I was taking a look at old probability qualifiers and this is from one of them:
Suppouse $\alpha$ is known in $X\sim Gamma(\alpha, \lambda)$. Find the UMVUE for $1/\lambda^3$.
It is well known that:
$$\mathbb{E}[X^k]=\frac{1}{\lambda^k}\frac{\Gamma(\alpha+k)}{\Gamma(\alpha)}$$
From this, we may find an unbiased estimator given by $X_1^3 \frac{\Gamma(\alpha)}{\Gamma(\alpha+3)} $. By Lehman-Scheffe we need only find an unbiased estimator which is a function of $\sum X_i$, because this comes from the exponential family. Rao-Blacwell teaches us how to find such an estimator, namely:
$$\mathbb{E}\left[\left.X_1^3 \frac{\Gamma(\alpha)}{\Gamma(\alpha+3)}\right|\sum X_i\right]$$
Let me compute the conditional density function:
$$f_{X_1|\sum X_i=y}(x)= \frac{f_{X_1}(x)f_{\sum_{i>1} X_i}(y-x)}{f_{\sum X_i}(y)}=\frac{(\lambda^\alpha/\Gamma(\alpha)) e^{-\lambda x}x^{\alpha-1}(\lambda^{(n-1)\alpha}/\Gamma((n-1)\alpha)) e^{-\lambda (y-x)}(y-x)^{(n-1)\alpha-1}} {(\lambda^{n\alpha}/\Gamma(n\alpha)) e^{-\lambda y}y^{n\alpha-1}}$$
$$\therefore f_{X_1| \sum X_i=y}(x)=\frac{\Gamma(n \alpha)}{\Gamma(\alpha)\Gamma((n-1)\alpha)}\frac{x^{\alpha-1}(y-x)^{(n-1)\alpha-1}}{y^{n\alpha-1}}$$
I believe, but I am not completely sure this enables us to write:
$$ \mathbb{E}\left[\left.X_1^3 \frac{\Gamma(\alpha)}{\Gamma(\alpha+3)}\right|\sum X_i\right]=\int_0^\infty x^3 \frac{\Gamma(\alpha)}{\Gamma(\alpha+3)} f_{X_1| \sum X_i=y}(x)dx =$$
$$ \frac{\Gamma(\alpha)}{\Gamma(\alpha+3)}\frac{\Gamma(n \alpha)}{\Gamma(\alpha)\Gamma((n-1)\alpha)}\int_0^y\frac{x^{\alpha+2}(y-x)^{(n-1)\alpha-1}}{y^{n\alpha-1}}dx=$$
$$\frac{1}{\Gamma(\alpha+3)}\frac{\Gamma(n \alpha)}{\Gamma((n-1)\alpha)}\int_0^1\frac{{(y\tilde{x})}^{\alpha+2}(y-(y\tilde{x}))^{(n-1)\alpha-1}}{y^{n\alpha-1}}yd\tilde{x}=$$
$$\frac{(\sum X_i)^3}{\Gamma(\alpha+3)}\frac{\Gamma(n \alpha)}{\Gamma((n-1)\alpha)}\int_0^{1} \tilde{x}^{\alpha+2}(1-\tilde{x})^{(n-1)\alpha-1}d\tilde{x}$$
I am having trouble computing this integral and I am not really not sure if I am on the right track.
Best Answer
Note that by IID-ness we have $E[e^{i\xi (X_1+...+X_n)}]=(E[e^{i\xi X_1}])^n=(1-i\xi/\lambda)^{-n\alpha}$ so $X_1+...+X_n\sim \textrm{Gamma}(n\alpha,\lambda)$. At this point we then have $E[(X_1+...+X_n)^\ell]=\lambda^{-\ell}\frac{\Gamma(n\alpha+\ell)}{\Gamma(n\alpha)}$, and we can conclude.