# How do you calculate the expectation of $\left(\sum_{i=1}^n {X_i} \right)^2$?

If $X_i$ is exponentially distributed $(i=1,...,n)$ with parameter $\lambda$ and $X_i$'s are mutually independent, what is the expectation of

$$\left(\sum_{i=1}^n {X_i} \right)^2$$

in terms of $n$ and $\lambda$ and possibly other constants?

Note: This question has gotten a mathematical answer on http://math.stackexchange.com/q/12068/4051. The readers would take a look at it too.

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The two copies of this question reference each other and, appropriately, the stats site (here) has a statistical answer and the math site has a mathematical answer. It seems like a good division: let it stand! – whuber♦ Mar 4 '11 at 21:59

## 3 Answers

If $x_i \sim Exp(\lambda)$, then (under independence), $y = \sum x_i \sim Gamma(n, 1/\lambda)$, so $y$ is gamma distributed (see wikipedia). So, we just need $E[y^2]$. Since $Var[y] = E[y^2] - E[y]^2$, we know that $E[y^2] = Var[y] + E[y]^2$. Therefore, $E[y^2] = n/\lambda^2 + n^2/\lambda^2 = n(1+n)/\lambda^2$ (see wikipedia for the expectation and variance of the gamma distribution).

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+1 Nice answer! – whuber♦ Nov 27 '10 at 17:10
Thanks. A very neat way of answering the question (leading to the same answer) was also provided on math.stackexchange (link above in the question) a few minutes ago. – Wolfgang Nov 27 '10 at 17:24

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The math answer computes the integrals using linearity of expectation. In some ways it's simpler. But I like your solution because it exploits statistical knowledge: because you know a sum of independent Exponential variables has a Gamma distribution, you're done. – whuber♦ Nov 27 '10 at 21:19

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I enjoyed it quite a bit and I am by no means a statistician or a mathematician. – Kortuk Nov 29 '10 at 18:04
very elegant answer. – Cyrus S Nov 30 '10 at 16:44
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The answer above is very nice and completely answers the question but I will, instead, provide a general formula for the expected square of a sum and apply it to the specific example mentioned here.

For any set of constants $a_1, ..., a_n$ it is a fact that

$$\left( \sum_{i=1}^{n} a_i \right)^2 = \sum_{i=1}^{n} \sum_{j=1}^{n} a_{i} a_{j}$$

this is true by the Distributive property and becomes clear when you consider what you're doing when you calculate $(a_1 + ... + a_n) \cdot (a_1 + ... + a_n)$ by hand.

Therefore, for a sample of random variables $X_1, ..., X_n$, regardless of the distributions,

$$E \left( \left[ \sum_{i=1}^{n} X_i \right]^2 \right) = E \left( \sum_{i=1}^{n} \sum_{j=1}^{n} X_i X_j \right) = \sum_{i=1}^{n} \sum_{j=1}^{n} E(X_i X_j)$$

provided that these expectations exist.

In the example from the problem, $X_1, ..., X_n$ are iid ${\rm exponential}(\lambda)$ random variables, which tells us that $E(X_{i}) = 1/\lambda$ and ${\rm var}(X_i) = 1/\lambda^2$ for each $i$. By independence, for $i \neq j$, we have

$$E(X_i X_j) = E(X_i) \cdot E(X_j) = \frac{1}{\lambda^2}$$

There are $n^2 - n$ of these terms in the sum. When $i = j$, we have

$$E(X_i X_j) = E(X_{i}^{2}) = {\rm var}(X_{i}) + E(X_{i})^2 = \frac{2}{\lambda^2}$$

and there are $n$ of these term in the sum. Therefore, using the formula above,

$$E \left( \sum_{i=1}^{n} X_i \right)^2 = \sum_{i=1}^{n} \sum_{j=1}^{n} E(X_i X_j) = (n^2 - n)\cdot\frac{1}{\lambda^2} + n \cdot \frac{2}{\lambda^2} = \frac{n^2 + n}{\lambda^2}$$

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This problem is just a special case of the much more general problem of 'moments of moments' which are usually defined in terms of power sum notation. In particular, in power sum notation:

$$s_1 = \sum_{i=1}^{n} X_i$$

Then, irrespective of the distribution, the original poster seeks $E[s_1^2]$ (provided the moments exist). Since the expectations operator is just the 1st Raw Moment, the solution is given in the mathStatica software by:

[ The '___ToRaw' means that we want the solution presented in terms of raw moments of the population (rather than say central moments or cumulants). ]

Finally, if $X$ ~ Exponential($\lambda$) with pdf $f(x)$:

````f = Exp[-x/λ]/λ;      domain[f] = {x, 0, ∞} && {λ > 0};
````

then we can replace the moments $\mu_i$ in the general solution `sol` with the actual values for an Exponential random variable, like so:

All done.

P.S. The reason the other solutions posted here yield an answer with $\lambda^2$ in the denominator rather than the numerator is, of course, because they are using a different parameterisation of the Exponential distribution. Since the OP didn't state which version he was using, I decided to use the standard distribution theory textbook definition Johnson Kotz et al … just to balance things out :)

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