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I am self-studying statistical inference, and I got confused by an example about a biased Bernoulli estimator. Please do not refer to measure theory when answering this question, because it is not supposed to be at that level. I really appreciate it! Here is my question:

My Question

Let $X_1,\dots,X_n$ be a random sample from a Bernoulli$(\theta)$ population. Consider the estimator $\hat{\theta}_{n+2}$ of the parameter $\theta$ defined by $$ \hat{\theta}_{n+2}=\frac{1}{n+2}\sum_{i=1}^nX_i. $$ The note then claimed than the sampling distribution of $\hat{\theta}_{n+2}$ is Binomial with mean $\frac{n\theta}{n+2}$ and variance $\frac{n\theta(1-\theta)}{(n+2)^2}$. Thus, the bias is $$ \text{Bias}_{\theta}\left(\hat{\theta}_{n+2}\right)=E_{\theta}\left[\hat{\theta}_{n+2}\right]-\theta=-\frac{2\theta}{n+2}. $$

Where I Got Stuck

I have problem deriving the sampling distribution of $\hat{\theta}_{n+2}$. Here is what I have tried: The range of $\hat{\theta}_{n+2}$ is $\left\{0,\frac{1}{n+2},\frac{2}{n+2},\dots,\frac{n}{n+2}\right\}$. Then $$ \text{Pr}\left\{\hat{\theta}_{n+2}=k|\theta\right\}=f(k|\theta)={n\choose k(n+2)}\theta^{k(n+2)}(1-\theta)^{n-k(n+2)}. $$ But I don't know what I should do next to show that the sampling distribution of $\hat{\theta}_{n+2}$ is Binomial with mean $\frac{n\theta}{n+2}$ and variance $\frac{n\theta(1-\theta)}{(n+2)^2}$. Could someone please help me out? Thanks a lot in advance!

Update

As my conversation with @tkw shown below, it could be the case that the sampling distribution of $\hat{\theta}_{n+2}$ is approximated by a binomial distribution. But how could we derive that? (If it is necessary to invoke measure theory, then please feel free to do that.)

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  • $\begingroup$ As my conversation with @tkw shown below, it could be the case that the sampling distribution of $\hat{\theta}_{n+2}$ is approximated by a binomial distribution. But how could we derive that? (If it is necessary to invoke measure theory, then please feel free to do that.) $\endgroup$ Commented Oct 15, 2024 at 1:50

1 Answer 1

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We have, $$ (n + 2)\widehat{\theta}_{n + 2} = \sum_{k = 1}^n X_k $$ Since $X_k \sim \text{Ber}(\theta)$ or equivalently $X_k \sim \text{Binomial}(1, \theta)$, and $X_1, \ldots, X_k$ are independent, this means $$ \sum_{k = 1}^n X_k \sim \text{Binomial}(n, \theta) $$ So, $$ \widehat{\theta}_{n + 2} \sim \dfrac{1}{n + 2}\text{Binomial}(n, \theta) $$ I think this is what the original note means when it said "the sampling distribution of $\widehat{\theta}_{n + 2}$ is Binomial"

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  • $\begingroup$ Do you mean the following: the sampling distribution of $\hat{\theta}_{n+2}$ is, in fact, a binomial distribution scaled by $\frac{1}{n+2}$. Then, if we denote by $Y$ the random variable that has a Binomial$(n,\theta)$ distribution, we would have the (sampling distribution) mean of the estimator $\hat{\theta}_{n+2}$ being $$E\left[\hat{\theta}_{n+2}\right]=E\left[\frac{1}{n+2}Y\right]=\frac{1}{n+2}E[Y]=\frac{n\theta}{n+2}$$ and the variance of the estimator being $$\text{Var}(\hat{\theta}_{n+2})=\text{Var}(\frac{1}{n+2}Y)=\frac{1}{(n+2)^2}\text{Var}(Y)=\frac{n\theta(1-\theta)}{(n+2)^2}.$$ $\endgroup$ Commented Oct 15, 2024 at 2:26
  • $\begingroup$ @Beerus Yes, that is correct $\endgroup$ Commented Oct 15, 2024 at 2:28
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    $\begingroup$ Thank you so much! $\endgroup$ Commented Oct 15, 2024 at 2:29

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