Variance for Conditional and Marginal Distributions

Study guide: Exam P-1

The Variance for Conditional and Marginal Distributions measures spread at two different levels: the marginal variance $\text{Var}(X)$ describes the spread of $X$ across the whole population, while the conditional variance $\text{Var}(X \mid Y=y)$ measures the spread of $X$ within the subpopulation where $Y=y$ . The two are tied together by the law of total variance.

$\text{Var}(X \mid Y=y) = E[X^2 \mid Y=y] - \bigl(E[X \mid Y=y]\bigr)^2$

$E[X] = E\bigl[E[X \mid Y]\bigr]$

$\text{Var}(X) = E\bigl[\text{Var}(X \mid Y)\bigr] + \text{Var}\bigl(E[X \mid Y]\bigr)$

The second identity is the law of total expectation (double expectation); the third is the law of total variance. Both are the fastest route to a marginal moment when the problem hands you a conditional distribution and a distribution for its parameter.
Watch the notation carefully: $E[X \mid Y=y]$ is a number, but $E[X \mid Y]$ is a random variable — a function of $Y$ . The outer $E[\cdot]$ and $\text{Var}(\cdot)$ in the identities average over $Y$ 's distribution.
Read the decomposition as (average within-group variance) + (variance of the group means). Both terms are non-negative, so conditioning can never raise the average spread: $E[\text{Var}(X \mid Y)] \leq \text{Var}(X)$ .
A common slip is to compute $\text{Var}(E[X\mid Y])$ and stop — that is only the between-group piece, never the whole marginal variance.
Conditional distributions come from the Conditional Probability Function (discrete) or the Joint Probability Density Function (continuous). See also Moments for Joint Distributions.

Variance of Claims Given Policy Type Joint PMF: $p(0,1)=0.3$ , $p(1,1)=0.2$ , $p(0,2)=0.1$ , $p(1,2)=0.4$ where $X$ = claims and $Y$ = policy type. Find $\text{Var}(X \mid Y=2)$ .

Answer Since $p_Y(2)=0.5$ : $p_{X|Y}(0\mid 2)=0.2$ and $p_{X|Y}(1\mid 2)=0.8$ . Then $E[X \mid Y=2] = 0(0.2)+1(0.8) = 0.8$ and $E[X^2 \mid Y=2] = 0.8$ . So: $\text{Var}(X \mid Y=2) = 0.8 - (0.8)^2 = 0.16$

Marginal Variance via the Law of Total Variance A driver's annual claim count is $N \mid \Lambda = \lambda \sim \text{Poisson}(\lambda)$ , and the risk parameter $\Lambda$ varies across the portfolio with $E[\Lambda] = 0.2$ and $\text{Var}(\Lambda) = 0.05$ . Find $E[N]$ and $\text{Var}(N)$ .
Answer For a Poisson, $E[N \mid \Lambda] = \Lambda$ and $\text{Var}(N \mid \Lambda) = \Lambda$ . By double expectation:
 $\begin{align*} E[N] &= E\bigl[E[N \mid \Lambda]\bigr] \\ &= E[\Lambda] \\ &= 0.2 \end{align*}$ 
By the law of total variance:
 $\begin{align*} \text{Var}(N) &= E\bigl[\text{Var}(N \mid \Lambda)\bigr] + \text{Var}\bigl(E[N \mid \Lambda]\bigr) \\ &= E[\Lambda] + \text{Var}(\Lambda) \\ &= 0.2 + 0.05 \\ &= 0.25 \end{align*}$ 
$\text{Var}(N) > E[N]$ : heterogeneous risk parameters make the portfolio's claim counts overdispersed relative to a single Poisson.

Both Terms Matter A loss $X$ is uniform on $(0, Y)$ , where $Y$ takes the values 10 and 20 with equal probability. Find $\text{Var}(X)$ .
Answer For $X \mid Y = y \sim \text{Uniform}(0,y)$ : $E[X \mid Y] = Y/2$ and $\text{Var}(X \mid Y) = Y^2/12$ . With $Y \in \{10, 20\}$ equally likely, $E[Y] = 15$ , $E[Y^2] = \tfrac{100 + 400}{2} = 250$ .

Within-group piece:
 $\begin{align*} E\bigl[\text{Var}(X \mid Y)\bigr] &= \frac{E[Y^2]}{12} \\ &= \frac{250}{12} \approx 20.83 \end{align*}$ 
Between-group piece:
 $\begin{align*} \text{Var}\bigl(E[X \mid Y]\bigr) &= \text{Var}\!\left(\frac{Y}{2}\right) \\ &= \frac{1}{4}\left(250 - 15^2\right) \\ &= \frac{25}{4} = 6.25 \end{align*}$ 
$\text{Var}(X) = 20.83 + 6.25 \approx 27.08$ Reporting only 6.25 (the between-group term) would understate the spread by a factor of more than four ( $27.08/6.25 \approx 4.3$ ).