Linear Combinations of Random Variables

Study guides: Exam P-1, Exam MAS-I

Linear Combinations of Random Variables are expressions of the form L = c_1 X_1 + c_2 X_2 + \cdots + c_n X_n for constants c_i, collapsing several random variables into a single quantity — an aggregate loss, a portfolio value, a sample mean.

E[L] = \sum_{i=1}^{n} c_i\,E[X_i]

\text{Var}(L) = \sum_{i=1}^{n} c_i^2\,\text{Var}(X_i) + 2\sum_{i < j} c_i c_j\,\text{Cov}(X_i, X_j)

• Expectation is always linear — E[L] needs no independence assumption at all. The variance formula holds in general too; the Covariance terms drop out whenever the X_i are pairwise uncorrelated. Independent variables are always uncorrelated, but zero covariance does not by itself imply independence. See Moments for Linear Combinations.
• Coefficients enter the variance squared, so \text{Var}(X - Y) = \text{Var}(X) + \text{Var}(Y) for independent X, Y — a difference has a larger variance than either term, never a smaller one.
• A linear combination of independent normals is exactly normal, which is what makes Probabilities for Linear Combinations computable in closed form.
• For n i.i.d. variables with mean \mu and finite variance \sigma^2, \bar{X} has mean \mu and variance \sigma^2/n, and for large n the Central Limit Theorem makes the sum X_1 + \cdots + X_n and the mean \bar{X} approximately normal, whatever their common distribution.

Mean and Variance of a Two-Policy Portfolio Independent losses have E[X] = 300, \text{Var}(X) = 2{,}500, E[Y] = 500, \text{Var}(Y) = 10{,}000. Find the mean and variance of L = 2X + 3Y.

Answer

\begin{align*}
E[L] &= 2(300) + 3(500) \\
     &= 2{,}100
\end{align*}

Independence kills the covariance term, and the coefficients square:

\begin{align*}
\text{Var}(L) &= 2^2(2{,}500) + 3^2(10{,}000) \\
              &= 10{,}000 + 90{,}000 \\
              &= 100{,}000
\end{align*}

A Difference of Correlated Losses X and Y have \text{Var}(X) = 9, \text{Var}(Y) = 16, and \text{Cov}(X,Y) = 6. Find \text{Var}(X - Y).

Answer Here c_1 = 1 and c_2 = -1, so the cross term picks up 2c_1c_2 = -2:

\begin{align*}
\text{Var}(X - Y) &= \text{Var}(X) + \text{Var}(Y) - 2\,\text{Cov}(X,Y) \\
                  &= 9 + 16 - 2(6) \\
                  &= 13
\end{align*}

Positive covariance shrinks the variance of a difference and inflates the variance of a sum — the sign of the cross term flips with the sign of c_2.

Sample Mean of 25 Claims Claim sizes are i.i.d. with mean 400 and standard deviation 100. Find the mean and standard deviation of the average of 25 claims.

Answer \bar{X} = \frac{1}{25}\sum_{i=1}^{25} X_i, so every c_i = 1/25:

\begin{align*}
E[\bar{X}] &= 25 \cdot \tfrac{1}{25}(400) \\
           &= 400 \\
\text{Var}(\bar{X}) &= 25 \cdot \left(\tfrac{1}{25}\right)^2 (100^2) \\
                    &= \frac{10{,}000}{25} \\
                    &= 400
\end{align*}

So \text{SD}(\bar{X}) = 20. Averaging leaves the mean alone but divides the standard deviation by \sqrt{25} = 5.