1. Moments of Gaussian Densities (10 points) Consider the one-dimensional Gaussian pdf
Use the fact that
and the identity
1 (x−m)2 p(x)=√2πσ2 exp− 2σ2 .
∞2π
exp−(αu ) du = α −∞
22d2
u exp−(αu ) du = −dα exp−(αu ) du
to show that the even central moments of the Gaussian density are E[(x−m)n] = (1·3·5·…·n−1)σn forneven .
Use symmetry arguments (hint: antisymmetric integrand over symmetric bounds) to show that the odd central moments are all zero
E[(x−m)n] = 0 for n odd 2. Conditional and Unconditional Variance (10 points)
In class we showed the relationship between conditional means and unconditional means. Specifically for random variables x ∈ RN and y ∈ RM , the conditional mean of x is
N E[x|y] = xp(x|y)d x
and the unconditional mean is
N M N
E[x] = xp(x)d x = x p(x|y)p(y)d y d x N M
= xp(x|y)d x p(y)d y = Ey[Ex[x|y]] .
The relationship between the conditional variance and the unconditional variance is a bit more interesting. For simplicity, take x ∈ R and y ∈ R (scalar random variables). The conditional variance is
2
var(x|y) = (x−E[x|y]) p(x|y)dx . (1)
(Note that like the mean, the conditional variance is a function of x2.) Show that the uncon- ditional variance is related to the condition variance by
2
var(x) = (x−E[x]) p(x)dx = Ey[varx(x|y)] + vary(E[x|y]). (2)
Your derivation must show explicitly what vary(E[x|y]) means in terms of integral averages over quantities.
(Hint: Rewrite
(x − E[x])2 =
= (x − E[x|y])2 + (E[x|y] − E[x])2
+ 2 (x − E[x|y])(E[x|y] − E[x])
)
3. A Maximum likelihood estimation (5 points)
This problem has an interesting practical origin, that I’ll explain after you hand your solution
back.
I have a bag filled with m plastic balls, numbered consecutively 1, 2, . . . , m. I don’t tell you
what the value of m is; I want you to make a (statistically informed) guess.
So I give you one piece of data. I reach into the bag and pull out one of the balls at random
(i.e. with probability 1/m) and hand it to you. It has the value “19” printed on it.
Let’s compute the maximum likelihood estimate of the total number of balls m. Mathemat- ically, this is the value of m that maximizes p(x = 19|m). Start by building a likelihood function — since there is one ball with each number 1,2,3,…,19,…,m, any number on a ball in the range 1 − m can be observed with equal probability
p(1|m) = p(2|m) = ··· = p(m|m) = 1/m .
Note also that it’s not possible to observe a number on a ball greater than (the unknown) m
p(n|m)=0forn>m .
These two pieces of information fix the likelihood function p(x|m). Given this information, what is the value of m that maximizes the likelihood of the data p(19|m)?
(x − E[x] + E[x|y] − E[x|y])2 = (x − E[x|y] + E[x|y] − E[x])2