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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)\\ =(1+2E[X]+E[X]^2) Var(Y)+(1+2E[Y]+E[Y]^2)^2 Var(X)+Var(X)Var(Y)\\ \text{ }\\ \text{if E[X] = 0 and E[Y] = 0, then } Var(X)=E[X]^2 \text{ and } Var(Y)=E[Y]^2\\ =(1+Var(X)) Var(Y) + (1+Var(Y)) Var(X) + Var(X)Var(Y)\\ \text{ }\\ \end{equation} Now, I got this far. Still I don't know is this very promising and what would happen with out assumption of zero expectations.\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}

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)\\ =(1+2E[X]+E[X]^2) Var(Y)+(1+2E[Y]+E[Y]^2)^2 Var(X)+Var(X)Var(Y)\\ \text{ }\\ \text{if E[X] = 0 and E[Y] = 0, then } Var(X)=E[X]^2 \text{ and } Var(Y)=E[Y]^2\\ =(1+Var(X)) Var(Y) + (1+Var(Y)) Var(X) + Var(X)Var(Y)\\ \text{ }\\ \end{equation} Now, I got this far. Still I don't know is this very promising and what would happen with out assumption of zero expectations.

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}

Moved on based on the first comment
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Techonomist
Techonomist
  • 55
  • 3

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)\\ =(1+2E[X]+E[X]^2) Var(Y)+(1+2E[Y]+E[Y]^2)^2 Var(X)+Var(X)Var(Y)\\ \text{ }\\ \text{if E[X] = 0 and E[Y] = 0, then } Var(X)=E[X]^2 \text{ and } Var(Y)=E[Y]^2\\ =(1+Var(X)) Var(Y) + (1+Var(Y)) Var(X) + Var(X)Var(Y)\\ \text{ }\\ \end{equation} Now, I got this far. Still I don't know is this very promising and what would happen with out assumption of zero expectations.

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?

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)\\ =(1+2E[X]+E[X]^2) Var(Y)+(1+2E[Y]+E[Y]^2)^2 Var(X)+Var(X)Var(Y)\\ \text{ }\\ \text{if E[X] = 0 and E[Y] = 0, then } Var(X)=E[X]^2 \text{ and } Var(Y)=E[Y]^2\\ =(1+Var(X)) Var(Y) + (1+Var(Y)) Var(X) + Var(X)Var(Y)\\ \text{ }\\ \end{equation} Now, I got this far. Still I don't know is this very promising and what would happen with out assumption of zero expectations.

typo
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Techonomist
Techonomist
  • 55
  • 3

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 tothe solution simplify if expected values of the variables are zero?

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 to solution simplify if expected values of the variables are zero?

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?

Source Link
Techonomist
Techonomist
  • 55
  • 3