0
$\begingroup$

I am having trouble finding out and verifying estimators properties ( like unbiased, consistency sufficiency, efficiency ) but in this particular problem given below I have to find a constant such that the given statistic will be an unbiased estimator

let $X_1,\dots,X_n$ be I.I.d. $N(0,\theta^2)$ random variables, $\theta>0$. Then the value of $k$ for which the estimator $\left(k\sum_{i=1}^3\left|X_i\right|\right)$ is an unbiased estimator of $\theta$ is?

What is the significance of modulus even if modulus is not there then the given statistic is not at all an estimator of $\theta$ or is it? As I know by examples in lessons I took previously the unbiased estimator of variance in normal distribution $\left(\frac{1}{n-1}\sum_{i=1}^n\left(X_i-\sum_{i=1}^n\frac{X_i}{n}\right)^2\right)$ sorry but I have no idea what to do this one as I am not able to relate it with the things I have learned. Edit: need to find unbiased estimator of standard deviation $\theta$ and after reading this page, found out $\hat\sigma = \sqrt{ \frac{1}{n-1.5} \sum_{i=1}^n(x_i - \bar{x})^2}$ can be used to find estimator of standard deviation which has very low bias( for this sample n=3 is 1.3%) so, but I still can't compare the given formula and the expression in question, can I take $\bar{x}=0$, How to obtain relation between $\sqrt{\sum_{i=1}^3(X_i)^2}\text{with}\sum_{i=1}^3\left|X_i\right|$

$\endgroup$
5
  • 2
    $\begingroup$ Did you try taking the expectation of $|X|$? $\endgroup$
    – JohnK
    Jan 26, 2016 at 18:25
  • $\begingroup$ Yeah, but I don't know about what to do with modulus as I know the random variables are taken from a symmetric distribution and every negative one will have it's equal positive counter part taking this in mind I thought that if there may any one of the three is negative it will be change to its positive counter part due to modulus and then its expectation will be 0 due to given distribution$N(0,\theta^2)$ so no benefit. $\endgroup$
    – Onix
    Jan 26, 2016 at 18:34
  • $\begingroup$ If the "modulus" (absolute value) were removed from the expression, then the expectation of the estimator would obviously be $0$--and it would have a $1/2$ chance of being non-positive, no matter what value $k$ is. That doesn't seem like a very good estimator of the standard deviation, does it? $\endgroup$
    – whuber
    Jan 26, 2016 at 18:41
  • $\begingroup$ That's the same thing I was thinking about but this question is given in my practice questions booklet. And it got options should I add them $\endgroup$
    – Onix
    Jan 26, 2016 at 18:47
  • 1
    $\begingroup$ In contrast to your example, the question does not ask for an unbiased estimator of the variance of a Normal with unknown mean, it asks for an unbiased estimator of the standard deviation of a Normal with known mean. $\endgroup$ Jan 26, 2016 at 19:16

1 Answer 1

4
$\begingroup$

As JohnK suggested all you have to do is take an expectation. $|X_i|$ follows a "half normal" distribution (https://en.wikipedia.org/wiki/Half-normal_distribution) which has mean $\theta \sqrt{2 / \pi}$ and so

\begin{align} \text{E} \left ( k \sum_{i=1}^{3} |X_i| \right ) &= 3 k \text{E} \left ( |X_1| \right ) \\ &= 3 k \theta \sqrt{\frac{2}{\pi}} . \end{align}

For what value of $k$ does this equal $\theta$?

$\endgroup$
3
  • 1
    $\begingroup$ The expectation may also be computed with the LOTUS, avoiding any reference to the half-normal distribution. $\endgroup$
    – JohnK
    Jan 26, 2016 at 23:46
  • $\begingroup$ Oh man I don't understand how much I have to memorize :p thanks @dsaxton $\endgroup$
    – Onix
    Jan 27, 2016 at 15:56
  • $\begingroup$ Yeah found it using LOTUS don't mind it THANKS@JohnK $\endgroup$
    – Onix
    Jan 27, 2016 at 16:09

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.