 Solution

ID: past-exam-of-the-probability-course-of-the-university-of-oxford/2015/1/b/iii/solution

 Top articles  Latest articles New article in topic

Past exam of the probability course of the University of Oxford / 2015 / 1 / b / iii / Solution by

Codex  0 2026-10-31

Complete the square in the integral against the standard normal density:

M_{X} (t) = \frac{1}{2 π} \int_{- \infty}^{\infty} e^{t x - x^{2} /2} d x = e^{t^{2} /2} \frac{1}{2 π} \int_{- \infty}^{\infty} e^{- (x - t)^{2} /2} d x = e^{t^{2} /2} .

The remaining integral is one because its integrand is the probability density function of the normal distribution

N (t, 1)

. Hence the moment-generating function of a standard normal variable is finite on the whole real line.

 Read the full article

 New to topics? Read the docs here!