Let $\varphi_i$ be a Gaussian random variable such that $$\varphi_i \sim N(0,\sigma^2), \quad i = 1,2,\ldots,n.$$ What's the expectation: $$E\left(\left | \sum_{i=1}^n e^{j \varphi_i} \right |\right) $$ where $\left | \cdot \right |$ is the absolute value operation and $j = \sqrt{-1}$.
1
-
$\begingroup$ I don't know if there is a nice closed form for the absolute value but for the square of the absolute value the expectation is $\frac{n (n-1)}{\exp \left(\sigma ^2\right)}+n$. $\endgroup$– JimBJan 1, 2019 at 19:55
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
8
The expectation you are interested in is equivalent to
$$E\left[\sqrt{n+2\sum_{j=1}^{n-1}\sum_{k=j+1}^n (\cos \varphi_j \cos \varphi_k+\sin\varphi_j \sin\varphi_k)}\right]$$
I don't believe there is a nice closed-form solution for that expectation. So I think you'll need to perform simulations or numerically integrate to get solutions.
If you were interested in
$$E\left[n+2\sum_{j=1}^{n-1}\sum_{k=j+1}^n (\cos \varphi_j \cos \varphi_k+\sin\varphi_j \sin\varphi_k)\right]$$
then there is a closed-form solution:
$$n+n(n-1)e^{-\sigma^2}$$
-
$\begingroup$ hi, since X is always positive, do we have $E[\sqrt{X}] = \sqrt{ E[X] }$? If it holds, the first expectation can have a closed-form solution, right? $\endgroup$– VicNov 16, 2020 at 20:32
-
$\begingroup$ @Vic No. If that were the case, then by your argument you'd have $E[X^2]=E[X]^2$ which would imply that all variances are identically zero. $\endgroup$– JimBNov 16, 2020 at 21:09
-
$\begingroup$ that’s right. BTW, Jim, it seems that there is closed form solution for $$E\left(\left | \sum_{i=1}^n e^{j \varphi_i} \right |^{2}\right). $$ I am curious whether there is always closed-form solution or formula of expansion for $$E\left(\left | \sum_{i=1}^n e^{j \varphi_i} \right |^{2k}\right),$$ where $k$ is some integer. $\endgroup$– VicNov 16, 2020 at 22:36
-
-
$\begingroup$ Yes, that sounds a good idea. I have raised a question and considered a more general case. Jim, can you also have a look math.stackexchange.com/questions/3910426/… $\endgroup$– VicNov 16, 2020 at 23:12