Question

If we observe y0 as the value for a geometric random variable Y, P(Y = y0)...

If we observe y0

as the value for a geometric random variable

Y, P(Y = y0)

is maximized when p = 1/Y0.

The maximum likelihood estimator for p is 1/Y
(note that Y is the geometric random variable, not a particular value of it). Derive E(1/y).

E (1/y) =

0 0
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