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Probability
by Samuel S. Watson
Introduction
Counting
Probability Models
Random Variables
Probability Distributions
Joint Distributions
Conditional Probability
Independence
Expectation and Variance
Covariance
Continuous Distributions
Conditional Expectation
Common Distributions
Central Limit Theorem

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ProbabilityConditional Expectation

Tiempo de leer: ~10 min

The conditional expectation of Y given \{X=x\} is defined to be the expectation of Y calculated with respect to its conditional distribution given \{X=x\}. For example, if X and Y are continuous random variables, then

\begin{align*}E[Y | X = x] = \int_{-\infty}^{\infty}y f_{Y | \{X = x\}} (y) \, \mathrm{d} y.\end{align*}

Example
Suppose that f is the function which returns 2 for any point in the triangle with vertices (0,0), (1,0), and (0,1) and otherwise returns 0. If (X,Y) has joint pdf f, then the conditional density of Y given \{X = x\} is the mean of the uniform distribution on the segment [x,1], which is \frac{1+x}{2}.

The conditional variance of Y given \{X = x\} is defined to be the variance of Y with respect to its conditional distribution of Y given \{X=x\}.

Example
Continuing with the example above, the conditional density of Y given \{X = x\} is the variance of the uniform distribution on the segment [x,1], which is \frac{(1-x)^2}{12}.

We can regard the conditional expectation of Y given X as a random variable, denoted \mathbb{E}[Y | X] by coming up with a formula for \mathbb{E}[Y | \{X = x\}] for each x \in \mathbb{R}, and then substituting X for x. And likewise for conditional variance.

Example
With Y and X as defined above, we have \mathbb{E}[Y | X] = \frac{1+X}{2} and \operatorname{Var}[Y | X] = \frac{(1-X)^2}{12}.

Exercise
Find the conditional expectation of Y given X where the pair (X,Y) has density x + y on [0,1]^2.

Solution. We calculate the conditional density as

\begin{align*}\frac{f_{X,Y}(x,y)}{f_X(x)} = \frac{x + y}{x + \frac{1}{2}},\end{align*}

which means that

\begin{align*}\mathbb{E}[Y | X = x] = \int_0^1 \frac{y(x+y)}{x + \frac{1}{2}} \, \mathrm{d} x = \frac{3x+2}{6(x+\frac{1}{2})}.\end{align*}

So \mathbb{E}[Y | X] = \frac{3X+2}{6(X+\frac{1}{2})}

The tower law

Conditional expectation can be helpful for calculating expectations, because of the tower law.

Theorem (Tower law of conditional expectation)
If X and Y are random variables defined on a probability space, then

\begin{align*}\mathbb{E}[\mathbb{E}[Y | X]] = \mathbb{E}[Y].\end{align*}

Exercise
Consider a particle which splits into two particles with probability p \in (0,1) at time t=1. At time t = 2, each extant particle splits into two particles independently with probability p.

Find the expected number of particles extant just after time t = 2. Hint: define X to be 1 or 0 depending on whether the particle splits at time t = 1, and use the tower law with X.

Solution. If Y is the number of particles and X is the indicator of the event that the particle split at time 1, then

\begin{align*}\mathbb{E}[Y | \{X = 0\}] = 2(p) + 1(1-p) = 1+p\end{align*}

while

\begin{align*}\mathbb{E}[Y | \{X = 1\}] = 2(1+p) = 2 + 2p.\end{align*}

Therefore, \mathbb{E}[Y | \{X = x\}] = (1+p)(1+X). By the tower law, we have

\begin{align*}\mathbb{E}[Y] = \mathbb{E}[\mathbb{E}[Y | X]] = (1+p)(1+\mathbb{E}[X]) = (1+p)^2.\end{align*}

Bruno
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