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}

  • $\begingroup$ I'd start the first way. You can simplify that substantially. $\endgroup$ – Glen_b Jan 19 '18 at 12:58
  • 1
    $\begingroup$ You may find this post useful stats.stackexchange.com/questions/15978/… $\endgroup$ – Matt Barstead Jan 19 '18 at 13:54
  • $\begingroup$ Thanks a lot both. I'll look at the post, and maybe get back to it $\endgroup$ – user31383 Jan 19 '18 at 14:06

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}


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:

$$\begin{array}\\ Var((X+1)(Y+1)) &= Var(X'Y') \\ &= Var(X')Var(Y') + E(X')^2 Var(Y') + E(Y')^2 Var(X') \\ &= Var(X)Var(Y) + \cdots\end{array}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.