CDF of Z=XY with X~Uniform(0.5,1.5) and Y~Uniform(0.8,1.5)

Question

I am looking for the CDF of the product of two independent random variables (X and Y) with uniform distributions. Both random variables uniform distributions have interval boundaries (upper and lower boundary of the uniform distribution) greater 0. Furthermore, the upper boundary is the same for both random variables . For simplicity let's assume the first random variable X to be Uniform(0.5,1.5) and the second random variable to be Uniform(0.8,1.5).

It is clear to me that the support of the product of the two independent random variables (X and Y) is [0.4,2.25] but I struggle to derive the cdf of Z=XY.

Many thanks in advance

Answer 1

The solution is straightforward but messy using a standard formula, such as obtained in the first half of the Wikipedia derivation. The interest lies in simplifying the work.

I propose starting with a seemingly strange expression for the uniform distribution on the square $[a,b]\times [c,d],$ where (to avoid wrangling too heavily with negative numbers and special cases) I will assume all endpoints are non-negative. Its density is equal to $$f(x,y; a,b,c,d)=\frac{1}{(b-a)(d-c)}$$ throughout this square--that is, provided $a\le x\le b$ and $c\le y\le d$--and otherwise is zero. I will write that more formally and rigorously in the form

$$f(x,y;a,b,c,d) = \frac{1}{(b-a)(d-c)}\left[\mathcal{I}_{a,c}(x,y) - \mathcal{I}_{a,d}(x,y) - \mathcal{I}_{b,c}(x,y) + \mathcal{I}_{c,d}(x,y)\right]$$

where $\mathcal{I}$ is the indicator function of the quadrant bounded at the lower left by the point $(u,v),$

$$\mathcal{I}_{u,v}(x,y) = \left\{\array{ 1 & \text{if}\quad x\ge u, y\ge v \\ 0 & \text{otherwise.} }\right.$$

(For motivation, please see the illustrations and discussion under "Intuition from Geometry" at https://stats.stackexchange.com/a/43075/919, which uses the same idea to compute sums of uniform variables.)

To find the chance that $xy \le z$ for any $z$ all we need to do is integrate expressions of the form

$$F_{u,v}(z) = \int_{x y \le z} \mathcal{I}_{u,v}(x,y) dx dy.$$

These are elementary to compute. If $z\le uv$, this integral must be zero; otherwise, it is

$$\int_{x y \le z} \mathcal{I}_{u,v}(x,y) dx dy = \int_u^{\frac{z}{v}} \left(\frac{z}{x} - v\right) dx dy = z(\log(z) - \log(u) - \log(v)) - z + u v.$$

Therefore

$$\eqalign{\Pr(xy \le z) &= \int_{x y \le z}f(x,y;a,b,c,d) dx dy \\ &= \frac{F_{a,c}(z) - F_{a,d}(z) - F_{b,c}(z) + F_{c,d}(z)}{(b-a)(d-c)}.}$$

For instance, here is a Mathematica implementation of the basic integral (here written g):

g[z_, a_, b_] := Boole[z >= a b] (a b - z + z (Log[z] - Log[a] - Log[b]))

To illustrate its use, we may plot the CDF:

With[{a = 8/10, b = 3/2, c = 1/2, d = 3/2},
 Plot[(g[z, a,c] - g[z, a,d] - g[z, b,c] + g[z, b,d] )/((b-a)(d-c)), {z, a c, b d}]]