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I am required to computed following Expectations : $E[\displaystyle\sum\limits_{i=1}^N \frac{X_{i}}{N}] = \bar{x}_{0}$ & $E[\displaystyle(\sum\limits_{i=1}^N \frac{X_{i}}{N})^2] = \frac{(N\bar{x}_{0}^2 + \sigma_{x}^2)}{N}$ with $ X_{i} \sim N(\bar{x}_{0},\sigma_{x}^2) \forall i = 1,2,\cdots,N$ & $X_{i} \bot X_{j} \forall i\not=j$ I'm looking to solve these expectation in symbolic as all these variable are unknown. Thanks for your help

Alexandre

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  • $\begingroup$ Looks more like a question for math.SE right now. Do you have a Mathematica angle here? $\endgroup$
    – Yves Klett
    Commented Jun 11, 2014 at 20:54
  • $\begingroup$ Hi, sorry I've should have precised that I know the answer. This question is related to a larger system that require this computation to work $\endgroup$ Commented Jun 11, 2014 at 21:30
  • $\begingroup$ @Alex - If you know the answer: Why don't you just post it (preferably in Mathematica InputForm) ??? $\endgroup$
    – eldo
    Commented Jun 11, 2014 at 21:58
  • $\begingroup$ @eldo I know how to calculate this expectation by hand but my goal is to have it done by Mathematica but I can't manage to make it work. I've tried meanH = Sum[Subscript[h, i], {i, 1, a}]/a; Expectation[meanH*meanH, Subscript[h, i] [Distributed] NormalDistribution[\!(TraditionalForm` *SubscriptBox[ OverscriptBox[(H), (_)], (0)]), Subscript[[Sigma], h]]]; out : Subscript[[Sigma], h]^2 + Subscript[OverBar[H], 0]^2 But this is not the right answer $\endgroup$ Commented Jun 12, 2014 at 1:54
  • $\begingroup$ @Alex Can you please clarify 3 aspects of your notation. FIRST, you have capital $N$ and lower case $n$ ... I presume that both of these should be the same $n$. SECOND, you specify that $X_i \sim N(\mu, \sigma)$. I presume you intend: $N(\mu, \sigma^2)$ i.e. the variance is $\sigma^2$. THIRD, you specify $i = 1, \dots,n$, but your sums run from 0 to $n$. Presumably you intend them to run from 1 to $n$. $\endgroup$
    – wolfies
    Commented Jun 12, 2014 at 8:06

1 Answer 1

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This is a problem known as finding 'moments of moments'.

Notation Define the power sum $s_r$:

$$s_r=\sum _{i=1}^n X_i^r$$

Your problem only involves $s_1$.

The Problem

Let $\left(X_1,\ldots,X_n\right)$ denote a random sample of size $n$ from a population random variable $X$.

The problem is to find:

$$ E\Big [\Big (\frac1n\sum_{i=1}^n X_i\Big)^2\Big ] = E\Big [\big(\frac{s_1}{n}\big)^2\Big]$$

i.e. we seek the expectation of $\big(\frac{s_1}{n}\big)^2$ ... i.e. the 1st Raw Moment of $\big(\frac{s_1}{n}\big)^2$ ... so the solution (expressed ToCentral moments of the population) is:

enter image description here

where:

  • RawMomentToCentral is a function from the mathStatica package for Mathematica,

  • $\acute{\mu}_1$ denote the 1st raw moment of random variable $X$ (i.e. the mean of $X$) and

  • $\mu_2$ denotes the 2nd central moment of random variable $X$ (i.e. the variance of $X$).

In your case, $X \sim N(\bar{x}_{0}, \sigma^2)$, so $\acute{\mu}_1 = \bar{x}_{0}$ and $\mu_2 = \sigma^2$. Substituting these values into Out[1]= yields: $$\frac{\sigma^2}{n} + \bar{x}_{0}^2 \quad \quad \text{(as required)}$$ All done.

More detail

There is an extensive discussion of moments of moments in Chapter 7 of our book:

  • Rose and Smith, "Mathematical Statistics with Mathematica", Springer, NY

A free download of the chapter is available here:

http://www.mathstatica.com/book/Rose_and_Smith_2002edition_Chapter7.pdf

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