# What is $E[1/\|x\|^4]$ where $x\sim$ Gaussian($0,C\cdot$diag($1,1/2,1/3,1/4,\ldots$))?

• Let $$\Sigma_d$$ be a diagonal matrix with diagonal $$1,1/2,1/3,\ldots,1/d$$
• Let $$E_d$$ denote expectation w.r.t normally distributed $$x$$ which is centered at 0 with covariance matrix $$\frac{\Sigma_d}{\operatorname{Tr}\Sigma_d}$$

What is the value of the following limit? $$\lim_{d\to \infty} E_d\left[\frac{1}{\|x\|^4}\right]$$

You can show that $$\lim_{d\to\infty} E_d[\|x\|^4]=1$$ by using Gaussian fourth-moment formula $$E[\|x\|^4]=(\operatorname{Tr}\Sigma)^2+2\operatorname{Tr}(\Sigma^2)$$

Numerical simulation suggests the limit in question might be equal to $$1$$, any tips?

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Closely related issue was discussed on stats.SE and on Mathoverflow in the last month. Question is still open.

Let $$\frac{\Sigma_d}{\operatorname{Tr}\Sigma_d} = \mathrm{diag}(a_1, a_2, \cdots, a_d).$$

By Jensen's inequality, we have $$\mathbb{E}\left[\frac{1}{\|x\|^4}\right] \ge \frac{1}{(\mathbb{E}[\|x\|^2])^2} = 1. \tag{1}$$

Using the known identity ($$q > 0$$) $$\frac{1}{q^2} = \int_0^\infty t\, \mathrm{e}^{-tq} \, \mathrm{d} t,$$ we have \begin{align*} \mathbb{E}\left[\frac{1}{\|x\|^4}\right] &= \mathbb{E}\left[ \int_0^\infty t\,\mathrm{e}^{-t\|x\|^2} \,\mathrm{d} t \right]\\ &= \int_0^\infty t\, \mathbb{E}[\mathrm{e}^{-t\|x\|^2}]\, \mathrm{d} t \\ &= \int_0^\infty t\, \prod_{i=1}^d \mathbb{E}[\mathrm{e}^{-tx_i^2}]\, \mathrm{d} t\\ &= \int_0^\infty t\, \prod_{i=1}^d \frac{1}{\sqrt{1 + 2a_i t}} \,\mathrm{d} t \\ &= \int_0^\infty t\, \mathrm{e}^{-\frac12\sum_{i=1}^d \ln (1 + 2a_i t)} \,\mathrm{d} t \tag{2} \end{align*} where we use $$\mathbb{E}[\mathrm{e}^{-tx_i^2}] = \int_{-\infty}^\infty \mathrm{e}^{-ty^2}\cdot \frac{1}{\sqrt{2\pi a_i}}\mathrm{e}^{-\frac{y^2}{2a_i}}\,\mathrm{d} y = \frac{1}{\sqrt{1 + 2a_i t}}$$.

Fact 1: $$\ln(1 + 2a_i t) \ge \frac{1}{i}\ln (1 + 2a_1 t)$$ for all $$i$$ and all $$t \ge 0$$.
(Proof: Note that $$a_i = a_1/i$$. Let $$f(t) = \ln(1 + 2a_i t) - \frac{1}{i}\ln (1 + 2a_1 t)$$. We have $$f'(t) = \frac{4a_1^2 t (i - 1)}{i(2a_1t + i)(2a_1 t + 1)} \ge 0$$. Also, $$f(0) = 0$$. The desired result follows.)

Denote $$H_d = \sum_{i=1}^d \frac{1}{i}$$. By Fact 1, we have $$-\frac12\sum_{i=1}^d \ln (1 + 2a_i t) \le -\frac12 \ln(1 + 2a_1 t) \cdot \sum_{i=1}^d \frac{1}{i} \le - \frac12 H_d\ln(1 + 2t/H_d). \tag{3}$$

From (3), we have $$\int_0^\infty t\, \mathrm{e}^{-\frac12\sum_{i=1}^d \ln (1 + 2a_i t)} \,\mathrm{d} t \le \int_0^\infty t\, \mathrm{e}^{- \frac12 H_d\ln(1 + 2t/H_d)} \,\mathrm{d} t. \tag{4}$$

From (1) and (2) and (4), we have $$1\le \mathbb{E}\left[\frac{1}{\|x\|^4}\right] \le \int_0^\infty t\, \mathrm{e}^{- \frac12 H_d\ln(1 + 2t/H_d)} \,\mathrm{d} t.$$

By the Dominated Convergence Theorem, we have $$\lim_{d\to \infty} \int_0^\infty t\, \mathrm{e}^{- \frac12 H_d\ln(1 + 2t/H_d)} \,\mathrm{d} t = 1.$$ (Note: Note that $$u \mapsto u \ln (1 + 2t/u)$$ is non-decreasing on $$u > 0$$. Let $$f_d(t) = t\, \mathrm{e}^{- \frac12 H_d\ln(1 + 2t/H_d)}$$. Then $$f_2(t) \ge f_3(t) \ge \cdots$$, and $$\lim_{d\to \infty} f_d(t) = t\mathrm{e}^{-t}$$, and $$\int_0^\infty f_m(t)\,\mathrm{d} t < \infty$$ for some $$m$$. See: Monotone Convergence theorem for decreasing sequence)

Thus, we have $$\lim_{d\to \infty} \mathbb{E}\left[\frac{1}{\|x\|^4}\right] = 1.$$

We are done.

• Thanks! The identity under "Using the identity (q>0)", is this a well-known identity? Does it have a name? Oct 14, 2022 at 17:11
• @YaroslavBulatov Not really : once you multiply both sides by $q$, you'll see that it's just the formula for the expectation of an "exponential" random variable with parameter $q$. Oct 14, 2022 at 17:21
• @YaroslavBulatov It is quite known. And it is easy to prove from RHS to LHS using $t = s/q$ and $\int_0^\infty s e^{-s} ds = 1$. Oct 15, 2022 at 0:46