Y$.

We want to know the distribution. Which is $ F_Z(z) =Pr(Z<z) =Pr(X+Y<z) $

Now note that because of independence we can say: $ Pr(X<x) = \int_0^2 Pr (X <x) f_Y(y) dy $

Furthermore we can say: $ F_Z(z) = Pr(X+Y<z)=\int_0^2 Pr(X<z-y) f_Y(y) dy =\int_0^2 \int_0^{z-y} f_X(x) f_Y(y)dx dy $

asosnovsky
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