Question

Let y have a uniform (0, 1) distribution. Show that u = -2 ln(y) has an exponential distribution with mean 2.

Answer 1

To prove that u = -2 ln(y) follows an exponential distribution with mean 2, we transformed the CDF of a uniform distribution and showed that the resulting CDF matches that of an exponential distribution with a rate parameter of 1/2, resulting in a mean of 2.

Step-by-step explanation:

To show that u = -2 ln(y) has an exponential distribution with mean 2, where y is uniformly distributed between 0 and 1, we can use the transformation technique. Starting with the cumulative distribution function (CDF) of y, since it's uniformly distributed:

• FY(y) = y for 0 ≤ y ≤ 1.

Next, we apply the transformation u = -2 ln(y). We want to find the CDF of u, FU(u), which is P(U ≤ u). So we find the equivalent value of y that makes this statement true and substitute:

• P(U ≤ u) = P(-2 ln(Y) ≤ u)
• P(ln(Y) ≥ -u/2) = P(Y ≥ e-u/2)
• Since Y is uniform, P(Y ≥ y) = 1 - y, so P(Y ≥ e-u/2) = 1 - e-u/2, for 0 ≤ e-u/2 ≤ 1.

Thus, the CDF of U is FU(u) = 1 - e-u/2 for u ≥0. This is the CDF of an exponential distribution with rate parameter λ = 1/2. The mean of an exponential distribution with rate λ is 1/λ, so the mean here is 2, as required.