# Upper bounds to high dimensional gaussian vectors

I'm trying to prove a claim given in Vershynin's Book - High-dimensional probability.

The notation used is $$g$$ as a vector, $$\sim$$ represents similarity. $$N(\mu, \sigma^2)$$ is a normal distribution with $$\mu$$ mean and $$\sigma^2$$ variance. $$t \in \mathbb{R}$$ and $$\mathbb{P}$$ is a notation for probability.

The author says:

Given $$g \sim N(0, I_{d \times d})$$ a standard Gaussian vector in d-dimensions then $$\begin{equation} \forall t \ge 0: \mathbb{P} \{ | \|g \|_2^2 - d | \ge t \} \le 2 \; \mathrm{exp}\left( - \frac{t^2}{C_1 d + C_2 t} \right) \end{equation}$$ for arbitrary $$C_1$$ and $$C_2$$.

Continuing the claim it extends to random matrices too. As stated as:

Given a random matrix $$\Gamma \in \mathbb{R}^{d \times k}$$ i.i.d. $$N(0,1)$$ For any $$v \in \mathbb{R}^k$$, and $$\| v\|_2 = 1$$ (normalized) $$\begin{equation} \forall t \ge 0: \mathbb{P}\{ | \| \Gamma v \|_2^2 - d | \ge t\} \le 2 \; \mathrm{exp}\left( - \frac{t^2}{C_1 d + C_2 t} \right) \end{equation}$$

I think if I can prove the first one the second part is simple since $$\Gamma v \sim N(0, I_{d \times d})$$ given that a centered normal is invariant by rotation.

• What section is this? – J.G Jan 26 at 19:30
• A combination of en.wikipedia.org/wiki/Chernoff_bound with the MGF of multivariate normal may do the trick en.wikipedia.org/wiki/Multivariate_normal_distribution. You will then need to optimize over the added parameter of the chernoff bound – P. Quinton Jan 26 at 19:32
• I think chapter 2. (or I saw in some class notes regarding Vershynin's book. I got this from personal anotations. – Lin Jan 26 at 19:58

Write $$\|g\|_2^2=\sum_{i=1}^d g_i^2$$ and note that the $$g_i$$ are i.i.d with expectation $$1$$.

Right tail

Chernoff's bound yields for $$s\geq 0$$

$$P\left(\sum_{i=1}^d [g_i^2-E(g_i^2)]\geq t\right)\leq e^{-st} E(e^{s(g_1^2-1)})^d$$

Note that $$\displaystyle E(e^{s(g_1^2-1)}) = \frac{e^{-s}}{\sqrt{2\pi}}\int e^{-x^2(\frac 12 -s)} dx$$ which is finite only if $$s<\frac 12$$, and in that case a substitution proves that $$\displaystyle E(e^{s(g_1^2-1)}) = \frac{e^{-s}}{\sqrt{1-2s}}$$ hence $$P\left(\sum_{i=1}^d [g_i^2-E(g_i^2)]\geq t\right)\leq e^{-st} \frac{e^{-sd}}{(1-2s)^{d/2}}$$ This is minimized in $$s$$ for $$s=\frac 12 \frac t{t+d}$$ thus $$P\left(\sum_{i=1}^d [g_i^2-E(g_i^2)]\geq t\right)\leq e^{-t/2}\left(1+\frac td \right)^{d/2}$$

This is not exactly the bound you're looking for, and it is tighter as $$t\to \infty$$ provided that $$C_2>2$$. To get the wanted bound, note that $$e^{-t/2}\left(1+\frac td \right)^{d/2} = \exp\left(-t^2\left[\frac 1{2t}-\frac{d}{2t^2}\log\left(1+\frac td\right) \right] \right)$$ and $$\frac 1{2t}-\frac{d}{2t^2}\log\left(1+\frac td\right) = \frac 1{2t}\left(1-\frac dt \log\left(1+\frac td\right)\right)$$

Numerically, I observe (and this inequality is referenced on Wikipedia) that $$\forall x\geq 0,\; 1-\frac 1x \log\left(1+ x\right)\geq \frac{1}{2\left(1+\frac 1x \right)}$$ hence $$\frac 1{2t}\left(1-\frac dt \log\left(1+\frac td\right)\right)\geq \frac{1}{4t\left(1+\frac dt \right)} = \frac{1}{4t+4d}$$ thus finally

$$P\left(\sum_{i=1}^d [g_i^2-E(g_i^2)]\geq t\right)\leq \exp\left(-\frac{t^2}{4t+4d} \right)$$

Left tail

Since $$P\left(\sum_{i=1}^d [g_i^2-E(g_i^2)]\leq -t\right) = P\left(\sum_{i=1}^d [-g_i^2+E(g_i^2)]\geq t\right)$$ and $$\displaystyle \sum_{i=1}^d [-g_i^2+E(g_i^2)] \leq d$$ we have $$t>d \implies P\left(\sum_{i=1}^d [g_i^2-E(g_i^2)]\leq -t\right) = 0$$ so it suffices to get an upper bound when $$t\leq d$$.

Chernoff's bound yields $$P\left(\sum_{i=1}^d [g_i^2-E(g_i^2)]\leq -t\right)\leq e^{t/2}\left(1-\frac td \right)^{d/2}$$ and numerically $$\forall x\in [0,1),\; 1+\frac 1x \log\left(1- x\right)\leq \frac{1}{2\left(1+\frac 1x \right)}$$ thus $$t\leq d \implies P\left(\sum_{i=1}^d [g_i^2-E(g_i^2)]\leq -t\right)\leq \exp\left(-\frac{t^2}{4t+4d} \right)$$

Finally, $$P\left(|\|g\|_2^2-d|\geq t\right)\leq 2\exp\left(-\frac{t^2}{4t+4d} \right)$$