Answer to Question #280372 in Statistics and Probability for Sana

Question

Let 𝑋1, … , 𝑋𝑛 be iid Gamma(1, 1/𝛼) RVs

(a) Show that the estimator T(𝑋1, … , 𝑋𝑛) = (n-1)/n𝑋̅ is the UMVUE for 𝛼 with 

variance 𝛼²/(𝑛 − 2)

(b)Show that the minimum variance from CR inequality is 𝛼²/𝑛

Answer 1

Solution:

(a):

$X_i\sim G(1,1/\alpha)$ are iid for all i=1,..,n

$f_{x_i}(x_i)=\alpha^{-x_i \alpha} ; x_i>0 \\ f( \underline{X})=\Pi_{i=1}^nf_{x_i}(x_i) \\=\alpha^ne^{-\Sigma x_i \alpha_i} \\=A(\alpha)h(\underline x)e^{\eta(\alpha)T(\underline x)} \\A(\alpha)=\alpha^n, h(\underline x)=1,\eta(\alpha)=-\alpha,T'(\underline x)=\Sigma x_i$

$\therefore f( \underline{X})$ is from exponential family.

Now, $\Sigma x_i\sim G(n,1/\alpha)$

$\therefore E(\dfrac1{\Sigma x_i})=\int_0^{\infty}\dfrac 1t. \dfrac{\alpha^nt^{n-1}e^{-t \alpha}}{\Gamma(n)}dt \\=\dfrac{\Gamma(n-1)}{\Gamma(n)}\alpha \\=\dfrac{\alpha}{n-1} \\\therefore E(\dfrac{n-1}{n\bar X})=\alpha$

So, $T(\underline X)=\dfrac{n-1}{n\bar X}$ is unbiased for $\alpha$.

$\therefore E(T(\underline X)|T'(\underline X)=\Sigma x_i)$ is UMVUE from Lehmann Scheffe theorem.

$\therefore T(\underline X)=\dfrac{n-1}{n\bar X}$ is UMVUE for $\alpha$.

$\therefore E(\dfrac1{(\Sigma x_i)^2})=\int_0^{\infty}\dfrac 1{t^2}. \dfrac{\alpha^nt^{n-1}e^{-t \alpha}}{\Gamma(n)}dt \\=\dfrac{\Gamma(n-2)}{\Gamma(n)}\alpha^2 \\=\dfrac{\alpha^2}{(n-1)(n-2)}$

$V(\dfrac{n-1}{n\bar X})=(n-1)^2V(\dfrac1{(\Sigma x_i)^2}) \\=(n-1)^2[E(\dfrac1{(\Sigma x_i)^2})-E^2(\dfrac1{\Sigma x_i})] \\=(n-1)^2[\dfrac{\alpha^2}{(n-1)(n-2)} -\dfrac{\alpha^2}{(n-1)^2} ] \\=\dfrac{\alpha^2}{n-1}$

(b):

$f_X(x)=\alpha e^{-\alpha x};x>0\ \& \ g(\theta)=\alpha \Rightarrow g'(\theta)=1 \\L(f_X(x))=\alpha^n e^{-\alpha \Sigma x_i} \\ \log L(f_X(x))=n \log \alpha-\alpha \Sigma x_i \\ \dfrac{\partial }{\partial \alpha}\log L(f_X(x))=\dfrac{n}{\alpha}-\Sigma x_i \\ \dfrac{\partial^2 }{\partial \alpha^2}\log L(f_X(x))=\dfrac{-n}{\alpha^2} \\ I(\theta)=-E[\dfrac{\partial^2 }{\partial \alpha^2}\log L(f_X(x))] \\=-E[\dfrac{-n}{\alpha^2}]=\dfrac{n}{\alpha^2}$

And CR inequality $=CR=\dfrac{[g'(\theta)]^2}{I(\theta)}$

Here $g'(\theta)=1$

So, $CR=\dfrac{1}{\dfrac n{\alpha^2}}=\dfrac{\alpha^2}{n}$

Hence, proved.