Can use averaging of multiple shots of an image to reduce variance due to noise
Proof that variance is reduced by averaging $N$ samples:
$\begin{array}{lll} \textrm{random variable} & X \\ \textrm{mean } & E(X) & = \mu & = \sum_i x_i \; P(X=x_i) \\ \textrm{variance } & V(X) & = \sigma^2 & = E( \; (X-\mu)^2 \; ) \\ \textrm{standard deviation } & \sqrt{ \; V(X)} & = \sigma \\ \end{array}$
$\begin{array}{rl} V( \textrm{average of N samples} ) & = V( {1 \over N} \sum_i x_i ) \\ & = E( ({1 \over N} \sum_i x_i - \mu)^2 ) \\ & = E( ({1 \over N} \sum_i x_i - {1 \over N} \sum_i \mu)^ 2 ) \\ & = E( ({1 \over N} (\sum_i( x_i - \mu )))^2 ) \\ & = E( {1 \over N^2} (\sum_i( x_i - \mu ))^2 ) \\ & = {1 \over N^2} \; E( \sum_i( x_i - \mu ) \times \sum_j( x_j - \mu ) ) \\ & = {1 \over N^2} \; E( \sum_i \sum_j (x_i-\mu) \times (x_j-\mu) ) \\ & = {1 \over N^2} \; \sum_i \sum_j E((x_i-\mu) \times (x_j-\mu)) \\ \\ & \textrm{when } i = j, \; E((x_i-\mu) \times (x_j-\mu)) = V(X) \\ & \textrm{when } i \neq j, \; E((x_i-\mu) \times (x_j-\mu)) = 0 \;\;\; \textrm{because the samples are independent (i.e. their covariance = 0)} \\ \\ & = {1 \over N^2} \; N \; V(X) \\ & = {1 \over N} \; V(X) \\ & = {1 \over N} \; \sigma^2 \\ \\ \textrm{stdev}( {1 \over N} \sum_i x_i ) & = \sqrt{{1 \over N} \; \sigma^2} \\ & = {1 \over \sqrt{N}} \; \sigma \\ \end{array}$
So the standard deviation diminishes by $1 \over \sqrt{N}$ for $N$ samples
This is slow: halving the standard deviation requires ${1 \over \sqrt{N}} = {1 \over 2}$. That is, $N = 4$ times the number of samples.
Example:
[Gonzales and Woods]