Final answer:
The distribution of X defined as -log(U) for U uniform on (0,1) follows an exponential distribution. For large n, log(Yn), where Yn is the geometric mean of an i.i.d. uniform (0,1) sample, resembles a normal distribution. W = e^X, when X is normally distributed, is lognormally distributed, and its CDF can be expressed using the standard normal CDF Φ.
Step-by-step explanation:
Understanding Probability Distributions
Let's explore these concepts step by step:
a. If U is uniform on (0,1), then by defining X = -log(U), X can take all positive values. The cumulative distribution function (CDF) of X is given by P(X ≤ x) = P(-log(U) ≤ x) = P(U ≥ e^{-x}) = 1 - P(U < e^{-x}) = 1 - (e^{-x} - 0)/(1 - 0) = 1 - e^{-x}, for x ≥ 0. X follows an exponential distribution with parameter λ = 1.
- b. For the product of an i.i.d. uniform (0,1) sample, as n becomes large, the Central Limit Theorem implies that the distribution of log(Yn) converges to a normal distribution, due to the additive property of logarithms of the product. Utilizing the result from Part a, we find the distribution resembles the normal distribution with the mean and variance obtained from the exponential distribution of -log(U).
- c. For X being normal with mean μ and variance σ^2, and W = e^X, the CDF of W is given by P(W ≤ w) = P(e^X ≤ w) = P(X ≤ log(w)) = Φ((log(w) - μ) / σ). This covers all values of W since X can span the entire real line.
- d. Approximating the distribution of Yn for large n, it is lognormal based on the behavior of log(Yn) and its relation to the normal distribution. The approximate CDF of Yn without using integrals is thus P(Yn ≤ y) = Φ((log(y) - μ) / σ), where μ and σ are the mean and standard deviation derived from the underlying normal distribution of log(Yn).