Discrete Random Variable

Study guide: Exam P-1

A Discrete Random Variable $X$ is a Random Variable whose possible values form a finite or countably infinite set — typically a count, such as the number of claims — so its distribution is given by a probability mass function $p(x) = P(X = x)$ .

$p(x) = P(X = x)$

$F(x) = \sum_{k \le x} p(k)$

A valid PMF has $p(x) \ge 0$ and $\sum_x p(x) = 1$ . The CDF is a step function that jumps by $p(x)$ at each possible value. The named families are in Discrete Univariate Distributions.
Endpoints matter. $P(X \le 3)$ and $P(X < 3)$ differ by $p(3)$ — the most common slip when moving from a continuous problem to a discrete one. Expectations are sums: $E[g(X)] = \sum_x g(x)\,p(x)$ .
With two variables, the Joint Probability Function $p(x,y)$ is a table. Marginals are its row and column totals; a conditional PMF is one row divided by its total, $p(y \mid x) = p(x,y)/p_X(x)$ .
Joint moments, conditional and marginal variances, Covariance and the Correlation Coefficient are all sums over the cells of that table.
The PMF of a sum of independent discrete variables is a convolution, $P(X + Y = s) = \sum_x p_X(x)\,p_Y(s - x)$ . Some families are closed under it — independent Poissons sum to a Poisson with mean $\lambda_1 + \lambda_2$ . See Probabilities for Linear Combinations.

The first two examples use one table. A policyholder's annual auto claims $X$ and home claims $Y$ have joint PMF $p(0,0) = 0.40$ , $p(0,1) = 0.10$ , $p(1,0) = 0.20$ , $p(1,1) = 0.15$ , $p(2,0) = 0.05$ , $p(2,1) = 0.10$ .

Auto Claims Given a Home Claim Using the joint PMF above, find the conditional PMF of $X$ given $Y = 1$ , then $E[X \mid Y = 1]$ and $\text{Var}(X \mid Y = 1)$ .
Answer The marginal is $p_Y(1) = 0.10 + 0.15 + 0.10 = 0.35$ . Dividing each $Y = 1$ cell by it gives $p(x \mid 1) = \tfrac{2}{7}, \tfrac{3}{7}, \tfrac{2}{7}$ for $x = 0, 1, 2$ .
 $\begin{align*} E[X \mid Y = 1] &= 0\left(\tfrac{2}{7}\right) + 1\left(\tfrac{3}{7}\right) + 2\left(\tfrac{2}{7}\right) \\ &= 1 \\ E[X^2 \mid Y = 1] &= 1\left(\tfrac{3}{7}\right) + 4\left(\tfrac{2}{7}\right) \\ &= 11/7 \\ \text{Var}(X \mid Y = 1) &= 11/7 - 1^2 \\ &= 4/7 \approx 0.571 \end{align*}$ 
Unconditionally $E[X] = 0.65$ ; a home claim raises the expected auto count to $1$ , so the two lines move together.

Covariance and Correlation of Two Lines Using the same joint PMF, find $\text{Cov}(X, Y)$ and the correlation coefficient $\rho$ .
Answer The marginals are $p_X = (0.50, 0.35, 0.15)$ on $\{0,1,2\}$ and $p_Y = (0.65, 0.35)$ on $\{0,1\}$ .
 $\begin{align*} E[X] &= 0.35 + 2(0.15) \\ &= 0.65 \\ \text{Var}(X) &= 0.35 + 4(0.15) - 0.65^2 \\ &= 0.5275 \end{align*}$ 
$Y$ is Bernoulli, so $E[Y] = 0.35$ and $\text{Var}(Y) = 0.35(0.65) = 0.2275$ . The only nonzero products $xy$ sit in cells $(1,1)$ and $(2,1)$ :
 $\begin{align*} E[XY] &= (1)(1)(0.15) + (2)(1)(0.10) \\ &= 0.35 \\ \text{Cov}(X,Y) &= 0.35 - (0.65)(0.35) \\ &= 0.1225 \\ \rho &= \frac{0.1225}{\sqrt{0.5275 \times 0.2275}} \\ &= \frac{0.1225}{0.3464} \\ &= 0.354 \end{align*}$ 
A moderate positive correlation — worth knowing before pricing a bundled auto-and-home discount.

Total Claims from Two Independent Policies Policy A has claim count $X$ with $P(X = 0, 1, 2) = 0.7, 0.2, 0.1$ . Policy B, independent, has $Y$ with $P(Y = 0, 1) = 0.6, 0.4$ . Find the PMF of $S = X + Y$ and $P(S \ge 2)$ .

Answer Convolve, summing over the ways each total can occur:

$P(S = 0)$ : $(0.7)(0.6) = 0.42$

$P(S = 1)$ : $(0.7)(0.4) + (0.2)(0.6) = 0.40$

$P(S = 2)$ : $(0.2)(0.4) + (0.1)(0.6) = 0.14$

$P(S = 3)$ : $(0.1)(0.4) = 0.04$

The four probabilities sum to 1, and $P(S \ge 2) = 0.14 + 0.04 = 0.18$ : an 18% chance the pair produces two or more claims in the year.