Let $Y$ be a $n$-dimensional Gaussian random vector with expectation $0$ and covariance matrix $\Sigma$ (which explicitely is not diagonal). Let $\gamma$ be a vector in $\mathbb{R}^n$, let $\| \cdot \|$ denote the Euclidean norm in $\mathbb{R}^n$ and let $\langle \cdot, \cdot \rangle$ denote the Euclidean scalar product. My question is if it is possible to find a vector $\gamma$ such that it holds that $$ \mathbb{E} \left[ \frac{| \langle \gamma, Y \rangle |}{\| \gamma\| \| Y \| } \right] \leq \frac{1}{n}$$ or if this inequality is false for every $\gamma$. Obviously, Cauchy-Schwarz gives the upper bound $1$ for any $\gamma$, but I would be content with finding just a single $\gamma$ that fulfills the stricter bound.


What you want to show is not true. Either the integrand has to be squared or the right-hand side needs to be $\frac 1 {\sqrt n}$.

To see this, choose a basis in which $\Sigma$ is diagonal and order its eigenvalues: $0\le\lambda_1\le\dots\le \lambda_n$. Let $\gamma=(1,0,\dots,0)$ (in this basis). Then for standard independent normals $N_i$ we have: $$E_n^2=E\left(\frac{| \langle \gamma, Y \rangle |}{\| \gamma\| \| Y \| }\right)^2\le E\left(\frac{| \langle \gamma, Y \rangle |^2}{\| \gamma\|^2 \| Y \|^2 }\right) =E\left(\frac {{\lambda_1}N_1^2} {\sum_{i=1}^{n}\lambda_iN_i^2}\right)\le E\left(\frac {N_1^2}{\sum_{i=1}^n N_i^2}\right)=\frac 1 n$$ The last equality is achieved iff all eigenvalues are equal. However the first inequality is not sharp if $n>1$. Clearly it's the worst if all eigenvalues are equal so assume that. When $n=2$ we get:

$$\sqrt 2E_2=\sqrt 2 E\left(\frac {|N_1|}{\sqrt {N_1^2+N_2^2}}\right)=\sqrt 2E\left(\frac 1 {\sqrt {1+C^2}}\right)=\frac {2\sqrt 2} \pi\approx {0.90032}$$ where $C=\frac {N_2}{N_1}$ is Cauchy - already better than $1$. Asymptotically by SLLN we get: $$\sqrt n E_n=E\left(\frac {\frac{\sum_1^n |N_i|}n}{\sqrt{\frac{\sum_1^n N_i^2}n}}\right)\to \frac {E(|N_1|)}{\sqrt{E(N_1^2)}}=\sqrt{\frac 2 \pi}\approx 0.79788$$ which is better than the original simple bound of $1$.

  • $\begingroup$ Thanks for your answer! Even if my original conjecture is wrong, this version is also very useful. Could you elaborate on the final step, however, i. e. how to obtain the bound $\frac{1}{n}$? If I use $\lambda_1 \leq \lambda_i$ for $i \neq 1$ in the denominator, I'm left with a sum of standard normals, which is a chi squared distribution - same in the numerator, but the two are dependent - is this even the correct way to go and what am I missing to obtain the bound? $\endgroup$ – bhz Nov 18 '15 at 10:32
  • $\begingroup$ @bhz $E(\frac {N_1^2}{\sum_1^n N_i^2})=\frac 1 n$ by symmetry. $\endgroup$ – A.S. Nov 18 '15 at 10:38
  • $\begingroup$ Talk about thinking too complicated - I got it now. Thanks again. $\endgroup$ – bhz Nov 18 '15 at 13:23
  • $\begingroup$ @bhz Good. I also update the answer with some improved asymptotic estimate for the quantity you actually asked about. There might be some analytical way to track $E_n$ - either through chi-square distribution or transforming into Cauchy (I think they'll be independent), but I don't immediately see simplifications. $\endgroup$ – A.S. Nov 18 '15 at 13:45
  • $\begingroup$ Those Cauchys certainly wouldn't be independent. $\endgroup$ – A.S. Nov 18 '15 at 14:03

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