Postscript version of this page

STAT 380 Week 2

Summary

• Review expectation

• Conditional distributions, conditional expectation.

• Summarize review.

Example 3: Mean values

$Z_n$ = total number of sons in generation $n$.

$Z_0=1$ for convenience.

Compute ${\rm E}(Z_n)$.

Recall definition of expected value:

If $X$ is discrete then

$${\rm E}(X) = \sum_x x P(X=x)$$

If $X$ is absolutely continuous then

$${\rm E}(X) = \int_{-\infty}^\infty x f(x) dx$$

Theorem: If $Y=g(X)$, $X$ has density $f$ then

$${\rm E}(Y) = {\rm E}(g(X)) =\int g(x) f(x) dx$$

Key properties of ${\rm E}$:

1: If $X\ge 0$ then ${\rm E}(X) \ge 0$. Equals iff $P(X=0)=1$.

2: ${\rm E}(aX+bY) = a{\rm E}(X) +b{\rm E}(Y)$.

3: If $0 \le X_1 \le X_2 \le \cdots$ then

$${\rm E}(\lim X_n) = \lim {\rm E}(X_n)$$

4: ${\rm E}(1) = 1$.

Conditional Expectations

If $X$, $Y$, two discrete random variables then

$${\rm E}(Y\vert X=x) = \sum_y y P(Y=y\vert X=x)$$

Extension to absolutely continuous case:

Joint pmf of $X$ and $Y$ is defined as

$$p(x,y) = P(X=x,Y=y)$$

Notice: The pmf of $X$ is

$$p_X(x) = \sum_y p(x,y)$$

Analogue for densities: joint density of $X,Y$ is

$$f(x,y) dx dy \approx P(x \le X \le x+dx, y \le Y \le y+dy)$$

Interpretation is that

$$P(X \in A, Y \in B) = \int_A \int_B f(x,y) dy dx$$

Property: if $X,Y$ have joint density $f(x,y)$ then $X$ has density

$$f_X(x) = \int_y f(x,y) dy$$

Sums for discrete rvs are replaced by integrals.

Example:

$$f(x,y) = \begin{cases}x+y & 0 \le x,y \le 1 \\ 0 & \text{otherwise} \end{cases}$$

is a density because

$$\iint f(x,y)dx dy = \int_0^1\int_0^1 (x+y) dy dx$$

$$= \int_0^1 x dx + \int_0^1 y dy$$

$$= \frac{1}{2} + \frac{1}{2} = 1$$

The marginal density of $X$ is, for $0 \le x \le 1$.

$$f_X(x) = \int_{-\infty}^\infty f(x,y)dy$$

$$= \int_0^1 (x+y) dy$$

$$= \left.(xy+y^2/2)\right\vert _0^1 = x+\frac{1}{2}$$

For $x$ not in $[0,1]$ the integral is 0 so

$$f_X(x) = \begin{cases}x+\frac{1}{2} & 0 \le x \le 1 \\ 0 & \text{otherwise} \end{cases}$$

Conditional Densities

If $X$ and $Y$ have joint density $f_{X,Y}(x,y)$ then we define the conditional density of $Y$ given $X=x$ by analogy with our interpretation of densities. We take limits:

$$\begin{multline*} f_{Y\vert X}(y\vert x)dy \\ \approx \frac{ P(x \le X \le x+dx, y \le Y \le y+dy)}{P(x \le X \le x+dx)} \end{multline*}$$

in the sense that if we divide through by $dy$ and let $dx$ and $dy$ tend to 0 the conditional density is the limit

$$\frac{\lim_{dx, dy \to 0} \frac{ P(x \le X \le x+dx, y \le Y \le y+dy)}{(dx\,dy)}}{ \lim_{dx\to 0} \frac{P(x \le X \le x+dx)}{dx}}$$

Going back to our interpretation of joint densities and ordinary densities we see that our definition is just

$$f_{Y\vert X}(y\vert x) = \frac{f_{X,Y}(x,y)}{f_X(x)}$$

When talking about a pair $X$ and $Y$ of random variables we refer to $f_{X,Y}$ as the joint density and to $f_X$ as the marginal density of $X$.

Example: For $f$ of previous example conditional density of $Y$ given $X=x$ defined only for $0 \le x \le 1$:

$$f_{Y\vert X}(y\vert x) = \begin{cases} \frac{x+y}{x+\frac{1}{2}} ... ...y>1 \\ 0 & 0 \le x \le 1, y < 0 \\ \text{undefined} & otherwise \end{cases}$$

Example: $X$ a Poisson$(\lambda)$ random variable. Observe $X$ then toss a coin $X$ times. $Y$ is number of heads. $P(H) = p$

$$f_Y(y) = \sum_x f_{X,Y}(x,y)$$

$$= \sum_x f_{Y\vert X}(y\vert x) f_X(x)$$

$$= \sum _{x=0}^\infty \dbinom{x}{y} p^y(1-p)^{x-y} \times \frac{\lambda^x}{x!} e^{-\lambda}$$

WARNING: in sum $0 \le y \le x$ is required and $x$, $y$ integers so sum really runs from $y$ to $\infty$

$$f_Y(y) = \frac{(p\lambda)^ye^{-\lambda}}{y!} \sum_{x=y}^\infty \frac{\left[(1-p)\lambda\right]^{x-y}}{(x-y)!}$$

$$= \frac{(p\lambda)^ye^{-\lambda}}{y!}\sum_0^\infty \frac{\left[(1-p)\lambda\right]^{k}}{k!}$$

$$= \frac{(p\lambda)^ye^{-\lambda}}{y!}e^{(1-p)\lambda}$$

$$= e^{-p\lambda} (p\lambda)^y/y!$$

which is a Poisson($p\lambda$) distribution.

Conditional Expectations

If $X$ and $Y$ are continuous random variables with joint density $f_{X,Y}$ we define:

$$E(Y\vert X=x) = \int y f_{Y\vert X}(y\vert x) dy$$

Key properties of conditional expectation

1: If $Y\ge 0$ then ${\rm E}(Y\vert X=x) \ge 0$. Equals iff $P(Y=0\vert X=x)=1$.

2: ${\rm E}(A(X)Y+B(X)Z\vert X=x) = A(x){\rm E}(Y\vert X=x) +B(x){\rm E}(Z\vert X=x)$.

3: If $Y$ and $X$ are independent then

$${\rm E}(Y\vert X=x) = {\rm E}(Y)$$

4: ${\rm E}(1\vert X=x) = 1$.

Example:

$$f(x,y) = \begin{cases}x+y & 0 \le x,y \le 1 \\ 0 & \text{otherwise} \end{cases}$$

has conditional of $Y\vert X$:

$$f_{Y\vert X}(y\vert x) = \begin{cases} \frac{x+y}{x+\frac{1}{2}} ... ...y>1 \\ 0 & 0 \le x \le 1, y < 0 \\ \text{undefined} & otherwise \end{cases}$$

so, for $0 \le x \le 1$,

$${\rm E}(Y\vert X=x) = \int_0^1 y \frac{x+y}{x+\frac{1}{2}} dy$$

$$= \frac{x/2 +1/3}{x+1/2}$$

Computing expectations by conditioning:

Notation: ${\rm E}(Y\vert X)$ is the function of $X$ you get by working out ${\rm E}(Y\vert X=x)$, getting a formula in $x$ and replacing $x$ by $X$. This makes ${\rm E}(Y\vert X)$ a random variable.

Properties:

1: ${\rm E}(A(X)Y+B(X)Z\vert X) = A(X){\rm E}(Y\vert X) +B(X){\rm E}(Z\vert X)$.

2: If $Y$ and $X$ are independent then

$${\rm E}(Y\vert X) = {\rm E}(Y)$$

3: ${\rm E}(1\vert X) = 1$.

4: ${\rm E}\left[{\rm E}(Y\vert X)\right] = {\rm E}(Y)$ (compute average holding $X$ fixed first, then average over $X$).

In example:

$${\rm E}(Y\vert X) = \frac{X+2/3}{2X+1}$$

Application to last names problem. Put $m={\rm E}(X)$

$${\rm E}(Z_n) = {\rm E}\left[{\rm E}(Z_n\vert X)\right]$$

$$= {\rm E}\left[ X{\rm E}(Z_{n-1})\right]$$

$$= {\rm E}(X){\rm E}(Z_{n-1})$$

$$= m {\rm E}(Z_{n-1})$$

$$= m^2 {\rm E}(Z_{n-2})$$

$$\quad \vdots$$

$$= m^{n-1}{\rm E}(Z_1)$$

$$= m^n$$

For $m < 1$ expect exponential decay. For $m>1$ exponential growth (if not extinction).

Summary of Probability Review

We have reviewed the following definitions:

• Probability Space (or Sample Space): $\Omega =$ set of possible outcomes, $\omega$.

• Events, subsets of $\Omega$.

• Probability: $P$ function defined for events, values in $[0,1]$ satisfying:
  1. $P(\emptyset)=0$ and $P(\Omega)=1$.

  2. Countable additivity: $A_1,A_2,\cdots$ pairwise disjoint ($j\neq k$ $A_j\cap A_k=\emptyset$)
     $$P(\cup_{i=1}^\infty A_i) = \sum_{i=1}^\infty P(A_i)$$

• Joint, marginal, conditional pmfs: $X,Y$ discrete:

  
  

  • Joint pmf
    $$p_{X,Y}(x,y) = P(X=x,Y=y).$$

    
    

  • Marginal pmf
    $$p_X(x) = P(X=x) = \sum_y p_{X,Y}(x,y)$$

    
    

  • Conditional pmf
    $$p_{Y\vert X}(y\vert x) =p_{X,Y}(x,y)/p_X(x).$$

  
  

• Joint, marginal, conditional densities: $X,Y$ continuous:

  
  

  • Joint density $f_{X,Y}(x,y)$; probabilities are double integrals

    
    

  • Marginal density
    $$f_X(x) = \int f_{X,Y}(x,y) dy$$

    
    

  • Conditional density
    $$f_{Y\vert X}(y\vert x) =f_{X,Y}(x,y)/f_X(x).$$

• Expected value:
  $${\rm E}(g(X)) = \sum g(x) p_X(x)$$
  or
  $${\rm E}(g(X)) = \int g(x) f_X(x) dx$$

• Conditional expectation
  $${\rm E}(g(Y)\vert X=x) = \sum g(y) p_{Y\vert X}(y\vert x)$$
  or
  $${\rm E}(g(Y)\vert X=x) = \int g(y) f_{Y\vert X}(y\vert x) dy$$

• ${\rm E}(g(Y)\vert X)$ is previous with $x$ replaced by $X$ after getting formula.

Tactics:

• Convert English expressions to set notation:
  • ``Or'' means union (remember inclusive or).

  • ``And'' means intersection.

  • ``not'' means complement.

• Compute prob something happens as 1-prob it doesn't.

• Break up event into disjoint pieces and add up probabilities.

• Find independent events and write event as intersection.

• Find $A$, $B$ for which $P(A\vert B)$ is known.

• Use Bayes rule: if $B_1,\ldots,B_p$ is a partition then
  $$P(B_1\vert A) = \frac{P(A\vert B_1)P(B_1)}{\sum P(A\vert B_j)P(B_j)}$$

• Use first step analysis. (See family names example, craps game, alternating dice tossing on asst 1.)

Tactics for expected values:

• Use linearity:
  $${\rm E}(\sum a_j X_j) = \sum a_j {\rm E}(X_j)$$

• Condition on another variable and use
  $${\rm E}\left[{\rm E}(Y\vert X)\right] = {\rm E}(Y)$$

• Use first step analysis. (Special case of previous.)