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Let $a$ and $\phi$ be independent random variables, each with a known probability density function. Furthermore, in the sequences of observations $(a_{1},a_{2},...,a_{N})$ all $a_{i}$ are independent and identically distributed ($iid$). Similarly, in the sequence of observations $(\phi_{1},\phi_{2},...,\phi_{N})$ all $\phi_{i}$ are independent and identically distributed.

The goal is to calculate $E[P]$ where $P$ is given by:

$$P=\sum_{i=1}^{N}\sum_{j=1}^{N}a_{i}a_{j}\cos\left(\phi_{i}-\phi_{j}\right)$$

Here's what I came up with so far:

$$P = \sum_{i=1}^{N}a_{i}^{2} + \sum_{i=1}^{N}\sum_{j=1,j\neq i}^{N}a_{i}a_{j}\cos\left(\phi_{i}-\phi_{j}\right)$$

$$E[P]=\sum_{i=1}^{N} E[a_{i}^{2}] + \sum_{i=1}^{N}\sum_{j=1,j\neq i}^{N}E[a_{i}a_{j}\times \cos(\phi_{i}-\phi_{j})]$$

because $a$ and $\phi$ are independent RVs we can write:

$$E[P] = \sum_{i=1}^{N} E[a_{i}^{2}] + \sum_{i=1}^{N}\sum_{j=1,j\neq i}^{N}E[a_{i}a_{j}] E[\cos(\phi_{i}-\phi_{j})]$$

because the $a$ sequence is $iid$ we can write:

$$E[P] = \sum_{i=1}^{N} E[a_{i}^{2}] + \sum_{i=1}^{N}\sum_{j=1,j\neq i}^{N}E[a_{i}]E[a_{j}] E[\cos(\phi_{i}-\phi_{j})]$$

and finally: $$E[P] = N E[a^{2}] + N(N-1) E[a]^{2} E[\cos(\Delta\phi)]$$

where, since the $\phi$ sequence is $iid$, we have $E[\cos(\Delta\phi)] = \int\int p_{\phi}^{2}(\phi)\cos(\phi_{i}-\phi_{j})d\phi_{i}d\phi_{j}$.

Does this make sense?

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    $\begingroup$ Evidently $a=(a_1,a_2,\ldots, a_N)$ and $\phi=(\phi_1,\phi_2,\ldots, \phi_n)$ are vector-valued random variables. What can we assume about their joint distribution? Given you haven't said anything about their individual distributions, what kind of answer are you hoping for that would consist of anything more than "apply the definition"? $\endgroup$
    – whuber ♦
    Commented Dec 5, 2017 at 16:57
  • $\begingroup$ It is amazing that the only letter you could think of to denote your expression is $P$ which is so often used for probability when discussing random variables and the like. $\endgroup$
    – Dilip Sarwate
    Commented Dec 5, 2017 at 16:57
  • $\begingroup$ The new approach is correct up until the very end, where the summations disappear but somehow its indices remain attached to the phi's, just as the Cheshire Cat's smile magically remained after the cat itself vanished. $\endgroup$
    – whuber ♦
    Commented Dec 8, 2017 at 15:26
  • $\begingroup$ Yep, I agree, thanks. How about now? $\endgroup$
    – unkown
    Commented Dec 8, 2017 at 21:13
  • $\begingroup$ It still makes no sense, because you still have "$i$" and "$j$" in the answer. Set $N=2$ and write out the double sum to see what the problem is and how to resolve it. $\endgroup$
    – whuber ♦
    Commented Dec 8, 2017 at 22:16

2 Answers 2

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$$ E(P) = \sum_{i=1}^{N} \sum_{j=1}^{N} E(a_{i}a_{j}\cos\left(\phi_{i}-\phi_{j}\right) ) = \sum_{i \neq j} E(a_{i}a_{j})E(\cos\left(\phi_{i}-\phi_{j})\right) + \sum_{i=j} {\rm var}(a_i)$$

$$ E(a_i a_j) = \int_{A_{ij}} xy p(x,y) $$

$$ E(a_j) = \int_{A_j} x p(x) $$

$$ {\rm var}(a_j) = \int_{A_j} (x - E(a_j))^2 p(x) $$

$$ E(\cos\left(\phi_{i}-\phi_{j})\right) = \int_{\Phi} \cos( \phi_i - \phi_j) p(\phi_i, \phi_j)$$

Where $p$ generically refers to the density of the given variable, and $\Phi, A_{ij}, A_j$ refer to the relevant domains.

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  • $\begingroup$ This solution implicitly assumes all the components of $a$ have no variance! After all, among other things, for each $i$ it is replacing $E(a_i^2) = E(a_ia_i)$ by $E(a_i)E(a_i)=E(a_i)^2$, implying $\operatorname{Var}(a_i) = E(a_i^2) - E(a_i)^2 = 0.$ $\endgroup$
    – whuber ♦
    Commented Dec 5, 2017 at 17:03
  • $\begingroup$ Forgot about the part where $i=j$. $\endgroup$
    – within_person
    Commented Dec 5, 2017 at 17:11
  • $\begingroup$ OK--but please don't forget to mention the crucial assumption you have made; namely, that the $a_i$ are uncorrelated. Since you distinguish $A_i$ and $A_j$, you imply the $a_i$ don't even have common distributions, so it is important to be explicit about what you're assuming and what you're not. Your notation, too, is heavily overloaded: you use "$p$" for all density functions. That's potentially confusing. $\endgroup$
    – whuber ♦
    Commented Dec 5, 2017 at 17:45
  • $\begingroup$ Thanks a lot for you help. Still, I'm a bit confused. Perhaps my question wasn't clear. Let me rephrase it: how can one calculate $E\left[\sum_{i=1}^{N}\sum_{j=1}^{N}a_{i}a_{j}\cos(\phi_{i}-\phi_{j})\right]$ given that $E[a]$, $var[a]$, $E[\phi]$ and $var[\phi]$ are know and $a$ is independent of $\phi$? $\endgroup$
    – unkown
    Commented Dec 6, 2017 at 11:14
  • $\begingroup$ What you ask isn't generally possible. In order to make any progress you need to know $\operatorname{Cov}(a)$ and you need to know much more about the distribution of $\phi$ itself. When $a$ is independent, this answer outlines an approach. $\endgroup$
    – whuber ♦
    Commented Dec 6, 2017 at 14:42
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Ok, here's what I came up with so far:

$$P=\sum_{i=1}^{N}\sum_{j=1}^{N}a_{i}a_{j}\cos\left(\phi_{i}-\phi_{j}\right) = \sum_{i=1}^{N}a_{i}^{2} + \sum_{i=1}^{N}\sum_{j=1,j\neq i}^{N}a_{i}a_{j}\cos\left(\phi_{i}-\phi_{j}\right)$$

$$E[P]=N E[a^{2}] + \sum_{i=1}^{N}\sum_{j=1,j\neq i}^{N}E[a_{i}a_{j}\times \cos(\phi_{i}-\phi_{j})]$$

because $a$ and $\phi$ are independent RVs we can write:

$$E[P] = N E[a^{2}] + \sum_{i=1}^{N}\sum_{j=1,j\neq i}^{N}E[a_{i}a_{j}] E[\cos(\phi_{i}-\phi_{j})]$$

because the $a$ sequence is $iid$ we can write:

$$E[P] = N E[a^{2}] + \sum_{i=1}^{N}\sum_{j=1,j\neq i}^{N}E[a]E[a] E[\cos(\phi_{i}-\phi_{j})]$$

and finally: $$E[P] = N E[a^{2}] + N(N-1) E[a]^{2} E[\cos(\phi_{i}-\phi_{j})]$$

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    $\begingroup$ I have spotted a mistake: assuming $a_i$ and $a_j$ are both r.v. (having the same marginal distribution as a generic $a$ - same holds for $\phi_i$), your second line must read $$E(P) = NE(a^2) + (N-1)\sum_{i=1}^N E(a_ia_j)E(cos(\phi_i-\phi)),$$ where I've already made use of the independence of the $a_i$ with the $\phi_i$. otherwise we are not provided with enough information to give an explicit answer. $\endgroup$
    – BloXX
    Commented Dec 6, 2017 at 13:24
  • $\begingroup$ Thank you for your comment. I'm sorry to say that I'm not convinced. Would it be more evident if, rather than writing $\sum_{i=1}^{N}\sum_{j=1,j\neq i}^{N}a_{i}a_{j}\cos(\phi_{i}-\phi_{j})$, I would write $\sum_{i=1}^{N}a_{i}\sum_{j=1,j\neq i}^{N}a_{j}\cos(\phi_{i}-\phi_{j})$? $\endgroup$
    – unkown
    Commented Dec 6, 2017 at 13:30
  • $\begingroup$ $a_i$ will still be a r.v. as $a_j$ is. in the second expression of the right hand side of your solution steps you handled $a_j$ as a r.v. but $a_i$ like a fixed number. besides that i missed the second sum in my comment :) $\endgroup$
    – BloXX
    Commented Dec 6, 2017 at 13:38
  • $\begingroup$ Thanks a lot for taking the time to review this. I appreciate your help. I rewrote my first expression to try and make my point more clear, in my humble opinion, of course. Does it make sense? $\endgroup$
    – unkown
    Commented Dec 6, 2017 at 13:50

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