Homework 1

12. (i) Suppose X has a normal distribution with mean µ and variance σ 2 . Use the last
exercise to compute the density of exp(X). (The answer is called the lognormal
distribution, and is important in finance).
(ii) Suppose X has a normal distribution with mean 0 and variance 1. Use the last
exercise to compute the density of X 2 . (The answer is called the chi-squared
distribution.)
13. Show that if F (x) = P(X ≤ x) is continuous then Y = F (X) has a uniform distribution
on (0, 1), that is P(Y ≤ y) = y for y ∈ (0, 1).
This gives you an extremely useful method of simulating random variables on the
computer. If you can compute F −1 , then you can ask the computer to generate a
uniform random variable Y for you (all computers have a way of doing this, but the
techniques behind it could take up an entire class), and then X = F −1 (Y ) will have F
as its distribution function.
14. If a distribution function is not continuous, it’s because there is at least one number
x 0 (there may be more) such that
lim F (x) =: F (x 0 +) > F (x 0 −) := lim F (x).
x↓x 0 x↑x0
Such numbers x 0 are called atoms of the distribution because there is a positive
probability of the number being picked. In fact
Convince yourself that
P(X = x 0 ) = F (x 0 +) − F (x 0 −).
(i) a distribution function with discontinuities does not have a density function,
(ii) a distribution function can have at most countably many discontinuities.
There’s no need to write anything for this problem, it’s for your own edification. Just
make sure you think about it for a little bit.
15. A distribution function can be continuous but still not have a density function. Such
distributions are called singularly continuous, and are kind of wicked. Here’s a
simple and classical example of how to construct one called the Cantor staircase:
(i) Start by constructing the Cantor middle-thirds set. Let C 0 = [0, 1], and to
construct C 1 remove the “middle-third interval” (1/3, 2/3) of C 0 . This leaves
C 1 = [0, 1/3] ∪ [2/3, 1]. In general construct C n+1 from C n by removing the
middle-third of each interval in C n . Draw a picture of C n for n = 0, 1, 2, 3. You
see that the set gets sparser and sparser very quickly. In the limit though there
is a non-empty set remaining, let’s call it C ∞ that is referred to as the Cantor
middle-thirds set. Look it up on Wikipedia if you haven’t seen it before.