Variance of X+Y+XY? Assuming that random variables X and Y are independent, what is $\displaystyle Var((1+X)(1+Y)-1)=Var(X+Y+XY)$?
Should I start as follows
\begin{equation}
Var((1+X)(1+Y)-1)\\
=Var((1+X)(1+Y))\\
=(E[(1+X)])^2 Var(1+Y)+(E[(1+Y])^2 Var(1+X)+Var(1+X)Var(1+Y)
\end{equation}
or maybe as follows
\begin{equation}
\\
Var((1+X)(1+Y)-1)\\
=Var(1+Y+X+XY-1)\\
=Var(X+Y+XY)\\
=Var(X)+Var(Y)+Var(XY)+2Cov(X,Y)+2Cov(X,XY)+2Cov(Y,XY)
\end{equation}
I'm considering could I express the problem in terms of covariances (and variances) between individual random variables. I would like to forecast the variance by individual covariances in my model if its possible. Does the solution simplify if expected values of the variables are zero?
Edit:
Moving on from the first alternative
\begin{equation}
=(E[(1+X)])^2 Var(1+Y)+(E[(1+Y])^2 Var(1+X)+Var(1+X)Var(1+Y)\\
=(E[(1+X)])^2 Var(Y)+(E[(1+Y])^2 Var(X)+Var(X)Var(Y)\\
=(1+E[X])^2 Var(Y)+(1+E[Y])^2 Var(X)+Var(X)Var(Y)\\
\text{ }\\
\text{if E[X] = 0 and E[Y] = 0, then }\\
=Var(Y) + Var(X) + Var(X)Var(Y)\\
\text{ }\\
\end{equation}
 A: Using the result mentioned by Matt Barstead in the comments ( a textbook result: https://en.wikipedia.org/wiki/Variance#Product_of_independent_variables ):
for uncorrelated X and Y you have
$$Var(XY) = Var(X)Var(Y) + E(X)^2 Var(Y) + E(Y)^2 Var(X)$$
You only have to substitute $X'=X+1$ and  $Y'=Y+1$, with $E(X')=E(X)+1$, $var(X') = var(X)$, $E(Y')=E(Y)+1$, $var(Y') = var(Y)$ leading to:
$$\begin{array}\\
Var((X+1)(Y+1)) &= Var(X'Y') \\
                  &= Var(X')Var(Y') + E(X')^2 Var(Y') + E(Y')^2 Var(X') \\ 
                  &= Var(X)Var(Y) +  \cdots\end{array}$$
A: For independent random variables $X$ and $Y$ with means $\mu_X$ and $\mu_Y$ respectively, and variances $\sigma_X^2$ and $\sigma_Y^2$ respectively,
\begin{align}\require{cancel}
\operatorname{var}(X+Y+XY) &= \operatorname{var}(X)+\operatorname{var}(Y)+\operatorname{var}(XY)\\
&\quad +2\cancelto{0}{\operatorname{cov}(X,Y)}+2\operatorname{cov}(X,XY)+\operatorname{cov}(Y,XY)\\
&=\sigma_X^2+\sigma_Y^2+\big(\sigma_X^2\sigma_Y^2+\sigma_X^2\mu_Y^2+\sigma_Y^2\mu_X^2\big)\\
&\quad +2\operatorname{cov}(X,XY)+\operatorname{cov}(Y,XY).
\end{align}
Now,
\begin{align}
\operatorname{cov}(X,XY) &= E[X\cdot XY] - E[X]E[XY]\\
&=E[X^2Y]-E[X]\big(E[X]E[Y]\big)\\
&= E[X^2]E[Y]-\big(E[X]\big)^2E[Y]\\
&= \sigma_X^2\mu_Y
\end{align}
and similarly, $\operatorname{cov}(Y,XY) = \sigma_Y^2 \mu_X$.
Consequently,
\begin{align}\operatorname{var}(X+Y+XY) &=\sigma_X^2+\sigma_Y^2+\sigma_X^2\sigma_Y^2+\sigma_X^2\mu_Y^2+\sigma_Y^2\mu_X^2 +2\sigma_X^2\mu_Y + 2\sigma_Y^2 \mu_X\\
&= \sigma_X^2\big(1 + \mu_Y^2 + 2\mu_Y\big) + \sigma_Y^2\big(1 + \mu_X^2 + 2\mu_X\big) + \sigma_X^2\sigma_Y^2\\
&= \sigma_X^2\big(1 + \mu_Y\big)^2 + \sigma_Y^2\big(1 + \mu_X\big)^2 + \sigma_X^2\sigma_Y^2. 
\end{align}
