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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.

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2 Answers 2

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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.

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  • $\begingroup$ Such a concise argument. I am not sure I follow how you computed $\mathbb{E}[e^{i\xi X_1}]$ though... For $E((X_1+...+X_n)^l)$ did you compute $l$ partial derivatives on $\mathbb{E}(e^{i\xi(X_1+...+X_n)})$ with respect to $\xi$? $\endgroup$
    – Kadmos
    Commented Oct 17, 2023 at 18:32
  • $\begingroup$ @Kadmos $\xi\mapsto E[e^{i\xi X_1}]$ is the characteristic function of the Gamma distribution, which is known. This uniquely identifies probability distributions. Then $Y=X_1+...+X_n$ is a Gamma rv with parameters $n\alpha,\lambda$ and thus $E[Y^\ell]=\lambda^{-\ell}\Gamma(\alpha n +\ell)/\Gamma(\alpha n)$. $\endgroup$
    – Snoop
    Commented Oct 17, 2023 at 18:35
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I write again the last equation the OP has deduced: $$\mathbb{E}\left[\left.X_1^3\frac{\Gamma(n\alpha)}{\Gamma(\alpha+3)}\right| \sum X_i\right]=\frac{(\sum X_i)^3}{\Gamma(\alpha+3)}\frac{\Gamma(n \alpha)}{\Gamma((n-1)\alpha)}\int_0^{1} \tilde{x}^{\alpha+2}(1-\tilde{x})^{(n-1)\alpha-1}d\tilde{x}$$

It turns out that this is a well known integral given by:

$$\int_0^1 \tilde{x}^{\alpha+3-1}(1-\tilde{x})^{(n-1)\alpha-1}d\tilde{x}=\mathcal{B}(\alpha+3,(n-1)\alpha)=\frac{\Gamma(\alpha+3)\Gamma((n-1)\alpha)}{\Gamma(\alpha n+3)}$$

Substituting this expession yields:

$$\mathbb{E}\left[\left.X_1^3\frac{\Gamma(n\alpha)}{\Gamma(\alpha+3)}\right| \sum X_i\right]=\frac{(\sum X_i)^3}{\Gamma(\alpha+3)}\frac{\Gamma(n \alpha)}{\Gamma((n-1)\alpha)}\frac{\Gamma(\alpha+3)\Gamma((n-1)\alpha)}{\Gamma(\alpha n+3)}=\frac{(\sum X_i)^3\Gamma(n\alpha)}{\Gamma(\alpha n+3)}$$

This agrees with Snoop's clever observation that we could compute the expectation of $(\sum X_i)^3$ directly and correct the bias afterwards.

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