May 18

Sampling from uniform distribution - example of convolution

Sampling from uniform distribution - example of convolution

For the intuition behind the uniform distribution see this post. More formally, let us fix an interval [a,b]. We say that a random variable X is uniformly distributed on [a,b] if any value in this segment is equally likely and any value outside is impossible. The "equally likely" part gives a constant density value in the segment and the "impossible" part gives zero density outside the segment. Thus the density is of the form:

f_X(x)=\frac{1}{b-a}\left\{\begin{array}{ll}1, &x\in[a,b]\\0, &\text{otherwise}\end{array}\right.=\frac{1}{b-a}1_{[a,b]}(x).

This is a step function divided by b-a to make the total area under the graph equal to one. It is easy to find the distribution function of f_X by considering three ranges of x:

F_X(x)=\left\{\begin{array}{ll}0,&x<a;\\  \frac{x-a}{b-a}&a\le x\le b;\\1,&b<x.\end{array}\right.

What is the density of a sum?

Suppose we have a sample of two independent observations X,Y from the same uniform distribution. This means that they have the same density:


By independence we can apply the convolution formula (see equation (5)). The sum X+Y has the density

(1) f_{X+Y}(z)=\int_Rf_X(x)f_Y(z-x)dx=\frac{1}{(b-a)^2}\int 1_{[a,b]}(x)1_{[a,b]}(z-x)dx.

Here 1_{[a,b]}(x) is different from zero only for

(2) x\in[a,b].

Similarly, 1_{[a,b]}(z-x) is different from zero only for z-x\in[a,b] or, equivalently, only for

(3) x\in[z-b,z-a].

(2) and (3) show that the integrand in (1) is different from zero only for x\in[z-b,z-a]\cap[a,b]. So (1) becomes

(4) f_{X+Y}(z)=\frac{1}{(b-a)^2}\int_{[z-b,z-a]\cap[a,b]}dx.

Next we consider several cases depending on the relationship between the segments [z-b,z-a],\ [a,b].

Relationship between segments

Figure 1. Relationship between segments

a) The two segments do not intersect: z-a<a or z-b>b (see Cases 1 and 2 in Figure 1). Equivalently, either z<2a or z>2b. In this case the integral in (4) is zero.

b) a\le z-a\le b or, equivalently, 2a\le z\le a+b, see Case 3 in Figure 1. In this case the integral in (4) is equal to the length of the intersection: z-2a.

c) Finally, a\le z-b\le b or, equivalently, a+b\le z\le 2b (Case 4). Again, the integral in (4) equals the length of the intersection: 2b-z.


(5) f_{X+Y}(z)=\frac{1}{(b-a)^2}\left\{\begin{array}{ll}0,&z<2a\ \text{or}\ z>2b;\\  z-2a,&2a\le z\le a+b;\\2b-z,&a+b\le z\le 2b.\end{array}\right.

Figure 2. Density of a sum of two observations

This is a triangular distribution, shown in Figure 2 from Mathematica with a=1, b=2. Note that it is smoother than the density of the uniform distribution: the density in Figure 2 is continuous and doesn't have jumps.

Density of a sum of three observations

Figure 3. Density of a sum of three observations

The density of the sum of three independent observations X,Y,Z from the same uniform distribution is even smoother, see Figure 3. This is a general fact: as the number of independent observations X_1,...,X_n (from the same population) increases, the density of the sum X_1+...+X_n more and more reminds that of the normal distribution.

It is convenient to denote U[a,b] the uniform distribution on the segment [a,b] and T[a,b] the triangular distribution defined in (5). We have proved that if X,Y are independent and distributed as U[a,b], then their sum is distributed as T[a,b].

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