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^n\left(\prod_{i=1}^n x_i\right)^{\theta-1}\mathbf1_{0<x_1,\ldots,x_n<1} \\&=\exp\left[(\theta-1)\sum_{i=1}^n\ln x_i+n\ln\theta+\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 $$T(X_1,\ldots,X_n)=\sum_{i=1}^n\ln X_i$$

Since $E(X_1)=\frac{\theta}{1+\theta}$, at first thought, $E(X_1\mid T)$ 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 $X_1\mid \sum_{i=1}^n\ln X_i$.

On the other hand, I considered the following approach:

We have $X_i\stackrel{\text{i.i.d}}{\sim}\text{Beta}(\theta,1)\implies -2\theta\ln X_i\stackrel{\text{i.i.d}}{\sim}\chi^2_2$, so that $-2\theta\, T\sim\chi^2_{2n}$.

So $r$th order raw moment of $-2\theta\,T$ about zero, as calculated using the chi-square pdf is $$E(-2\theta\,T)^r=2^r\frac{\Gamma\left(n+r\right)}{\Gamma\left(n\right)}\qquad ,\,n+r>0$$

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, $E\left(-\frac{T}{n}\right)=\frac{1}{\theta}$ and $E\left(\frac{1-n}{T}\right)=\theta$ 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 $$E\left(1+\frac{T}{n}+\frac{T^2}{n(n+1)}+\frac{T^3}{n(n+1)(n+2)}+\cdots\right)=1-\frac{1}{\theta}+\frac{1}{\theta^2}-\frac{1}{\theta^3}+\cdots$$

That is, $$E\left(\sum_{r=0}^\infty \frac{T^r}{n(n+1)...(n+r-1)}\right)=\frac{\theta}{1+\theta}$$

Is it possible that my required UMVUE is $\displaystyle\sum_{r=0}^\infty \frac{T^r}{n(n+1)...(n+r-1)}$ 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 $E(X_1\mid \sum\ln X_i=t)=E(X_1\mid \prod X_i=e^t)$, I had proceeded to find the conditional distribution $X_1\mid \prod X_i$. 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 $J=\left(\prod_{i=1}^{n-1}y_i\right)^{-1}$.

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 $$f_{Y_1,Y_n}(y_1,y_n)=\frac{\theta^n\,y_n^{\theta-1}}{y_1}\int_0^{y_{n-2}}\int_0^{y_{n-3}}\cdots\int_0^{y_1}\frac{1}{y_3y_4...y_{n-1}}\frac{\mathrm{d}y_2}{y_2}\cdots\mathrm{d}y_{n-2}\,\mathrm{d}y_{n-1}$$

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 $V=-\sum_{i=2}^n\ln X_i$.

It is immediately seen that $U\sim\text{Exp}(\theta)$ and $V\sim\text{Gamma}(n-1,\theta)$ are independent.

And indeed, $U+V\sim\text{Gamma}(n,\theta)$.

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 $U\mid U+V=z$ is $$f_{U\mid U+V}(u\mid z)=\frac{(n-1)(z-u)^{n-2}}{z^{n-1}}\mathbf1_{0<u<z}$$

Now, my UMVUE is exactly $E(e^{-U}\mid U+V=z)=E(X_1\mid \sum_{i=1}^n\ln X_i=-z)$, 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.

  • $\begingroup$ You should proceed with the first method, finding the conditional distribution of $X_{[1]} | \prod X_i$, which form of the sufficient statistic may be easier to work with in this application. $\endgroup$
    – jbowman
    Commented May 31, 2018 at 14:22
  • 1
    $\begingroup$ At the point (early on) where you introduce $T$ you should be minded to work in terms of the variables $Y_i=-\log X_i.$ It's almost immediate that they are proportional to $\Gamma(1)$ distributions, which quickly reduces your problem to considering the joint distribution of $(U,U+V)$ where $U\sim\Gamma(1)$ and $V\sim\Gamma(n-1).$ That will shortcut the remaining two pages of mathematics and give you a quick route to a solution. $\endgroup$
    – whuber
    Commented Jun 5, 2018 at 20:58
  • $\begingroup$ @whuber To be clear, are you suggesting that I find the density of $(-\ln X_1,-\ln X_1-\sum_{i=2}^n\ln X_i)$ first and from that find the density of $(X_1,\prod X_i)$? I had noticed that the $-\ln X_i$'s are exponential variables with rate $\theta$ (which is also a Gamma variable as you say), but hadn't thought of working with that. $\endgroup$ Commented Jun 6, 2018 at 10:34
  • 1
    $\begingroup$ @whuber But how would I get $E(X_1\mid ...)$ from $E(\ln X_1\mid ...)$ directly ? $\endgroup$ Commented Jun 6, 2018 at 14:29
  • 1
    $\begingroup$ @whuber Please have a look at my edit. I have almost done it but not sure what to do with that integral. I am fairly confident my calculations are correct. $\endgroup$ Commented Jun 6, 2018 at 16:36

2 Answers 2


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, $\text{Exp}(\theta)\equiv\text{Gamma}(1,\theta)$.

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

We note that $X_i\stackrel{\text{i.i.d}}{\sim}\text{Beta}(\theta,1)\implies-\ln X_i\stackrel{\text{i.i.d}}{\sim}\text{Exp}(\theta)\implies-T\sim\text{Gamma}(n,\theta)$.

Method I:

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

So, $\mathbb E(X_1\mid \sum_{i=1}^n\ln X_i=t)=\mathbb E(e^{-U}\mid U+V=-t)$.

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

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 $U\mid U+V=z$ is \begin{align}f_{U\mid U+V}(u\mid 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, $\displaystyle\mathbb E(e^{-U}\mid U+V=z)=\frac{n-1}{z^{n-1}}\int_0^z e^{-u}(z-u)^{n-2}\,\mathrm{d}u$.

That is, the UMVUE of $\frac{\theta}{1+\theta}$ is $\displaystyle\mathbb E(X_1\mid T)=\frac{n-1}{(-T)^{n-1}}\int_0^{-T} e^{-u}(-T-u)^{n-2}\,\mathrm{d}u\tag{1}$

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,

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

Using this equation we obtain unbiased estimators (and UMVUE's) of $1/\theta^r$ for every integer $r\ge1$.

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

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

That is, $$\mathbb E\left(\sum_{r=0}^\infty \frac{T^r}{n(n+1)...(n+r-1)}\right)=\frac{\theta}{1+\theta}$$

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

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.

  • $\begingroup$ In fact, the equality between $(1)$ and $(2)$ follows by writing the power series expansion of $e^{-u}$ in $(2)$ and then interchanging the integral and sum. $\endgroup$ Commented Jun 9, 2018 at 19:35

I think we can get to the more compact answer that you alluded to in regards to the upper incomplete gamma function. Using the first Method, I found the expression

$$ E \left[ X_1 | X_1X_2 \cdots X_n=e^T \right]=\left( n - 1 \right) \int_0^1 z^r \left( 1 - r \right)^{n-z}dr,$$ where $z=e^T.$

Wolfram Alpha integrates this to get $$E \left[ X_1 | X_1X_2 \cdots X_n=e^T \right] = \frac{e^T \left( n-1 \right)}{T^{n-1} } \left[ \left( n-2 \right) !-\Gamma \left(n-1,T \right) \right]$$

Now the incomplete gamma function term does have a closed form when $n$ is an integer. It is

$$\Gamma \left(n-1,T \right)=\frac{\Gamma \left(n-1 \right)}{e^T} \sum_{j=0}^{n-2} \frac{T^j}{j!} $$

Rewriting the expectation and simplifying, we find

$$E \left[ X_1 | X_1X_2 \cdots X_n=e^T \right] = \frac{\Gamma \left( n \right)}{T^{n-1}} \left[ e^T-\sum_{j=0}^{n-2} \frac{T^j}{j!} \right]$$

I don't have access to the software that will verify equivalence with your results $(1)$ and $(2)$, but hand calculations for $n=2$ and $n=3$ do match with your $(1)$.

  • $\begingroup$ Where you wrote $\operatorname E\left[ X_1\mid x_1x_2\cdots x_n=e^T \right],$ you ought to have $\operatorname E\left[ X_1\mid X_1X_2\cdots X_n=e^T \right]. \qquad$ $\endgroup$ Commented Nov 8, 2018 at 22:50

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