Probabilities for Linear Combinations

Study guide: Exam P-1

Probabilities for a linear combination $L = c_1 X_1 + c_2 X_2 + \cdots + c_n X_n$ of independent random variables can be found exactly in two cases. When the $X_i$ are independent normals, $L$ is itself normal and is handled by standardization. When the $X_i$ are independent and discrete, the distribution of $L$ is built by adding up, for each value of $L$ , the probabilities of the combinations of values that produce it.

$L = c_1 X_1 + \cdots + c_n X_n \sim N\!\left(\sum_i c_i\mu_i,\ \sum_i c_i^2\sigma_i^2\right)$

$\text{where } X_1, \ldots, X_n \text{ are independent normal random variables}$

$P(X + Y = s) = \sum_{x} P(X = x)\,P(Y = s - x) \qquad \text{for independent integer-valued } X, Y$

The discrete sum above is the convolution of the two probability functions; for more than two variables, convolve one at a time, and a coefficient only relabels values ( $2X$ takes the values $0, 2, 4, \ldots$ ).
Some families are closed under independent sums: independent Poisson variables sum to a Poisson with the means added, so no convolution is needed.
For the sum or mean of many i.i.d. non-normal variables with finite variance, the Central Limit Theorem gives an approximation instead — see Normal Approximation.

Probability That Portfolio Loss Exceeds a Threshold Two independent losses: $X_1 \sim N(100, 10^2)$ and $X_2 \sim N(200, 15^2)$ . Find $P(X_1 + X_2 > 340)$ .

Answer The sum $L = X_1 + X_2$ is normal with: $\mu_L = 100+200 = 300, \qquad \sigma_L = \sqrt{10^2+15^2} = \sqrt{325} \approx 18.03$ Standardising: $P(L > 340) = P\!\left(Z > \frac{340-300}{18.03}\right) = P(Z > 2.22) \approx 0.0132$

Total Claims from Two Independent Policies Policy A has $X$ claims with $P(X=0)=0.5$ , $P(X=1)=0.3$ , $P(X=2)=0.2$ . Policy B, independently, has $Y$ claims with $P(Y=0)=0.6$ , $P(Y=1)=0.4$ . Find $P(X + Y \ge 2)$ .
Answer Add the products over the pairs of values that give each total:
 $\begin{align*} P(X+Y=0) &= (0.5)(0.6) = 0.30 \\ P(X+Y=1) &= (0.5)(0.4) + (0.3)(0.6) = 0.38 \end{align*}$ 
So $P(X + Y \ge 2) = 1 - 0.30 - 0.38 = 0.32$ . Directly, $P(X+Y=2) = (0.3)(0.4) + (0.2)(0.6) = 0.24$ and $P(X+Y=3) = (0.2)(0.4) = 0.08$ , which also total $0.32$ .