Checking the consistency and Bias of $\frac{\sum X_i +\sqrt{n}/2}{n+\sqrt{n}}$

Question

Let $X_1,\ldots,X_n$ be i.i.d. $B(1,\theta)$ random variables, $0<\theta<1$. Then, as an estimator $\theta$, check if $T(X_1,\ldots,X_n)= \dfrac{\sum_{i=1}^n X_i +\sqrt{n}/2}{n+\sqrt{n}}$ is consistent and/or unbiased.

$$T=\frac{\frac{1}{n}\sum_{i=1}^n X_i +\frac{1}{2\sqrt{n}}}{1+\frac{1}{\sqrt{n}}}$$

$$T=\frac{\bar{X} +\frac{1}{2\sqrt{n}}}{1+\frac{1}{\sqrt{n}}}$$

$$E(T)=\frac{E(\bar{X} +\frac{1}{2\sqrt{n}})}{1+\frac{1}{\sqrt{n}}}=\dfrac{\theta +\frac{1}{2\sqrt{n}}}{1+\frac{1}{\sqrt{n}}}.$$

So, $T$ is biased.

Consistency: $\lim\limits_{n \to\infty}E(T)=\lim\limits_{n \to\infty}\dfrac{E(\theta +\frac{1}{2\sqrt{n}})}{1+\frac{1}{\sqrt{n}}}$

So $T$ is consistent.

But The given answer is "neither unbiased nor consistent"

Where did I go wrong ? Please advise.

Answer 1

Since you wrote $B(1,\theta)$ rather than $\mathrm{Bin}(1,\theta)$, I thought at first that you meant the Beta distribution, but it now appears you meant the binomial distribution. Assuming that, the expected value of each observation is $\theta$ and your proofs of bias and consistency are correct. Perhaps if you quote the source of what you call "the given answer" more completely, then I could shed more light.