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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 $\sqrt{\sum_{i=1}^3(X_i)^2}\text{with}\sum_{i=1}^3\left|X_i\right|$

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    $\begingroup$ Did you try taking the expectation of $|X|$? $\endgroup$
    – JohnK
    Commented 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
    Commented 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 ♦
    Commented 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
    Commented Jan 26, 2016 at 18:47
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    $\begingroup$ 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. $\endgroup$
    – Mark L. Stone
    Commented Jan 26, 2016 at 19:16

1 Answer 1

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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 $\theta$?

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

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