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I did some search here, but seems that no one asked about this.

I want to define a family of independent and identically distributed random variables, $x_1,...,x_n$, and then calculates the expected value of some expressions like $\sum_{i,j=1}^nx_ix_j$. The result would be some function depending on $n$. Is there a way to do this?

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up vote 3 down vote accepted

Here is a general solution for any distribution whose moments exist ...

Notation Define the power sum $s_r$:

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

The Problem

Let $\left(X_1,\ldots,X_n\right)$ denote $n$ iid random variables. This is the same problem as drawing a random sample of size $n$ from a population random variable $X$. The problem is to find:

$$E\Big(\sum_{i,j=1}^n X_i X_j\Big) = E\Big [\Big (\sum_{i=1}^n X_i\Big)^2\Big ] = E\Big [s_1^2\Big]$$

This is a problem known as finding moments of moments: they can be very difficult to solve by hand, but quite easy to solve with the help of a computer algebra system, for any arbitrary symmetric power sum. In this instance, we seek the expectation of $s_1^2$ ... i.e. the 1st raw moment of $s_1^2$ ... so the solution (expressed ToRaw moments of the population) is:

where RawMomentToRaw is a function from the mathStatica package for Mathematica, and where $\acute{\mu }_1$ and $\acute{\mu }_2$ denote the 1st and 2nd raw moments of random variable $X$, whatever its distribution (assuming they exist). 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:


Example 1: The Normal Distribution

If $X \sim N(\mu, \sigma^2)$, then: $$\acute{\mu }_1 = E[X] = \mu \quad \text{ and } \quad \acute{\mu }_2 = E[X^2] = \mu^2 + \sigma^2 $$

Substituting in $\acute{\mu }_1$ and $\acute{\mu }_2$ in Out[1] yields the solution: $$E\Big(\sum_{i,j=1}^n X_i X_j\Big) = n \left(n \mu ^2 + \sigma ^2\right)$$

Simple check: The Normal case with $n = 3$

In the case of $n = 3$, the joint pdf of $(X_1, X_2, X_3)$ is say $f(x_1, x_2, x_3)$:

The sum of products we are interested in is:

and the desired expectation is:

which matches perfectly the general $n$-Normal solution derived above, but with $n = 3$.

Example 2: The Uniform Distribution

If $X \sim Uniform(a,b)$ (as considered in both other answers), then: $$\acute{\mu }_1 = E[X] = \frac{a+b}{2} \quad \text{ and } \quad \acute{\mu }_2 = E[X^2] = \frac{1}{3} \left(a^2+a b+b^2\right)$$

Substituting in $\acute{\mu }_1$ and $\acute{\mu }_2$ in Out[1] yields the solution: $$E\Big(\sum_{i,j=1}^n X_i X_j\Big) =\frac{1}{3} n \left(a^2+a b+b^2\right)+\frac{1}{4} (n-1) n (a+b)^2$$

Again, this is different to the other answers posted - and much more complicated. Again, it is easy to perform a quick check:

Simple check: The Uniform case with $n = 3$

In the case of $n = 3$, the joint pdf of $(X_1, X_2, X_3)$ is say $g(x_1, x_2, x_3)$:

and the desired expectation is:

which matches perfectly our general $n$-Uniform solution derived above, with $n = 3$.

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Thanks for pointing out my mistake. – b.gatessucks Apr 26 '14 at 20:06
Thanks wolfies. I have to admit, I wasn't aware about what was the correct expectation and I just start to use Mathematica. That's why I couldn't know if the accepted answer was indeed correct. I just changed this, thank you very much for helping. – Integral Apr 26 '14 at 22:52
But still, you made manipulations with $n$ random variables to explain, but made the codes using only 3 random variables. I still don't know how to define $n$ random variables and calculates the expectation of some expression with them. Could you show the code to calculate $E (\sum_{i,j=1}^nX_iX_j)$ in the first example (and get as answer $n(n\mu^2+\sigma^2)$) ? Thanks. – Integral Apr 26 '14 at 23:12
I was looking for some symbolic calculations, in which $n$ is not assigned to a value. I want that Mathematica calculates some expectations in the same way you get the result $n(n\mu^2 + \sigma^2)$, doing only symbolic calculations. It is possible to get a lot of relevant results this way, and that is what I'm looking for. – Integral Apr 26 '14 at 23:58
You don't have to define $s_1$. You just type $s_1$, and the function knows what it is. The solution above is self-contained and complete. Out[1] is it. – wolfies Apr 27 '14 at 16:09

Here's an example, expected sum of n uniform random variates with specified minimums and maximums:

dt = TransformedDistribution[Sum[v1*v2, {x, 1, n}], 
       {v1 \[Distributed] UniformDistribution[{min1, max1}], 
        v2 \[Distributed] UniformDistribution[{min2, max2}]}];

exp=Expectation[x, x \[Distributed] dt]

(* 1/4 (max1+min1) (max2+min2) n *)

Check it with a simulation:

{min1, max1, min2, max2} = {1, 10, 20, 50};
v1sim = RandomVariate[UniformDistribution[{min1, max1}], {100000}];
v2sim = RandomVariate[UniformDistribution[{min2, max2}], {100000}];
v1sim*v2sim // Total // AccountingForm
exp /. {min1 -> 1, max1 -> 10, min2 -> 20, max2 -> 50, n -> 100000}




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Maybe I got this wrong, but looks to me that your n random variables are not i.i.d., they all depend on $v_1$ and $v_2$. – Integral Apr 26 '14 at 4:30
The names don't matter -- they are just symbols, and don't have any effect on the answer. It works just as well when written: dt = TransformedDistribution[Sum[x1*x2, {i, 1, n}], {x1 \[Distributed] UniformDistribution[{min1, max1}], x2 \[Distributed] UniformDistribution[{min2, max2}]}]; – m_goldberg Apr 26 '14 at 4:40
Im having trouble to define define $\sum_{i,j=1}^nx_ix_j$, in the case that the $x_i$ are $N(0,1)$. The expected value should be $n$ but Im always getting $0$. How do I exactly do that sum? – Integral Apr 26 '14 at 5:45
@Integral: If the N in your comment is the normal distribution, then that (0) is the expectation. – ciao Apr 26 '14 at 6:15
Im not sure about that, this sum will have squared terms, and $E(x_i^2) = 1$, because the variance of the $x_i$`s are 1. – Integral Apr 26 '14 at 6:17

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