9
$\begingroup$

Let$ X_1, ...,X_n$ be a random sample feom a distribution $Geometric(\theta)$ for $0<\theta<1$. I.e,

$$p_{\theta}(x)=\theta(1-\theta)^{x-1} I_{\{1,2,...\}}(x)$$

Find the unbiased estimator with minimum variance for $g(\theta)=\frac{1}{\theta}$

My attempt:

Since the geometric distribution is from the exponential family, the statistics $$\sum X_i $$ is complete and sufficient for $ \theta$. Also, if $$T(X)=X_1$$ is an estimator for $g(\theta)$, it is unbiased. Therefore, by the Rao-Blackwell theorem and Lehmann-Scheffé Theorem, $$W(X) = E[X_1|\sum X_i]$$ is the estimator we are looking for.

We have the following:

$W(X) = \sum_{i=1}^t i\, P(X_1=i|\sum X_i =t) = \sum_{i=1}^t i\, \frac{P(\sum_{i \geq 2} X_i =t-i)P(X_1=i)}{P(\sum_{i \geq 1}X_i =t)}$

Since the variables are iid geometric, the sums distributions are both negative binomials. But i am having troubles tosimplify the binomial coefficients and give a final answer with a better form, if it is possible.I wpuld be glad if I could get some help.

Thanks!

Edit: I dont think you guys understand my doubt: Ithink I made all the correct steps, maybe only forgot some indicator function. Here is what I did:

$$...=\sum_{i=1}^ti\frac{\binom{t-i-1}{n-2}\theta^{n-i}(1-\theta)^{t-i-n+1} \theta(1-\theta)^{i-1}}{\binom{t-1}{n-1}\theta^n(1-\theta)^{t-n}}=\sum_{i=1}^t i \frac{\binom{t-i-1}{n-2}}{\binom{t-1}{n-1}}$$

As i said, I am having troubles to simplify this and with the somatory index

$\endgroup$
4
$\begingroup$

Indeed for a Geometric ${\cal G}(\theta)$ variate, $X$, $$\mathbb{E}_\theta[X]=1/\theta=g(\theta)$$and the Rao-Blackwell theorem implies that $$\hat{\theta}(T)=\mathbb{E}_\theta\left[X_1\Bigg|\sum_{i=1}^n X_i=T\right]$$is the unique minimum variance unbiased estimator. But rather than trying to compute this conditional expectation directly, one could remark that $$\mathbb{E}_\theta\left[X_1\Bigg|\sum_{i=1}^n X_i=T\right]=\ldots=\mathbb{E}_\theta\left[X_n\Bigg|\sum_{i=1}^n X_i=T\right]$$ hence that $$\mathbb{E}_\theta\left[X_1\Bigg|\sum_{i=1}^n X_i=T\right]=\frac{1}{n}\sum_{i=1}^n \mathbb{E}_\theta\left[X_i\Bigg|\sum_{i=1}^n X_i=T\right]=\frac{T}{n}$$ Note, incidentally, that, since $\sum_{j\ge 2} X_j$ is a Negative Binomial $\cal{N}eg(n-1,\theta)$ $$\mathbb{P}\left(\sum_{j\ge 2} X_j=m\right)={m-1\choose n-2}\theta^{n-1}(1-\theta)^{m-n+1}\mathbb{I}_{m>n-1}$$ hence the final sum should be $$\sum_{i=1}^{t-n+1} i {\binom{t-i-1}{n-2}}\bigg/{\binom{t-1}{n-1}}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.