Question

Suppose that \(X\) and \(Y\) are independent random variables with probability density functions \(f_{X}\) and \(f_{Y}\). Determine a one-dimensional integral expression for \(P[X+\) \(Y

Short answer

The one-dimensional integral expression for the probability \(P[X+Y<x]\) is: \(P[X+Y<x] = \int_{-\infty}^{x} f_X(t) \cdot f_Y(x-t) \, dt\)

Step-by-step solution

Recall the formula for the probability of the sum of two independent random variables

Since X and Y are independent random variables with probability density functions f_X and f_Y, we can express the probability P[X+Y<x] as the convolution of the two density functions. That is: \(P[X+Y<x] = (f_X * f_Y)(x)\), where '*' denotes convolution.

Write down the convolution integral

The convolution of two functions can be calculated using the integral: \((f_X * f_Y)(x) = \int_{-\infty}^{\infty} f_X(t) \cdot f_Y(x-t) \, dt\) So, \(P[X+Y<x] = \int_{-\infty}^{\infty} f_X(t) \cdot f_Y(x-t) \, dt\)

Express the probability in terms of a one-dimensional integral

We can rewrite the probability P[X+Y<x] as an integral from -∞ to x instead of -∞ to ∞, because the only time events contribute is when \(X+Y < x\): \(P[X+Y<x] = \int_{-\infty}^{x} f_X(t) \cdot f_Y(x-t) \, dt\) This gives us the desired one-dimensional integral expression for the probability.