Method of moments estimator and Delta Method I’ve been a couple of days trying to figure out a satisfactory solution to this exercise but I’m stuck. By how the exercise is phrased I think it wants me to use the moments method and I also think it is a Delta Method exercise.
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 $\theta$
$$T_n=\frac{1}{\overline X_n-1}-1$$
supposing n is big enough so that $\overline X_n$ is not 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 $T_n$.
 A: Your pmf can be rewritten in the following way
$$P(X=x)=\frac{1}{\theta+1}\theta^{2-x}$$
$X=1,2$ that is the following rv
$$ X =
\begin{cases}
\frac{\theta}{\theta+1},  & \text{if $X=1$} \\
\frac{1}{\theta+1},  & \text{if $X=2$}
\end{cases}$$
Now we can transform it into
$$Y= X -1=
\begin{cases}
\frac{\theta}{\theta+1},  & \text{if $Y=0$} \\
\frac{1}{\theta+1},  & \text{if $Y=1$}
\end{cases}$$
That is
$$Y= 
\begin{cases}
1-p,  & \text{if $Y=0$} \\
p,  & \text{if $Y=1$}
\end{cases}$$
Concluding... $Y$ is bernulli $B(p)$.
A CSS (complete and sufficient statistic) for $p$ is
$$\Sigma_i Y_i=(\Sigma_i X_i) -n$$
this is a nice estimator for $p$. Now you can calculate the confidence interval for $p$ as an approximate interval using CLT (if $n$ is greater enough) or also an exact confidence interval using, i.e. the Statistical Method and a binomial table (or any calculator).
here I calculated a similar exact CI with Excel (both for Poisson and Bernulli distribution, thus I think it can be  useful for you).


*

*Confidence interval for $\theta$
$\frac{1}{\theta+1}=p$ is a monotonic function, thus given the CI for $p$ you can easily derive also the CI for $\theta$.



*Distribution of $\hat{\theta}$
$\Sigma_i  Y_i$ is binomial distributed, thus also the distribution of $\hat{\theta}$ is binomial...only with a modified support.

Note that using this method of estimating $\theta$, you can derive the same estimator you found with MoM.
