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About LinkBacksMore specifically, if X and Y are Poisson-distributed, what is the variance of X/Y (assuming that X and Y are independent)?
Thanks!
To get the variance, you need the expectation, and the expectation of the square. Since the variables are independent, you needOriginally Posted by floats
and
. I guess you know how to get
and
(
and
since I know the variance is
![E\left[\frac{1}{Y+1}\right]=e^{-\lambda}\sum_{n=0}^\infty \frac{1}{n+1}\frac{\lambda^n}{n!}=e^{-\lambda}\sum_{n=0}^\infty \frac{\lambda^n}{(n+1)!}](http://latex.codecogs.com/png.latex?E\left[\frac{1}{Y+1}\right]=e^{-\lambda}\sum_{n=0}^\infty \frac{1}{n+1}\frac{\lambda^n}{n!}=e^{-\lambda}\sum_{n=0}^\infty \frac{\lambda^n}{(n+1)!})
; then let
in the sum to see that it equals
.
is more tricky. I'd let
and compute
, which we have computed above ; now integrate to get
, and multiply by
to get
.
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