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The problem is:

Show that $\frac{X+1}{n+2}$ is a biased estimator of the binomial parameter $\theta$. Is this estimator asymptotically unbiased?

My way of doing this is following:

Calculate the expected value/mean of the estimator $\frac{X+1}{n+2}$:
$E\Big[\frac{X+1}{n+2}\Big]=\Big({1\over n+2}\Big)E\Big[X+1\Big]=\Big({1\over n+2}\Big)\Big[E[X]+1\Big]=\Big({1\over n+2}\Big)(n\theta+1)={n\theta\over n+2}+{1\over n+2}$
Since this is not equal to $\theta$, this is a biased estimator.
Now, check: $\lim_{n\to \infty}E\Big[\frac{X+1}{n+2}\Big]=\lim_{n\to \infty}\Big({n\theta\over n+2}+{1\over n+2}\Big)=\theta$.
So, this is asymptotically unbiased.

I wonder whether I was doing right about this problem.

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Who does $X$ depend on the sample points? –  Sasha Jan 21 '13 at 14:11
1  
Your first "equality" has a typo, at the very least. The approach is ok. –  cardinal Jan 21 '13 at 14:23
    
@cardinal Got you. I have already fixed it. –  Scorpio19891119 Jan 21 '13 at 14:33
    
This estimator is UNbiased for one value of theta. –  Did Jan 21 '13 at 14:37
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@Did: I see the point you are raising, though, generally speaking, the definition for unbiasedness of a statistical estimator entails that it is true uniformly in $\theta$, i.e., $\delta(X)$ is an unbiased estimator of a parameter $\theta$ iff $\mathbb E_{\theta} \delta(X) = \theta$ for all $\theta \in \Theta$. See, for example, E. L. Lehmann and G. Casella, Theory of Point Estimation, 2nd ed., Springer, 1998, pg. 5. –  cardinal Jan 21 '13 at 15:40

1 Answer 1

You are. Alternatively, you could have shown that the bias of your estimator is not zero.

$\mathbb{E}(\frac{X+1}{n+2}-\theta)=\frac{1}{n+2}\mathbb{E}(X+1-\theta n -2\theta)=\frac{1-2\theta}{n+2}$ implying estimator is biased.

Working with the bias rather than with the estimator is useful in the more general setting where the parameter you are estimating is a random variable rather than a constant.

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