Examples on continuous distirbution conditional on discrete distribution; bivariate normal distribution. Not much for today.

Portal to all the other notes

• Lecture 01 - 2017.09.06
• Lecture 02 - 2017.09.13
• Lecture 03 - 2017.09.20
• Lecture 04 - 2017.09.27
• Lecture 05 - 2017.10.04
• Lecture 06 - 2017.10.11
• Lecture 07 - 2017.10.18
• Lecture 08 - 2017.10.25
• Lecture 09 - 2017.11.01
• Lecture 10 - 2017.11.08
• Lecture 11 - 2017.11.15
• Lecture 12 - 2017.11.20 -> This post
• Lecture 13 - 2017.11.29
• Lecture 14 - 2017.12.06

Lecture 12 - Nov 20 2017

Some notes

• Exponential distribution and geometry distribution are the only distributions which have the memoryless property (continuous and discrete).
• Geometry is not scalable. It only takes integer values. T=160Geom(p) is not a geometry distribution.

Exercise: see the textbook’s treatment of discrete mixtures of continuous r.v.’s.

Example 1

Let X∼Exp(λ) and Y∼Exp(μ), ε∼Bernoulli(12). They are all independent.

let Z=X if ε=0 and Z=−Y if ε=1.

Exercise: See book on `continuous mixtures’.

Example 2

X∼Poisson(λ). Assume λ itself is random. λ∼Γ(m,θ).

Easy to say, X is Poisson conditional on λ, but what is the unconditional distribution of X?

Answer is in the book. We just want to compute P(X=k) for k=0,1,2,….

Bivariate Normal

Let X∼N(μ,σ2). We know we can represent X as X=μ+σZ where Z∼N(0,1).

More generally, let X1,X2 be bivariate normal. It turns out that we can represent X2 using X1 and an independent component ε2 like this:

where a&b are constants. ε2 is a normal r.v. independent of X , with E(ε2).

We would like to compute a and b and Var(ε2). All we know is

To simplify, assume σ1=σ2, then we know, from linear prediction, that b=ρ. Then for Var(ε2):

Also note: by taking expectation of the whole model,

Therefore, a=μ2−ρμ2.

Going back to our work with densities for the multivariate normal. We find the following density for the pair X=(X1,X2):

where d=1−ρ2, and const =12πd. Notice that

Also note: the expression

is the quadratic form’’, which we encountered as the term 1/2xT[cov(X)]−1x. Go back and check this is true.

When ρ=0, const 12π.

This proves the independence (ρ=0).