Suppose that $X$ is a discrete random variable with $P(X=1)=θ$ and $P(X=2)=1−θ$. Three independent observations of $X$ are made: $x_1=1,x_2=2,x_3=2$.
Find the MOM estimator of $θ$. What is the MOM estimate and its variance?

So I have found the MOM estimator to be $\hat{\theta}=2-\bar{X}$ and based on these 3 observations, the MOM estimate is $\hat{\theta}=2-\frac{5}{3}=\frac{1}{3}.$ I'm pretty confident with this part.
For the question asking to find the SE of this estimate, my first thought is:

Hence $\mathrm{Var}(\hat{\theta})=\mathrm{Var}(2-\bar{X})=\mathrm{Var}(\bar{X})=\frac{\sigma^2}{n}=\frac{\theta^2(1-\theta)^2}{n}$. So plug in the estimated value of $\hat{\theta}=\frac{1}{3}$ and $n=3$, I got $\mathrm{Var}(\hat{\theta})=\frac{2}{27}$.

However, after reading through this post, I came up with another answer as suggested by the poster:

Take $\mathrm{Var}(\bar{X})$ to be the sample variance, which is $\frac{1}{2}((1-\frac{5}{3})^2+2(2-\frac{5}{3})^2)=\frac{1}{3}$, thus $\mathrm{Var}(\hat{\theta})=\mathrm{Var}(\bar{X})=\frac{1}{3}$.

Could anyone help me figure out which one is wrong? The first one makes more sense to me, but I can't seem to find a loop-hole in the poster's explanation. Thanks in advance!


1 Answer 1


The first part is correct. Ok. The last one is wrong. The variance of the estimator is always unknown. One can estimate is using $\hat{\theta}$ but then it is random and varies for evey sample. Moreover, the variance of the estimator should converge to 0 as $n\to \infty$ (if not, the estimator would not be consitent and that is really not desirable)

There is just a mistype: You got: $$\sigma^2 = E[X^2] - E[X]^2 = \theta (1-\theta)$$ which is correct. But then $$SE_{\hat{\theta}} = \sqrt{Var(\hat{\theta})} = \sqrt{Var(2-\overline{X})} = \sqrt{\frac{\theta (1-\theta)}{n}}$$ where we use that $X_1,\dots, X_n$ are independent identically distributed.

So here I denoted by $SE_{\hat{\theta}}$ the standard error for the estimator $\hat{\theta}$ of $\theta$. So as you can see $SE_{\hat{\theta}}$ is unknown since $\theta$ is unknown, but one can estimate $\theta$ by $\hat{\theta} = 2-\overline{X}$ (stochastic). With our sample we have $\hat{\theta} = 1/3$ so an estimate of the standard error would be $$\hat{SE}_{\hat{\theta}} = \sqrt{\frac{\frac{1}{3} (1-\frac{1}{3})}{3}} = \sqrt{\frac{\frac{2}{9}}{3}} = \sqrt{2/27}.$$

  • $\begingroup$ Thank you! That helps me a lot! $\endgroup$
    – drawar
    Feb 18, 2014 at 2:29

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .