# density of sum of two uniform random variables $[0,1]$

I am trying to understand an example from my textbook.

Let's say $Z = X + Y$, where $X$ and $Y$ are uniform random variables with range $[0,1]$. Then the PDF is $$f(z) = \begin{cases} z & \text{for 0 < z < 1} \\ 2-z & \text{for 1 \le z < 2} \\ 0 & \text{otherwise.} \end{cases}$$

How was this PDF obtained?

Thanks

• There are a couple of ways. Have you done convolutions? Not the best way in my opinion, but certainly useful elsewhere. And there is a straightforward geometric approach. – André Nicolas Apr 11 '13 at 0:49
• what would be the bounds if i were to use convolutions? also, I don't quite understand the geometric approach. Can you direct me to an example? – Zhulu Apr 11 '13 at 1:19
• For convolution, you want $\int_{-\infty}^\infty f_Y(z-x)f_X(x)\,dx$. So since density is $0$ outside $(0,1)$, we need $0\le z-x\le 1$, or equivalently $x\le z\le x+1$. For $z\le 1$, the first bound is the one to use. For $1\lt z\le 2$, it is the second. – André Nicolas Apr 11 '13 at 1:33
• I sort of gave the bounds. For $0\le z\le 1$, integrate from $x=0$ to $x=z$. For $1\lz\le 2$, integrate from $z-1$$to 1. Will maybe write up answer. – André Nicolas Apr 11 '13 at 2:19 • The basic method is to my mind not convolution, but finding the cdf and differentiating. So draw the square on which the joint density lives. The probability that Z\le z is the probability (X,Y) lands in the part of the square "below" the line x+y=z. Imagine drawing lines x+y=z for various z. The geometry of the "part below" changes at z=1. For z\lt 1 the part below is a triangle. For 1\lt z\lt 2 it is the part above that is a triangle. – André Nicolas Dec 15 '14 at 5:36 ## 3 Answers If we want to use a convolution, let f_X be the full density function ofX, and let f_Y be the full density function of Y. Let Z=X+Y. Then$$f_Z(z)=\int_{-\infty}^\infty f_X(x)f_Y(z-x)\,dx.$$Now let us apply this general formula to our particular case. We will have f_Z(z)=0 for z\lt 0, and also for z\ge 2. Now we deal with the interval from 0 to 2. It is useful to break this down into two cases (i) 0\lt z\le 1 and (ii) 1\lt z\lt 2. (i) The product f_X(x)f_Y(z-x) is 1 in some places, and 0 elsewhere. We want to make sure we avoid calling it 1 when it is 0. In order to have f_Y(z-x)=1, we need z-x\ge 0, that is, x\le z. So for (i), we will be integrating from x=0 to x=z. And easily$$\int_0^z 1\,dx=z.$$Thus f_Z(z)=z for 0\lt z\le 1. (ii) Suppose that 1\lt z\lt 2. In order to have f_Y(z-x) to be 1, we need z-x\le 1, that is, we need x\ge z-1. So for (ii) we integrate from z-1 to 1. And easily$$\int_{z-1}^1 1\,dx=2-z.$$Thus$f_Z(z)=2-z$for$1\lt z\lt 2$. Another way: (Sketch) We can go after the cdf$F_Z(z)$of$Z$, 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$z=1$. • Thank very much for writing this up! This really helped me fully understand the concept of convolution. – Zhulu Apr 11 '13 at 3:45 • Does the second mehod of calculating areas only work in this case since we are using uniform distributions? – F.Webber May 30 '16 at 18:16 • @F.Webber: In the form that I used it, yes, we are reduced to finding area because of uniformity. But first going after the cdf is a general procedure. In the non-uniform case, we are finding an integral. The geometry is still useful in determining the bounds on the integration. – André Nicolas May 30 '16 at 18:41 • So, in general, once we have determined such area, the function we have to integrate in that region would be the joint probability density function? – F.Webber May 30 '16 at 18:46 • @F.Webber: Yes, for many problems a double integral, but integration in$n\$-dimensional space also comes up. – André Nicolas May 30 '16 at 20:52

Here's why we need to break the convolution into cases. The integral we seek to evaluate for each $$z$$ is $$f_Z(z):= \int_{-\infty}^\infty f(x)f(z-x)\,dx.\tag1$$ (On the RHS of (1) I'm writing $$f$$ instead of $$f_X$$ and $$f_Y$$ since $$X$$ and $$Y$$ have the same density.) Here the density $$f$$ is the uniform density $$f(x)$$, which equals $$1$$ for $$0, and is zero otherwise. The integrand $$f(x)f(z-x)$$ will therefore have value either $$1$$ or $$0$$. Specifically, the integrand is $$1$$ when $$0 and equals zero otherwise. To evaluate (1), which is an integral over $$x$$ (with $$z$$ held constant), we need to find the range of $$x$$-values where the conditions listed in (2) are satisfied. How does this range depend on $$z$$? Plotting the region defined by (2) in the $$(x,z)$$ plane, we find:

and it's clear how the limits of integration on $$x$$ depend on the value of $$z$$:

1. When $$0, the limits run from $$x=0$$ to $$x=z$$, so $$f_Z(z)=\int_0^z 1dx=z.$$

2. When $$1, the limits run from $$x=z-1$$ to $$x=1$$, so $$f_Z(z)=\int_{z-1}^11dx=2-z.$$

3. When $$z<0$$ or $$z>2$$, the integrand is zero, so $$f_Z(z)=0$$.

• very nice explanation! Thank you! – garej Apr 30 at 20:16

By the hint of jay-sun, consider this idea, if and only if $$f_X (z-y) = 1$$ when $$0 \le z-y \le 1$$. 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 (i) if $$z-1 \le 0$$ then $$-1 \le z-1 \le 0$$ that is $$z \in [0, 1]$$, we get the range of $$y \in [0, z]$$ since $$z \in [0, 1]$$. And we get $$\int_{-\infty}^{\infty}f_X(z-y)dy = \int_0^{z} 1 dy=z$$ if $$z \in [0, 1]$$.

Consider (ii) if $$z-1 \ge 0$$ that is $$z \in [1, 2]$$, so we get the range of $$y \in [z-1, 1]$$, and $$\int_{-\infty}^{\infty}f_X(z-y)dy = \int_{z-1}^{1} 1 dy = 2-z$$ if $$z \in [1, 2]$$.

To sum up, consider to clip the range in order to get $$f_X (z-y) = 1$$.