How Do You Calculate E[(x+1)^2(y-1)^2] for Independent Variables?

[Samwise_geegee]

Homework Statement

Let x and y be independent random variables with E[x]=1, E[y]=-1, var[x]=1/2, var[y]=2
Calculate E[(x+1)²(y-1)²]

Homework Equations

E[x]=1=μ
E[y]=-1=μ
var[x]=1/2 =E[(x-μ)²]
var[y]=2=E[(x-μ)²]

The Attempt at a Solution

Since x and y are independent,
E[(x+1)²(y-1)²]=E[(x+1)²]*E[(y-1)²]

var[x]=1/2=E[(x-1)²]

var[y]=2=E[(y+1)²

The signs in the equation I need to solve are throwing me off. I feel like I'm missing something simple. Any help is appreciated!

 

[Ray Vickson]

Homework Statement

Let x and y be independent random variables with E[x]=1, E[y]=-1, var[x]=1/2, var[y]=2
Calculate E[(x+1)²(y-1)²]

Homework Equations

E[x]=1=μ
E[y]=-1=μ
var[x]=1/2 =E[(x-μ)²]
var[y]=2=E[(x-μ)²]

The Attempt at a Solution

Since x and y are independent,
E[(x+1)²(y-1)²]=E[(x+1)²]*E[(y-1)²]

var[x]=1/2=E[(x-1)²]

var[y]=2=E[(y+1)²

The signs in the equation I need to solve are throwing me off. I feel like I'm missing something simple. Any help is appreciated!

Sometimes the easiest approach is to use the standard result
\text{Var}(Y) = E(Y^2) - (E Y)^2,
which is true for any random variable having finite mean and variance. (At some point in your life, you should prove it.) You can expand out $(Y-1)^2$ and go on from there.

 

[Samwise_geegee]

Thanks Ray!

Thank you for the hint! Does this look right?

E[(x+1)²]*E[(y-1)²]

=(E[X²]+2E[X]+E[1])(E[Y²]-2E[Y]+E[1])

=(Var[X]+E[X]²+2E[X]+E[1])(Var[Y]+E[Y]²-2E[Y]+E[1])

=(.5+1+2+1)(2+1+1+1)=18