Variance and Standard Deviation

Variance $\sigma^2$ and Standard Deviation $\sigma$ both measure how far a Random Variable spreads around its mean. Variance is the expected squared deviation; standard deviation is its square root, restored to the original units of $X$ .

$\sigma^2 = \text{Var}(X) = E\left[(X - \mu)^2\right]$

$= E[X^2] - \mu^2$

$\sigma = \text{SD}(X) = \sqrt{\text{Var}(X)}$

$\mu = E[X]$ . The second form $E[X^2] - \mu^2$ is the one to use in practice — it needs only two moments and avoids expanding a square.
$\text{Var}(aX + b) = a^2\,\text{Var}(X)$ and $\text{SD}(aX + b) = |a|\,\text{SD}(X)$ — a shift $b$ moves the distribution without changing its spread.
For a sum, $\text{Var}(X + Y) = \text{Var}(X) + \text{Var}(Y) + 2\,\text{Cov}(X,Y)$ ; the Covariance term drops out whenever $X$ and $Y$ are uncorrelated, $\text{Cov}(X,Y) = 0$ — for example when they are independent. Zero covariance does not imply independence. See Moments for Linear Combinations.
Standard deviations do not add: $\text{SD}(X+Y) \neq \text{SD}(X) + \text{SD}(Y)$ . Always add variances, then take the square root at the very end.
Dividing $\sigma$ by $\mu$ gives the unitless Coefficient of Variation.

Variance of a Discrete Claim Count The number of claims $N$ on a policy has $P(N=0) = 0.6$ , $P(N=1) = 0.3$ , $P(N=2) = 0.1$ . Find $\text{Var}(N)$ and $\text{SD}(N)$ .
Answer First both moments:
 $\begin{align*} E[N] &= 0(0.6) + 1(0.3) + 2(0.1) \\ &= 0.5 \\ E[N^2] &= 0^2(0.6) + 1^2(0.3) + 2^2(0.1) \\ &= 0.7 \end{align*}$ 
Then the variance and standard deviation:
 $\begin{align*} \text{Var}(N) &= E[N^2] - (E[N])^2 \\ &= 0.7 - 0.25 \\ &= 0.45 \\ \text{SD}(N) &= \sqrt{0.45} \\ &\approx 0.67 \end{align*}$ 

Variance from a Continuous Density Losses have density $f(x) = 3x^2$ on $0 < x < 1$ . Find $\text{SD}(X)$ .

Answer

 $\begin{align*} E[X] &= \int_0^1 x \cdot 3x^2\,dx \\ &= \left[\tfrac{3}{4}x^4\right]_0^1 \\ &= 0.75 \\ E[X^2] &= \int_0^1 x^2 \cdot 3x^2\,dx \\ &= \left[\tfrac{3}{5}x^5\right]_0^1 \\ &= 0.60 \end{align*}$

 $\begin{align*} \text{Var}(X) &= 0.60 - 0.75^2 \\ &= 0.0375 \\ \text{SD}(X) &\approx 0.194 \end{align*}$

Why Standard Deviations Cannot Be Added Two independent policies each have loss standard deviation $300. Find the standard deviation of their combined loss.
Answer Add the variances, not the standard deviations:
 $\begin{align*} \text{Var}(X_1 + X_2) &= 300^2 + 300^2 \\ &= 180{,}000 \\ \text{SD}(X_1 + X_2) &= \sqrt{180{,}000} \\ &\approx 424.26 \end{align*}$ 
The answer is $424.26, not $600. Pooling independent risks grows total standard deviation by $\sqrt{n}$ , not $n$ — the mathematical basis for diversification.