Please refer to the question in image

enter image description here

I have tried to find $ E(x) $ but i ended up with $\overline x $ = $\frac{\theta + 1}{\theta} $ which statisfies no option , i also tried to find $ E(x-1)^2 $ but then it gives $\frac{\sum (x-1)^2}{n}$= $\frac{\theta + 2}{\theta^2}$.

Please suggest the correct method.

  • $\begingroup$ Redo your calculation of $E[(x-1)^2],$ because the result you quote is incorrect. $\endgroup$ – whuber May 5 '19 at 15:20
  • $\begingroup$ How did you end up with $\frac{\theta+1}{\theta}$? $\endgroup$ – gunes May 5 '19 at 16:02
  • $\begingroup$ Please try to type out the questions and if this is self-study, please add the tag. $\endgroup$ – StubbornAtom May 5 '19 at 19:15

Since the first order (population) moment $1=E\left(\frac{1}{n}\sum\limits_{i=1}^n X_i\right)$ is independent of $\theta$, we can consider the second order raw moment $\frac{1}{\theta}+1=E\left(\frac{1}{n}\sum\limits_{i=1}^n X_i^2\right)$.

By method of moments,

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

So a valid method of moments estimator of $\theta$ is simply $$\hat\theta(X_1,\ldots,X_n) =\frac{1}{\frac{1}{n}\sum\limits_{i=1}^n X_i^2-1}$$

Since $E\left[\frac{1}{n}\sum\limits_{i=1}^n (X_i-1)^2\right]=\frac{1}{\theta}$, we again have by method of moments

$$\frac{1}{n}\sum\limits_{i=1}^n (X_i-1)^2=\frac{1}{\theta}$$

Thus giving the estimator $$\hat\theta'(X_1,\ldots,X_n)=\frac{n}{\sum\limits_{i=1}^n (X_i-1)^2}$$

Here we equated sample variance with population variance, i.e. considering central moments instead of raw moments. Looking at the options, this seems to be the convention followed in the question.

| cite | improve this answer | |
  • 1
    $\begingroup$ Couldn't you also proceed by noting that $$E\left[\frac{1}{n}\sum_{i=1}^n(X_i-1)^2\right]=\frac{1}{\theta}?$$ And after "If instead of," just as before you appear to intend that expectations should be taken--but then how do you obtain $1/\theta$ as the expectation, given that this is a biased estimator of the variance? $\endgroup$ – whuber May 5 '19 at 19:19
  • $\begingroup$ @whuber Isn't $1/\theta=E(X_1-1)^2$ the population variance? The sample variance with divisor $n$ is biased for $1/\theta$, but why is this important in method of moments? Your first equation is of course true. But is $\frac{1}{n}\sum (X_i-1)^2$ the sample variance instead of $\frac{1}{n}\sum (X_i-\overline X)^2$ (do we estimate $\overline X$ by $1$?)? $\endgroup$ – StubbornAtom May 5 '19 at 19:31
  • 1
    $\begingroup$ You know $E[\bar X]=1;$ there's no need to estimate it. Bias is unimportant, but computing the expectation correctly is; and the bias of the expression you write on the left hand side demonstrates that its expectation is not $1/\theta$: it must be $(n-1)/(n\theta).$ $\endgroup$ – whuber May 5 '19 at 19:45
  • $\begingroup$ @whuber I misunderstood your first comment. Thanks. $\endgroup$ – StubbornAtom May 5 '19 at 20:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.