Concentration bound on product of random Gaussian vector and random unit-sum vector

Question

Suppose that $X = (X_1, \dots, X_n) \in [0,1]^n$ is a random vector with $\Vert X \Vert_1 = 1$. Let $Q = (Q_1, \dots, Q_n)$ be a Gaussian random vector, where $Q_i$'s are i.i.d. from $N(0, \sigma^2)$. Suppose that $X$ and $Q$ are independent.

What can be said about the following probability ? $$\mathbb{P}\left(\left\vert \sum_{i=1}^n X_i Q_i \right\vert \ge t\right)$$

Answer 1

Denoting $\mathbb P_X$ the law of $X$, one has, by independence between $X$ and $Q$, that $$ \mathbb{P}\left(\left\vert \sum_{i=1}^n X_i Q_i \right\vert \geqslant t\right) =\int \mathbb{P}\left(\left\vert \sum_{i=1}^n x_i Q_i \right\vert \geqslant t\right)d\mathbb P_X(x_1,\dots,x_n). $$ For fixed $x_1,\dots,x_n$, the random variable $\sum_{i=1}^n x_i Q_i $ has a Gaussian distribution, with mean $0$ and variance $\sigma^2\sum_{i=1}^n x_i^2$. Consequently, denoting by $N$ a random variable having standard normal distribution, $$ \mathbb{P}\left(\left\vert \sum_{i=1}^n x_i Q_i \right\vert \ge t\right) =\mathbb P\left(\sigma\left(\sum_{i=1}^n x_i^2\right)^{1/2}N \geqslant t\right). $$ We need this inequality only for $(x_i)_{i=1}^n$ satisfying $0\leqslant x_i\leqslant 1$ and $\sum_{i=1}^n x_i=1$ hence for such $x_i$, $$ \mathbb{P}\left(\left\vert \sum_{i=1}^n x_i Q_i \right\vert \ge t\right) \leqslant \mathbb P(\sigma N\geqslant t), $$ for which classical bounds are available.

Answer 2

Cauchy-Schwarz inequality and the fact that $\|\cdot\|_2\le\|\cdot\|_1$ imply

$$|\langle X, Q\rangle|\le \|X\|_2\|Q\|_2\le \|X\|_1\|Q\|_2 =\|Q\|_2. $$

Therefore, since $\{|\langle X, Q\rangle|\ge t\}\subseteq\{\|Q\|_2\ge t\}$, this implies

$$\begin{align} \mathbb P\left(|\langle X, Q\rangle|\ge t\right)&\le \mathbb P\left(\|Q\|_2\ge t\right)\\ &=\mathbb P\left(\|Q\|_2^2\ge t^2\right)\\ &=\mathbb P\left(\sum_{i=1}^n(Q_i/\sigma)^2\ge (t/\sigma)^2\right)\\ &= 1 - F_{\chi_n^2}(t^2/\sigma^2)\\ &= 1 -\frac{1}{\Gamma(n/2)}\gamma\left(\frac n2,\frac{t^2}{2\sigma^2}\right)\tag1 \end{align}$$ Where $F_{\chi_n^2}$ denotes the CDF of the $\chi^2$ distribution with $n$ degrees of freedom.

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I believe bound $(1)$ is quite tight, but if you're not looking for something so precise, note that $(Q_i/\sigma)^2 $ is a sub-exponential random variable with parameter $(4,4)$ and so $\sum_{i=1}^n(Q_i/\sigma)^2$ is sub-exponential with parameter $(4n,4)$, and therefore it satifies the tail bound : $$\mathbb P\left(\sum_{i=1}^n(Q_i/\sigma)^2\ge (t/\sigma)^2\right) =\begin{cases}\exp\left(-\frac{(t/\sigma)^4}{8n}\right) &\text{ if } 0\le (t/\sigma)^2\le n\\ \exp\left(-\frac{(t/\sigma)^2}{8}\right) &\text{ if } (t/\sigma)^2> n \end{cases}\tag 2 $$ You can use these notes by Peter Bartlett as reference for the above claims.