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=\frac{1}{60}Geom(p) $ is not a geometry distribution.

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

Example 1

Let $ X\sim Exp(\lambda) $ and $ Y\sim Exp(\mu) $, $ \varepsilon \sim Bernoulli(\frac{1}{2}) $. They are all independent.

let $ Z=X $ if $ \varepsilon=0 $ and $ Z=-Y $ if $ \varepsilon = 1 $.

Exercise: See book on `continuous mixtures’.

Example 2

$ X\sim Poisson(\lambda) $. Assume $ \lambda $ itself is random. $ \lambda\sim \Gamma(m,\theta) $.

Easy to say, $ X $ is Poisson conditional on $ \lambda $, 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 \sim N(\mu,\sigma^2)$. We know we can represent $ X $ as $ X=\mu+\sigma Z $ where $ Z\sim N(0,1) $.

More generally, let $ X_1, X_2 $ be bivariate normal. It turns out that we can represent $ X_2 $ using $ X_1 $ and an independent component $ \varepsilon_2 $ like this:

where $ a \& b $ are constants. $ \varepsilon_2 $ is a normal r.v. independent of $ X $ , with $ E(\varepsilon_2) $.

We would like to compute $ a $ and $ b $ and $ Var(\varepsilon_2) $. All we know is

To simplify, assume $ \sigma_1=\sigma_2 $, then we know, from linear prediction, that $ b=\rho $. Then for $ Var(\varepsilon_2)$:

Also note: by taking expectation of the whole model,

Therefore, $ a=\mu_2-\rho\mu_2 $.

Going back to our work with densities for the multivariate normal. We find the following density for the pair $ X=(X_1,X_2) $:

where $ d=1-\rho^2 $, and const $ =\frac{1}{2\pi d} $. Notice that

Also note: the expression

is the ``quadratic form’’, which we encountered as the term $ 1/2 x^T[cov(X)]^{-1}x $. Go back and check this is true.

When $ \rho=0 $, const $\frac{1}{2\pi}$.

This proves the independence ($\rho=0$).