Let $(X_1,X_2,ldots,X_n)$ be a random sample from the density $$f_{theta}(x)=theta x^{theta-1}mathbf1_{0<x<1}quad,,theta>0$$

I am trying to find the UMVUE of $frac{theta}{1+theta}$.

The joint density of $(X_1,ldots,X_n)$ is

begin{align}

f_{theta}(x_1,cdots,x_n)&=theta^nleft(prod_{i=1}^n x_iright)^{theta-1}mathbf1_{0<x_1,ldots,x_n<1}

\&=expleft[(theta-1)sum_{i=1}^nln x_i+nlntheta+ln (mathbf1_{0<x_1,ldots,x_n<1})right],,theta>0

end{align}

As the population pdf $f_{theta}$ belongs to the one-parameter exponential family, this shows that a complete sufficient statistic for $theta$ is $$T(X_1,ldots,X_n)=sum_{i=1}^nln X_i$$

Since $E(X_1)=frac{theta}{1+theta}$, at first thought, $E(X_1mid T)$ would give me the UMVUE of $frac{theta}{1+theta}$ by the Lehmann-Scheffe theorem. Not sure if this conditional expectation can be found directly or one has to find the conditional distribution $X_1mid sum_{i=1}^nln X_i$.

On the other hand, I considered the following approach:

We have $X_istackrel{text{i.i.d}}{sim}text{Beta}(theta,1)implies -2thetaln X_istackrel{text{i.i.d}}{sim}chi^2_2$, so that $-2theta, Tsimchi^2_{2n}$.

So $r$th order raw moment of $-2theta,T$ about zero, as calculated using the chi-square pdf is $$E(-2theta,T)^r=2^rfrac{Gammaleft(n+rright)}{Gammaleft(nright)}qquad ,,n+r>0$$

So it seems that for different integer choices of $r$, I would get unbiased estimators (and UMVUEs) of different integer powers of $theta$. For example, $Eleft(-frac{T}{n}right)=frac{1}{theta}$ and $Eleft(frac{1-n}{T}right)=theta$ directly give me the UMVUE's of $frac{1}{theta}$ and $theta$ respectively.

Now, when $theta>1$ we have $frac{theta}{1+theta}=left(1+frac{1}{theta}right)^{-1}=1-frac{1}{theta}+frac{1}{theta^2}-frac{1}{theta^3}+cdots$.

I can definitely get the UMVUE's of $frac{1}{theta},frac{1}{theta^2},frac{1}{theta^3}$ and so on. So combining these UMVUE's I can get the required UMVUE of $frac{theta}{1+theta}$. Is this method valid or should I proceed with the first method? As UMVUE is unique when it exists, both should give me the same answer.

To be explicit, I am getting $$Eleft(1+frac{T}{n}+frac{T^2}{n(n+1)}+frac{T^3}{n(n+1)(n+2)}+cdotsright)=1-frac{1}{theta}+frac{1}{theta^2}-frac{1}{theta^3}+cdots$$

That is, $$Eleft(sum_{r=0}^infty frac{T^r}{n(n+1)…(n+r-1)}right)=frac{theta}{1+theta}$$

Is it possible that my required UMVUE is $displaystylesum_{r=0}^infty frac{T^r}{n(n+1)…(n+r-1)}$ when $theta>1$?

For $0<theta<1$, I would get $g(theta)=theta(1+theta+theta^2+cdots)$, and so the UMVUE would differ.

Having been convinced that the conditional expectation in the first approach could not be found directly, and since $E(X_1mid sumln X_i=t)=E(X_1mid prod X_i=e^t)$, I had proceeded to find the conditional distribution $X_1mid prod X_i$. For that, I needed the joint density of $(X_1,prod X_i)$.

I used the change of variables $(X_1,cdots,X_n)to (Y_1,cdots,Y_n)$ such that $Y_i=prod_{j=1}^i X_j$ for all $i=1,2,cdots,n$. This lead to the joint support of $(Y_1,cdots,Y_n)$ being $S={(y_1,cdots,y_n): 0<y_1<1, 0<y_j<y_{j-1} text{ for } j=2,3,cdots,n}$.

The jacobian determinant turned out to be $J=left(prod_{i=1}^{n-1}y_iright)^{-1}$.

So I got the joint density of $(Y_1,cdots,Y_n)$ as $$f_Y(y_1,y_2,cdots,y_n)=frac{theta^n, y_n^{theta-1}}{prod_{i=1}^{n-1}y_i}mathbf1_S$$

Joint density of $(Y_1,Y_n)$ is hence $$f_{Y_1,Y_n}(y_1,y_n)=frac{theta^n,y_n^{theta-1}}{y_1}int_0^{y_{n-2}}int_0^{y_{n-3}}cdotsint_0^{y_1}frac{1}{y_3y_4…y_{n-1}}frac{mathrm{d}y_2}{y_2}cdotsmathrm{d}y_{n-2},mathrm{d}y_{n-1}$$

Is there a different transformation I can use here that would make the derivation of the joint density less cumbersome? I am not sure if I have taken the correct transformation here.

Based on some excellent suggestions in the comment section, I found the joint density of $(U,U+V)$ instead of the joint density $(X_1,prod X_i)$ where $U=-ln X_1$ and $V=-sum_{i=2}^nln X_i$.

It is immediately seen that $Usimtext{Exp}(theta)$ and $Vsimtext{Gamma}(n-1,theta)$ are independent.

And indeed, $U+Vsimtext{Gamma}(n,theta)$.

For $n>1$, joint density of $(U,V)$ is $$f_{U,V}(u,v)=theta e^{-theta u}mathbf1_{u>0}frac{theta^{n-1}}{Gamma(n-1)}e^{-theta v}v^{n-2}mathbf1_{v>0}$$

Changing variables, I got the joint density of $(U,U+V)$ as

$$f_{U,U+V}(u,z)=frac{theta^n}{Gamma(n-1)}e^{-theta z}(z-u)^{n-2}mathbf1_{0<u<z}$$

So, conditional density of $Umid U+V=z$ is $$f_{Umid U+V}(umid z)=frac{(n-1)(z-u)^{n-2}}{z^{n-1}}mathbf1_{0<u<z}$$

Now, my UMVUE is exactly $E(e^{-U}mid U+V=z)=E(X_1mid sum_{i=1}^nln X_i=-z)$, as I had mentioned right at the beginning of this post.

So all left to do is to find $$E(e^{-U}mid U+V=z)=frac{n-1}{z^{n-1}}int_0^z e^{-u}(z-u)^{n-2},mathrm{d}u$$

But that last integral has a closed form in terms of incomplete gamma function according to *Mathematica*, and I wonder what to do now.

**Contents**hide

#### Best Answer

*It turns out that both approaches (my initial attempt and another based on suggestions in the comment section) in my original post give the same answer. I will outline both methods here for a complete answer to the question.*

Here, $text{Gamma}(n,theta)$ means the gamma density $f(y)=frac{theta^n}{Gamma(n)}e^{-theta y}y^{n-1}mathbf1_{y>0}$ where $theta,n>0$, and $text{Exp}(theta)$ denotes an exponential distribution with mean $1/theta$, ($theta>0$). Clearly, $text{Exp}(theta)equivtext{Gamma}(1,theta)$.

Since $T=sum_{i=1}^nln X_i$ is complete sufficient for $theta$ and $mathbb E(X_1)=frac{theta}{1+theta}$, by the Lehmann-Scheffe theorem $mathbb E(X_1mid T)$ is the UMVUE of $frac{theta}{1+theta}$. So we have to find this conditional expectation.

We note that $X_istackrel{text{i.i.d}}{sim}text{Beta}(theta,1)implies-ln X_istackrel{text{i.i.d}}{sim}text{Exp}(theta)implies-Tsimtext{Gamma}(n,theta)$.

**Method I:**

Let $U=-ln X_1$ and $V=-sum_{i=2}^nln X_i$, so that $U$ and $V$ are independent. Indeed, $Usimtext{Exp}(theta)$ and $Vsimtext{Gamma}(n-1,theta)$, implying $U+Vsimtext{Gamma}(n,theta)$.

So, $mathbb E(X_1mid sum_{i=1}^nln X_i=t)=mathbb E(e^{-U}mid U+V=-t)$.

Now we find the conditional distribution of $Umid U+V$.

For $n>1$ and $theta>0$, joint density of $(U,V)$ is

begin{align}f_{U,V}(u,v)&=theta e^{-theta u}mathbf1_{u>0}frac{theta^{n-1}}{Gamma(n-1)}e^{-theta v}v^{n-2}mathbf1_{v>0}\&=frac{theta^n}{Gamma(n-1)}e^{-theta(u+v)}v^{n-2}mathbf1_{u,v>0}end{align}

Changing variables, it is immediate that the joint density of $(U,U+V)$ is $$f_{U,U+V}(u,z)=frac{theta^n}{Gamma(n-1)}e^{-theta z}(z-u)^{n-2}mathbf1_{0<u<z}$$

Let $f_{U+V}(cdot)$ be the density of $U+V$. Thus conditional density of $Umid U+V=z$ is begin{align}f_{Umid U+V}(umid z)&=frac{f_{U,U+V}(u,z)}{f_{U+V}(z)}\&=frac{(n-1)(z-u)^{n-2}}{z^{n-1}}mathbf1_{0<u<z}end{align}

Therefore, $displaystylemathbb E(e^{-U}mid U+V=z)=frac{n-1}{z^{n-1}}int_0^z e^{-u}(z-u)^{n-2},mathrm{d}u$.

That is, the UMVUE of $frac{theta}{1+theta}$ is $displaystylemathbb E(X_1mid T)=frac{n-1}{(-T)^{n-1}}int_0^{-T} e^{-u}(-T-u)^{n-2},mathrm{d}utag{1}$

**Method II:**

As $T$ is a complete sufficient statistic for $theta$, any unbiased estimator of $frac{theta}{1+theta}$ which is a function of $T$ will be the UMVUE of $frac{theta}{1+theta}$ by the Lehmann-Scheffe theorem. So we proceed to find the moments of $-T$, whose distribution is known to us. We have,

$$mathbb E(-T)^r=int_0^infty y^rtheta^nfrac{e^{-theta y}y^{n-1}}{Gamma(n)},mathrm{d}y=frac{Gamma(n+r)}{theta^r,Gamma(n)},qquad n+r>0$$

Using this equation we obtain unbiased estimators (and UMVUE's) of $1/theta^r$ for every integer $rge1$.

Now for $theta>1$, we have $displaystylefrac{theta}{1+theta}=left(1+frac{1}{theta}right)^{-1}=1-frac{1}{theta}+frac{1}{theta^2}-frac{1}{theta^3}+cdots$

Combining the unbiased estimators of $1/theta^r$ we obtain $$mathbb Eleft(1+frac{T}{n}+frac{T^2}{n(n+1)}+frac{T^3}{n(n+1)(n+2)}+cdotsright)=1-frac{1}{theta}+frac{1}{theta^2}-frac{1}{theta^3}+cdots$$

That is, $$mathbb Eleft(sum_{r=0}^infty frac{T^r}{n(n+1)…(n+r-1)}right)=frac{theta}{1+theta}$$

So assuming $theta>1$, the UMVUE of $frac{theta}{1+theta}$ is $g(T)=displaystylesum_{r=0}^infty frac{T^r}{n(n+1)…(n+r-1)}tag{2}$

I am not certain about the case $0<theta<1$ in the second method.

According to *Mathematica*, equation $(1)$ has a closed form in terms of the incomplete gamma function. And in equation $(2)$, we can express the product $n(n+1)(n+2)…(n+r-1)$ in terms of the usual gamma function as $n(n+1)(n+2)…(n+r-1)=frac{Gamma(n+r)}{Gamma(n)}$. This perhaps provides the apparent connection between $(1)$ and $(2)$.

Using *Mathematica* I could verify that $(1)$ and $(2)$ are indeed the same thing.

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