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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 $T_n$.

Suriya
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1 Answers1

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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).

  1. 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 $\theta$.

  1. 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.

tommik
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  • Thanks for your answer! I have two doubts:

    1. If $\overline Y_n$ is an estimator of $\frac{1}{1+\theta}$, is it also an estimator of $\theta$? I guess the answer is yes, it’s just to be sure.

    2. Once we use the CLT we get $\sqrt{n}\frac{\overline Y_n-p}{\sqrt{p(1-p)}}\rightarrow N(0,1)$, how to I get the confidence interval for $p$? Should I replace the variance with the sample variance and use Slutsky’s theorem to conclude that it also converges to $N(0,1)$ and then calculate the confidence interval?

     – Suriya Jan 05 '21 at 13:06
  • @SaudiBombsYemen 1. It's a function of $\theta$ estimator

    $\frac{\Sigma_i Y_i-np}{\sqrt{np(1-p)}}\sim \Phi$ thus you can immediately calculate a CI for $p$. Being $p=\frac{1}{\theta+1}$ you immediately can have your CI for $\theta$. It is a standard application of CLT in the version of De Moivre - Laplace:

    https://en.wikipedia.org/wiki/De_Moivre%E2%80%93Laplace_theorem

     – tommik Jan 05 '21 at 13:19
  • A confidence interval for $p$ would be $P(-a<\frac{\sum Y_i -np}{\sqrt{np(1-p)}}<a) = 0.95$ right? My question is, how do I end up with something like $P(-c<p<d)=0.95$? I found an explicit formula, but I would like to do it without looking at that formula. Can I substitute the p in the variance for $\overline Y_n$ maybe? – Suriya Jan 05 '21 at 14:52
  • @SaudiBombsYemen : there are two procedures. The most common is to substitute standard dev of $SigmaY$ with its estimator. the other way (less common ) is to solve the double inequality in $p$ but the result is very nasty. In the link I wrote in my answer, there is the same problem I explained for the poisson – tommik Jan 05 '21 at 15:27
  • Great! Now I understand everything, thanks! With the estimator of the variance you mean the sample variance or substituting the estimator of p in p(1-p)? – Suriya Jan 05 '21 at 16:08