Distribution of Z X Y When X and Y Are Independent Uniform Random Variables on [0, 1]

Introduction

This article explores the distribution of the sum Z X Y, where X and Y are independent random variables uniformly distributed on the interval [0, 1]. The task of finding the distribution of their sum will be tackled using the concept of convolution of their probability density functions (PDFs).

Understanding the Problem

The two random variables X and Y are independent and each is uniformly distributed on the interval [0, 1]. This means that their individual probability density functions (PDFs) are given by:

PDFs of X and Y

The PDF of X and Y are identical and can be expressed as:

f_X(x) 1 for 0 ≤ x ≤ 1

f_Y(y) 1 for 0 ≤ y ≤ 1

Range of Z X Y

The sum Z X Y can take values ranging from 0, when both X and Y are 0, to 2, when both X and Y are 1. Therefore, Z will take values in the interval [0, 2].

Finding the PDF of Z

The PDF of Z can be determined using the convolution formula. According to the convolution formula, the PDF of Z is given by:

Convolution Formula

[f_Z(z) int_{-infty}^{infty} f_X(x) f_Y(z - x) dx]

Given that f_X(x) and f_Y(y) are zero outside the interval [0, 1], we can restrict the limits of integration based on the value of z.

For 0 ≤ z ≤ 1

[f_Z(z) int_0^z f_X(x) f_Y(z - x) dx int_0^z 1 cdot 1 dx z]

For 1 ≤ z ≤ 2

[f_Z(z) int_{z-1}^1 f_X(x) f_Y(z - x) dx int_{z-1}^1 1 cdot 1 dx 2 - z]

Combining the Results

Thus, the PDF of Z is given by:

[f_Z(z) begin{cases} z text{for } 0 leq z leq 1 2 - z text{for } 1 leq z leq 2 0 text{otherwise} end{cases}]

Summary

The distribution of Z X Y when X and Y are independent uniform random variables on [0, 1] follows a triangular distribution that peaks at z 1.

This distribution is characterized by the following PDF:

[f_Z(z) begin{cases} z text{for } 0 leq z leq 1 2 - z text{for } 1 leq z leq 2 0 text{otherwise} end{cases}]

Additional Insight: Probability that XY ≤ c

For a further understanding, we can calculate the probability that the product XY ≤ c for any value of c. The probability can be calculated as an integral over the two-dimensional domain defined by the uniform distribution on [0, 1]:

For 0 ≤ c ≤ 1

[P(XY leq c) int_0^c int_0^{c - x} dy dx frac{c^2}{2}]

For c ≥ 1

[P(XY leq c) c - 1 int_{c - 1}^1 int_0^{c - x} dy dx frac{c^2}{2} - c - 1^2]

The results indicate that the probability distribution for the product XY is not straightforward but can be computed using integration over the defined intervals.

Final Note: Expression of the Sawtooth Function

When expressing both X and Y with the density function y 1 when 0 ≤ x ≤ 1, the product X Y can be represented as:

Sawtooth Function Representation

[y x text{ when } 0 leq x leq 1]

[y 2 - x text{ when } 1 leq x leq 2]

This represents a triangular (sawtooth) function on the interval [0, 2].