# KarlRosaen

We look at a series of examples from All of Statistics that build off each other, starting with defining a conditional random variable, and then finding its expectation, and the expectation of a marginal.

### A Conditional Distribution

Example 2.39 (page 37) describes a conditional distribution

Suppose that $X \sim Uniform(0, 1$). After obtaining a value of $X$ we generate $Y|X = x \sim Uniform(x,1)$.

What is the marginal distribution of $Y$?

First, let's define note that with $X \sim Uniform(0, 1)$

$f_X(x) = \begin{cases} 1 & 0 \leq x \leq 1 \\ 0 & \text{otherwise} \end{cases}$

And that

$f_{Y|X}(y|x) = \begin{cases} \frac{1}{1-x} & 0 \leq x \leq y \leq 1 \\ 0 & \text{otherwise} \end{cases}$

Side note: I find this example interesting in that it describes the conditional distribution directly, in describing one value conidtioned on another, rather than describing a joint distribution asking to derive via the definition $f_{Y|X}(y|x) = \frac{f_{X,Y}(x, y)}{f_X(x)}$.

To find $F_Y(y)$ we first find the joint distribution

$f_{X,Y}(x,y) = f_{Y|X}(y|x) f_X(x) = \begin{cases} \frac{1}{1-x} & 0 \leq x \leq y \leq 1 \\ 0 & \text{otherwise} \end{cases}$

and then the marginal for $Y$

$f_Y(y) = \int_0^y f_{X,Y}(x,y)dx = \int_0^y \frac{dx}{1 - x} = -\int_1^{1-y} \frac{du}{u} = -log(1-y)$

### Conditional Expectation

Example 3.23 (page 54) continues with this conditional distribution and asks to compute $E(Y|X=x)$.

$E(Y|X=x) = \int_x^1 y f_{Y|X}(y|x) dy = \frac{1}{1-x} \int_x^1 y dy = \frac{1+x}{2}$

### Finding E(Y) with help of The Rule of Iterated Expectation

Example 3.25 (page 55) asks to compute $E(Y)$ and notes that, knowing $E(Y|X=x)$, this can be done easily with The Rule of Iterated Expectations, which states that $E[E(Y|X)] = E(Y)$. So,

$E(Y) = E[E(Y|X)] = E(\frac{1+x}{2}) = \frac{1 + E(X)}{2} = \frac{1 + 1/2}{2} = 3/4$