# Expectation of squared elements of Gaussian vector multiplied by sum-of-elements

Let $$\mathbf{n}\sim\mathcal{CN}(\mathbf{0}_N,\nu^2\mathbf{I}_N)$$ be a circularly symmetric complex Gaussian vector of size $$N$$. Note that the entries $$n_i$$ are i.i.d. Let $$q=\sum_{i=1}^N n_i$$ and $$\mathbf{u}=\frac{\overline{q}}{|q|}\mathbf{n}=\frac{\sum_{i=1}^N \overline{n_i}}{\left|\sum_{i=1}^N n_i\right|}\mathbf{n}\,.$$ I'm interested in computing $$\mathbb{E}\left\{\sum_{i=1}^N u_i^2\right\}$$, but it could also be useful to have any pointers for higher powers $$r\geq2$$ as well.

My attempt so far: developing the sum, \begin{align} \sum_{i=1}^N u_i^2 &=\frac{1}{\left|\sum_{j=1}^N n_j\right|^2} \sum_{i=1}^N \left(\sum_{j=1}^N \overline{n_j}\right)^2n_i^2=\frac{1}{\left|\sum_{j=1}^N n_j\right|^2} \sum_{i=1}^N \sum_{j=1}^N\sum_{k=1}^N \overline{n_j}\overline{n_k}n_i^2\\ &=\frac{1}{\left|\sum_{j=1}^N n_j\right|^2}\sum_{i=1}^N |n_i|^4 +\frac{1}{\left|\sum_{j=1}^N n_j\right|^2} \sum_{i=1}^N\sum_{\substack{j=1\\j\neq i}}^N\overline{n_j}^2n_i^2 \\ &\qquad\qquad+ \frac{1}{\left|\sum_{j=1}^N n_j\right|^2}\sum_{i=1}^N\sum_{\substack{j=1\\j\neq i}}^N\sum_{\substack{k=1\\k\neq i,j}}^N\overline{n_j}\overline{n_k}n_i^2 \end{align}

and here I am stuck when trying to compute expectation, in particular because of the ratio with the squared magnitude of the sum.

Additionally, the second and third terms look straightforward at first glance due to the independence of the $$n_i$$'s (which lead to think that we can factor out expectations of each factor in the numerator), but I am unsure if this is the case involving the ratio over squared magnitude. I tend to think that the factors of the summands are no longer independent, and I don't see a way to work around it.

The answer is $$\nu^2$$.

Let $$n = x + \mathrm{i} y$$. The PDF of $$x_1$$ is given by $$f(x_1) = \frac{1}{\sqrt{\pi \nu^2}}\mathrm{e}^{- \frac{x_1^2}{\nu^2}}, \quad x_1 \in \mathbb{R}.$$

Let $$r := \sum_j x_j$$ and $$s := \sum_j y_j$$. By symmetry, we have \begin{align*} \mathbb{E}\left[\sum_j u_j^2\right] &= N \mathbb{E}[u_1^2]\\[6pt] &= N \mathbb{E}\left[\frac{(r - \mathrm{i} s)^2(x_1 + \mathrm{i} y_1)^2}{r^2 + s^2}\right]\\[6pt] &= \mathrm{i}N \mathbb{E}\left[\frac{2x_1y_1(r^2 - s^2)}{r^2 + s^2}\right] + \mathrm{i}N \mathbb{E}\left[\frac{2rs(y_1^2 - x_1^2)}{r^2 + s^2}\right]\\[6pt] &\qquad + N\mathbb{E}\left[\frac{r^2x_1^2 - r^2y_1^2}{r^2 + s^2}\right] + N\mathbb{E}\left[\frac{s^2y_1^2 - s^2x_1^2}{r^2 + s^2}\right] + N\mathbb{E}\left[\frac{4rsx_1y_1}{r^2 + s^2}\right]\\[6pt] &= 2N\mathbb{E}\left[\frac{r^2x_1^2 - r^2y_1^2}{r^2 + s^2}\right] + \frac{1}{N}\mathbb{E}\left[\frac{4r^2s^2}{r^2 + s^2}\right] \tag{1} \end{align*} where we use (by symmetry) \begin{align*} \mathbb{E}\left[\frac{2x_1y_1(r^2 - s^2)}{r^2 + s^2}\right] &= 0, \\[6pt] \mathbb{E}\left[\frac{2rs(y_1^2 - x_1^2)}{r^2 + s^2}\right] &= 0, \\[6pt] \mathbb{E}\left[\frac{r^2x_1^2 - r^2y_1^2}{r^2 + s^2}\right] &= \mathbb{E}\left[\frac{s^2y_1^2 - s^2x_1^2}{r^2 + s^2}\right],\\[6pt] \mathbb{E}\left[\frac{4rsx_1y_1}{r^2 + s^2}\right] &= \mathbb{E}\left[\frac{4rs\cdot \frac{rs}{N^2}}{r^2 + s^2}\right]. \end{align*} Explanation:
i) Since $$x_1, \cdots, x_N, y_1, \cdots, y_N$$ are i.i.d., we have (swap $$x$$ and $$y$$) $$\mathbb{E}\left[\frac{2x_1y_1(r^2 - s^2)}{r^2 + s^2}\right] = \mathbb{E}\left[\frac{2y_1x_1(s^2 - r^2)}{s^2 + r^2}\right] = - \mathbb{E}\left[\frac{2x_1y_1(r^2 - s^2)}{r^2 + s^2}\right]$$ which results in $$\mathbb{E}\left[\frac{2x_1y_1(r^2 - s^2)}{r^2 + s^2}\right] = 0$$. The second one and the third are similar.

ii) We have $$\mathbb{E}\left[\frac{4rsx_1y_1}{r^2 + s^2}\right] = \mathbb{E}\left[\frac{4rsx_iy_j}{r^2 + s^2}\right]$$ for all $$i, j$$. Thus, $$\mathbb{E}\left[\frac{4rsx_1y_1}{r^2 + s^2}\right] = \mathbb{E}\left[\frac{4rs\cdot \frac{1}{N^2}\sum_{i,j} x_i y_j}{r^2 + s^2}\right] = \mathbb{E}\left[\frac{4rs\cdot \frac{rs}{N^2}}{r^2 + s^2}\right]$$.

Using the identity (for $$p > 0$$) $$\frac{1}{p} = \int_0^\infty \mathrm{e}^{-t p}\,\mathrm{d} t,$$ we have \begin{align*} &2N\mathbb{E}\left[\frac{r^2x_1^2 - r^2y_1^2}{r^2 + s^2}\right] + \frac{1}{N}\mathbb{E}\left[\frac{4r^2s^2}{r^2 + s^2}\right]\\[6pt] ={}& 2N\mathbb{E}\left[(r^2x_1^2 - r^2y_1^2)\int_0^\infty \mathrm{e}^{- t(r^2 + s^2)}\,\mathrm{d} t\right] + \frac{1}{N}\mathbb{E}\left[4r^2s^2 \int_0^\infty \mathrm{e}^{-t(r^2 + s^2)}\,\mathrm{d} t\right]\\[6pt] ={}& 2N\int_0^\infty \mathbb{E}\left[(r^2x_1^2 - r^2y_1^2) \mathrm{e}^{- t(r^2 + s^2)}\right]\,\mathrm{d} t + \frac{1}{N}\int_0^\infty \mathbb{E}\left[4r^2s^2 \mathrm{e}^{-t(r^2 + s^2)}\right]\,\mathrm{d} t. \tag{2} \end{align*}

Let $$r_1 := x_2 + x_3 + \cdots + x_N$$ and $$s_1 := y_2 + y_3 + \cdots + y_N$$. Then $$r_1, s_1$$ are i.i.d. with $$r_1, s_1 \sim \mathcal{N}(0, (N-1)\nu^2/2)$$. Also, $$x_1, y_1, r_1, s_1$$ are independent. We have \begin{align*} &\mathbb{E}\left[(r^2x_1^2 - r^2y_1^2) \mathrm{e}^{- t(r^2 + s^2)}\right]\\[6pt] ={}& \int_{\mathbb{R}^4} \Big((x_1 + r_1)^2x_1^2 - (x_1 + r_1)^2y_1^2\Big) \mathrm{e}^{- t(x_1 + r_1)^2 - t (y_1 + s_1)^2}\frac{1}{\sqrt{\pi \nu^2}}\mathrm{e}^{- \frac{x_1^2}{\nu^2}} \frac{1}{\sqrt{\pi \nu^2}}\mathrm{e}^{- \frac{y_1^2}{\nu^2}}\\[6pt] &\qquad \cdot \frac{1}{\sqrt{\pi (N - 1) \nu^2}}\mathrm{e}^{- \frac{r_1^2}{(N - 1)\nu^2}}\frac{1}{\sqrt{\pi (N - 1) \nu^2}}\mathrm{e}^{- \frac{s_1^2}{(N - 1)\nu^2}} \,\mathrm{d}x_1 \,\mathrm{d} y_1\, \mathrm{d} r_1\, \mathrm{d}s_1\\[6pt] ={}& \frac{v^4}{2(N t\nu^2 + 1)^3}.\tag{3} \end{align*} Thus, we have \begin{align*} &2N\int_0^\infty \mathbb{E}\left[(r^2x_1^2 - r^2y_1^2) \mathrm{e}^{- t(r^2 + s^2)}\right]\,\mathrm{d} t\\[6pt] ={}& 2N\int_0^\infty \frac{v^4}{2(N t\nu^2 + 1)^3}\,\mathrm{d} t \\[6pt] ={}& \frac12 \nu^2. \tag{4} \end{align*}

Note that $$r, s$$ are i.i.d. with $$r, s \sim \mathcal{C}(0, N\nu^2/2)$$. We have \begin{align*} &\mathbb{E}\left[4r^2s^2 \mathrm{e}^{-t(r^2 + s^2)}\right]\\[6pt] ={}& \int_{\mathbb{R}^2} 4r^2s^2 \mathrm{e}^{-t(r^2 + s^2)}\frac{1}{\sqrt{\pi N \nu^2}}\mathrm{e}^{- \frac{r^2}{N\nu^2}}\frac{1}{\sqrt{\pi N \nu^2}}\mathrm{e}^{- \frac{s^2}{N\nu^2}}\,\mathrm{d}r \, \mathrm{d} s\\[6pt] ={}& \frac{N^2 v^4}{(Nt \nu^2 + 1)^3}. \tag{5} \end{align*} Thus, we have $$\frac{1}{N}\int_0^\infty \mathbb{E}\left[4r^2s^2 \mathrm{e}^{-t(r^2 + s^2)}\right]\,\mathrm{d} t = \frac{1}{N}\int_0^\infty \frac{N^2 v^4}{(Nt \nu^2 + 1)^3}\,\mathrm{d} t = \frac12 \nu^2. \tag{6}$$

From (1), (2), (4) and (6), we have $$\mathbb{E}\left[\sum_j u_j^2\right] = \nu^2.$$

We are done.

• Thanks! Very neat use of the integral identity there: I could never come up with that myself. Could you explain a little about the symmetries that you exploit for reducing expectations? I understood the one you commented about in your comment, but I still can't see the use of symmetry in the other expectation terms. Dec 17, 2023 at 21:21
• @cjferes I added some explanation. Dec 18, 2023 at 0:01
• Perfect. Thank you so much! Dec 18, 2023 at 4:58
• @cjferes You are welcome. Dec 18, 2023 at 5:49