Question

5.3.12. Let X1, X2, ..., Xn be a random sample from a Poisson distribution with mean u. Thus, Y = {i=1 X; has a Poisson distribution with mean nu. Moreover, X = Y/n is approximately N(41, 4/n) for large n. Show that u(Y/n) = VY/n is a function of Y/n whose variance is essentially free of u.

Please explain how the answer below was found...no other way please...i dont understand how this solution was found!! thank you

Answer 1

$u(Y/n)=\sqrt{Y}/n=(1/\sqrt{n})(\sqrt{Y}/\sqrt{n})=(1/\sqrt{n})(\sqrt{Y/n})$

$u(k)=\sqrt{k/n}$

$var(u(k))=[u'(\mu)]^2*\sigma^2$

where, $\mu=E(k), \sigma^2=var(k)$

$u'(k)=1/2\sqrt{kn}, [u'(k)]^2=1/4kn$

put k = Y/n, E(Y/n) = $\mu$ , V(Y/n) = $\mu$ /n

$var(u(Y/n))=(1/4\mu n)*(\mu/n)=1/4n^2$

which is independent of $\mu$

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