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So if expanding on the last entry I got | So if expanding on the last entry I got | ||
| − | E[ | + | E[Y] + (E[X(X-N)]-E[X]E[X-N])/(Var(x)) * (X-E[X]) |
After some rearranging | After some rearranging | ||
| − | E[ | + | E[Y] +(E[X^2]-E[XN]+E[X]^2+E[X]E[N])/(E[X^2]+E[X]^2]) * (X-E[X]) |
| + | Indep. | ||
| + | E[Y] + (E[X^2]-E[X]^2)/(E[X^2]-E[X]^2) * (X-E[X]) | ||
| − | E[ | + | E[Y] + (X-E[X]) |
| − | + | Thats what i got so far you sould be able to plug in values assuming this is right..theres an example like in class btw. Maybe someone could check? | |
| − | + | ||
| − | Thats what i got so far you sould be able to plug in values assuming this is right..theres an example like in class btw | + | |
Do you use law of iterated expectation here? | Do you use law of iterated expectation here? | ||
Latest revision as of 15:44, 8 December 2008
The way I went about this problem is I set Y=X-N and substituted all over the place
So if expanding on the last entry I got
E[Y] + (E[X(X-N)]-E[X]E[X-N])/(Var(x)) * (X-E[X])
After some rearranging
E[Y] +(E[X^2]-E[XN]+E[X]^2+E[X]E[N])/(E[X^2]+E[X]^2]) * (X-E[X])
Indep.
E[Y] + (E[X^2]-E[X]^2)/(E[X^2]-E[X]^2) * (X-E[X])
E[Y] + (X-E[X])
Thats what i got so far you sould be able to plug in values assuming this is right..theres an example like in class btw. Maybe someone could check?
Do you use law of iterated expectation here?
