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Expected value equation

  • #1

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!
 

Answers and Replies

  • #2
Ray Vickson
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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
[tex] \text{Var}(Y) = E(Y^2) - (E Y)^2,[/tex]
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.
 
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  • #3
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
 
Last edited:

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