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The statement

I am given the following discrete distribution with $\theta>0$

$$p(x) = \left(\frac{\theta}{1+\theta}\right) ^{2-x}\left(\frac{1}{1+\theta}\right)^{x-1} \hspace{1cm} x=1,2$$

I need to calculate an estimator of $\theta$ (call it $T_n$) using a sample $x_1,x_2,...,x_n$, deduce its distribution and calculate a confidence interval for $\theta$.

What I did

If $X_1,...,X_n$ are n copies of $X$, the sample is $x_1=X_1(\omega),..., x_n=X_n(\omega)$

Using the method of moments we can relate the sample mean to the expectation

$$\overline X_n = E[X] = 1+\frac{1}{1+\theta}=\mu$$

and define the estimator of $\theta$

$$T_n=\frac{1}{\overline X_n-1}-1$$

I also calculated the variance of X: $Var(X)=\frac{\theta}{(1+\theta)^2}=\sigma^2$

By the Central Limit Theorem

$$\sqrt{n}\,(\overline X_n-\mu) \rightarrow N(0,\sigma^2)$$

We can apply the Delta Method with the function $g(t)=\frac{1}{t-1}-1$ to get

$$\sqrt{n}\,(g(\overline X_n)-g(\mu)) \rightarrow N(0,\sigma^2g’(\mu)^2)$$

i.e.

$$\sqrt{n}\,(T_n-\theta) \rightarrow N(0,\theta(1+\theta)^2)$$

My Problem

Now I need to determine the distribution of $T_n$ but I don’t know the value of $\theta.$ I could approximate $\theta$ by $T_n$ and use Slutsky’s theorem to conclude that

$$\sqrt{n}\,\frac{(T_n-\theta)}{\sqrt{T_n}(T_n+1)} \rightarrow N(0,1)$$

But again I don’t know how to obtain the distribution of $T_n$.

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  • $\begingroup$ $T_n$ isn't even a random variable, because it isn't completely defined: what should it equal when $\bar x_n=1$?? $\endgroup$ – whuber Jan 2 at 17:35
  • $\begingroup$ @whuber but if $\theta$ Is greater than 0, $\overline X_n$ Is greater than 1. Should I substitute $\overline x_n$ with $\overline X_n$? $\endgroup$ – SaudiBombsYemen Jan 2 at 17:39
  • $\begingroup$ I don't understand what distinction you are making with that notation. By definition, the estimator is a function of the sample, which always has a positive chance of consisting entirely of $n$ $1$s. $\endgroup$ – whuber Jan 2 at 18:16
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    $\begingroup$ You could consider the MLE of $\theta$ which also has an asymptotic normal distribution. $\endgroup$ – StubbornAtom Jan 2 at 21:18
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    $\begingroup$ The MLE is one solution. The other is to complete the definition of your estimator by stating what value of $\theta$ should be estimated when $\bar x_n=1.$ That would be a little complicated due to the need to analyze everything conditionally on this event, but it could be done. There are other solutions, too, because there exist plenty of good estimators of $\theta$ and not all of them need to be MLEs. $\endgroup$ – whuber Jan 2 at 21:57

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