Vtyjkyuk

Let $X_1,X_2,dots, X_n$ be rvs with pdf:
$$f(xmid theta)=frac12thetaI(-theta<x<theta)$$

Find UMVUE of $(i)dfractheta1+theta$ and $(ii)dfrace^thetatheta$.

Note that, $(X_(1),X_(n))$ is complete sufficient statistic. But now I have to find unbiased estimator of $(i),(ii)$ of the form $g(X_(1),X_(n))$, then $g$ will become UMVUE. But I could not find such $g$. Thanks for any help.

I tried to find $E(X_(1)/X_(n))$, but it came out a total mess.

Here $X_(1)=min(X_1,X_2,dots, X_n)$ and $X_(n)=max(X_1,X_2,dots, X_n)$.

You have a $U(-theta,theta)$ population where $thetainmathbb R^+$.

Joint density of the sample $mathbf X=(X_1,X_2,ldots,X_n)$ is

beginalign
f_theta(mathbf x)&=frac1(2theta)^nmathbf1_-theta < x_1, ldots, x_n < theta
\&=frac1(2theta)^nmathbf1_x_1
\&=frac1(2theta)^nmathbf1_max_1le ile n
endalign

It is clear from Factorization theorem that a sufficient statistic for $theta$ is $$T(mathbf X)=max_1le ile n|X_i|$$

One could verify that $|X_i|sim U(0,theta)$, so that the density of $T$ is $$g_theta(t)=fracntheta^nt^n-1mathbf1_0<t<theta$$

That $T$ is a complete statistic for $theta$ is well-known.

We simply have to find unbiased estimators of the parametric functions of $theta$ based on the complete sufficient statistic. This would give us the UMVUE by the Lehmann-Scheffe theorem.

As the support of the complete sufficient statistic here depends on the parameter $theta$, unbiased estimators can be directly obtained through differentiation.

Let $h_1(T)$ and $h_2(T)$ be unbiased estimators of $theta/(1+theta)$ and $e^theta/theta$ respectively, based on the complete sufficient statistic $T$.

That is, for all $theta>0$,

beginalign
qquadquadfracntheta^nint_0^thetah_1(t)t^n-1,dt&=fractheta1+theta
\implies int_0^thetah_1(t)t^n-1,dt &= fractheta^n+1n(1+theta)
endalign

Differentiating both sides wrt $theta$,

beginalign
h_1(theta)theta^n-1&=fractheta^n(ntheta+n+1)n(1+theta)^2
\implies h_1(theta) &=fractheta(ntheta+n+1)n(1+theta)^2
endalign

Hence, $$h_1(T)=fracT(nT+n+1)n(1+T)^2$$

Similarly for the second problem, for all $theta>0$,

beginalign
qquadquadfracntheta^nint_0^thetah_2(t)t^n-1,dt&=frace^thetatheta
\implies int_0^thetah_2(t)t^n-1,dt &= fractheta^n-1 e^thetan
endalign

Differentiating both sides wrt $theta$ yields

beginalign
h_2(theta)theta^n-1&=frace^thetatheta^n-2(theta+n-1)n
\implies h_2(theta) &=frace^theta(theta+n-1)ntheta
endalign

So, $$h_2(T)=frace^T(T+n-1)nT$$

In my initial answer, the following calculation for the UMVUE was rather unnecessary and complicated. Had the support not depended on the parameter, I might have tried this. I am keeping this part in the answer as I might be able to salvage the somewhat faulty argument on some further consideration :

For $k> -n$, we have

beginalign
E_theta(T^k)&=fracntheta^nint_0^theta t^k+n-1,dt\[8pt]
& = fracntheta^kn+k
endalign

This suggests that an unbiased estimator of $theta^k$ based on $T$ is $$left(fracn+knright)T^k$$

For the first problem, one could write

beginalign
fractheta1+theta&=
begincasesleft(1+frac1thetaright)^-1=1-frac1theta+frac1theta^2-frac1theta^3+cdots&,text if theta>1\\theta(1+theta+theta^2+cdots)&,text if 0<theta<1endcases
endalign

For $0<theta<1$, we have

$$E_thetaleft[left(fracn+1nright)T+left(fracn+2nright)T^2+cdotsright]=theta+theta^2+cdots$$

Or, $$E_thetaleft[sum_k=1^inftyleft(fracn+knright)T^kright]=fractheta1+theta$$

For $theta>1$,

$$E_thetaleft[1-left(fracn-1nright)frac1T+left(fracn-2nright)frac1T^2-cdotsright]=1-frac1theta+frac1theta^2-cdots$$

That is, $$E_thetaleft[sum_k=0^inftyleft(fracn-knright)frac(-1)^kT^kright]=fractheta1+theta$$

Hence by Lehmann-Scheffe theorem, UMVUE of $theta/(1+theta)$ is

beginalign
h_1(T)&=begincasesdisplaystylesum_k=1^inftyleft(fracn+knright)T^k&,text if 0<theta<1\\displaystylesum_k=0^inftyleft(fracn-knright)frac(-1)^kT^k&,text if thetage1 endcases
\\&=begincasesdisplaystylefracT(n+1-nT)n(T-1)^2&,text if 0<theta<1\\displaystylefracT(n+1+nT)n(T+1)^2&,text if thetage1endcases
endalign

However, upon verification of unbiasedness for some values of $n$, it looks like only $$h_1(T)=displaystylefracT(n+1+nT)n(T+1)^2$$ should be the correct answer for all $theta>0$. I am not quite sure why that happens.

For the second problem, we can use the power series expansion of $e^theta$ to obtain

$$E_thetaleft[sum_k=-1^inftyleft(fracn+knright)fracT^k(k+1)!right]=sum_j=0^infty fractheta^j-1j!=frace^thetatheta$$

So the UMVUE of $e^theta/theta$ is

beginalign
h_2(T)&=sum_k=-1^inftyleft(fracn+knright)fracT^k(k+1)!
\\&=frace^T(n-1+T)nT
endalign