(New page: 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[X]-E[N] + (E[X(X-N)]-E[X]E[X-N])/(Var(x) * (X-E[X]) ...) |
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| Line 3: | Line 3: | ||
So if expanding on the last entry I got | So if expanding on the last entry I got | ||
| − | E[X]-E[N] + (E[X(X-N)]-E[X]E[X-N])/(Var(x) * (X-E[X]) | + | E[X]-E[N] + (E[X(X-N)]-E[X]E[X-N])/(Var(x)) * (X-E[X]) |
After some rearranging | After some rearranging | ||
Revision as of 15:41, 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[X]-E[N] + (E[X(X-N)]-E[X]E[X-N])/(Var(x)) * (X-E[X])
After some rearranging
E[X]-E[N] +(E[X^2]-E[XN]+E[X]^2+E[X]E[N])/(E[X^2]+E[X]^2] *(X-E[X])
E[X]-E[N] + (E[X^2]-E[X]^2)/(E[X^2]-E[X]^2) *(X-E[X])
E[X]-E[N] + (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
Do you use law of iterated expectation here?
