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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.

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  • $\begingroup$ Redo your calculation of $E[(x-1)^2],$ because the result you quote is incorrect. $\endgroup$ – whuber May 5 at 15:20
  • $\begingroup$ How did you end up with $\frac{\theta+1}{\theta}$? $\endgroup$ – gunes May 5 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 at 19:15
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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.

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    $\begingroup$ 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 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 at 19:31
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    $\begingroup$ 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 at 19:45
  • $\begingroup$ @whuber I misunderstood your first comment. Thanks. $\endgroup$ – StubbornAtom May 5 at 20:03

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