Variance of random variable for normal distribution How do I find the variance for
$z_n=\prod_{i=1}^n(1-k_i e^{a_i x})$
where $x$ is the random variable with a normal distribution and is the same for all $i$ (which is a subscript for time dependency which is not dependent with $x$).
I have tried using the method by Goodman, but since I have same $x$ for all $i$'s, I don't know exactly how to proceed.
 A: Following the discussion in the comments, assume that there is one random variable, and that $n=2$.  Then
$$Z_2 = (1-k_1 e^{a_1 X})(1-k_2 e^{a_2 X}) = 1-k_2 e^{a_2 X}-k_1 e^{a_1 X}+k_1k_2 e^{(a_1+a_2) X}$$
And
$$Z_2^2 = (1-k_2 e^{a_2 X}-k_1 e^{a_1 X}+k_1k_2 e^{(a_1+a_2) X})(1-k_2 e^{a_2 X}-k_1 e^{a_1 X}+k_1k_2 e^{(a_1+a_2) X}) \\$$
$$\begin{align}& =1-k_2 e^{a_2 X}-k_1 e^{a_1 X}+k_1k_2 e^{(a_1+a_2) X} \\
&-k_2 e^{a_2 X}+k_2^2 e^{2a_2 X}+k_1k_2 e^{(a_1+a_2) X}-k_1k_2^2 e^{(a_1+2a_2) X}\\
&-k_1 e^{a_1 X}+k_1k_2 e^{(a_1+a_2) X}+k_1^2 e^{2a_1 X}-k_1^2k_2 e^{(2a_1+a_2) X}\\
&+k_1k_2 e^{(a_1+a_2) X}-k_1k_2^2 e^{(a_1+2a_2) X}-k_1^2k_2 e^{(2a_1+a_2) X} +k_1^2k_2^2 e^{(2a_1+2a_2) X}\end{align}  $$
$$ \begin{align}\Rightarrow Z_n^2 &= 1-2k_2 e^{a_2 X}-2k_1 e^{a_1 X}+k_2^2 e^{2a_2 X}+ k_1^2 e^{2a_1 X}\\
&+ 4k_1k_2 e^{(a_1+a_2) X} -2k_1^2k_2 e^{(2a_1+a_2) X}-2k_1k_2^2 e^{(a_1+2a_2) X}\\
&+k_1^2k_2^2 e^{(2a_1+2a_2) X} \end{align}$$
In general if $X\sim N(\mu,\sigma^2)$ then  $b_iX=Y \sim N(b_i\mu, b_i^2\sigma^2)$
Also $W=e^{X}$ is a log-normal random variable, with $E(W)=e^{\mu + \sigma^2/2}$.
Finally $E(cW) = cE(W)$.
We see therefore that both $Z_2$ and $Z_2^2$ are just sums of scaled log-normal random variables, of which we want to calculate the expected value, nothing more. This won't change as $n$ increases, so
$$\text{Var}(Z_n) = E(Z_n^2) - [E(Z_n)]^2$$ 
will be computable as a matter of increasingly and exceedingly tedious arithmetic. 
ADDENDUM
The above procedure will also work if we assume that there are many $X$'s, that are independent and jointly normal. In such a case, instead of, say, $(a_1+2a_2)X$ we will have $a_1X_1+2a_2X_2$ which will again be a normal random variable.
