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I'm interested in finding tail bound for $\sum_{i=1}^k X_i^2 Y_i^2$, where $X_i$ and $Y_i$ are independent standard normal random variables. It should be roughly as tight as the standard Chernoff bound, something like $e^{-\Omega(k\sqrt t)}$ would be nice.

My first instinct was to look at the mgf. $\exp(t X^2 Y^2)$, but naturally it doesn't exist. I looked at $\exp(i t X^2 Y^2) = e^{i/(8t)} K_0(i/(8t))/\sqrt{\pi i t}$, but I don't know how to get a tail bound using the characteristic function. I also considered moment bounds. We have $E(X^2 Y^2)^k=2^{2k}\Gamma(k+1/2)^2/\pi\le2(2k/e)^{2k}$, but to get a tail bound I need $E(\sum_i X_i^2 Y_i^2)^k$, which is of course a lot harder to estimate. I also considered the Cauchy Schwarz bound: $E(\sum_i X_i^2 Y_i^2)^k\le E(\sum_i X_i^4)^{k/2}(\sum_i Y_i^4)^{k/2}=E(\sum_i X_i^4)^{k}$, but even those moments seem pretty involved.

We have that $\Pr[\sum_{i=1}^kX_iY_i\ge tk]\le \exp\left(\frac{-t^2k}{2+t}\right)$ by Chernoff bounds, which is close to gaussian at least for small $t$. Perhaps we might also expect that $\sum X^2_iY^2_i$ is close to Chi-Squared for small $t$?

Does anyone know if there is a standard bound for this sum? Or if one of my approaches might be workable?

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  • $\begingroup$ Hm, I'm not sure you can do much better than the Markov inequality bound on high moments. $X$ is a sub-gaussian variable, so $X^2$ is a sub-exponential, which means $X^2Y^2$ is worse than that. As you said, the mgf doesn't exist so your best bet is to find bounds on high moments. It's probably unavoidable to expand things out and wade into the combinatorics of gaussian moments. $\endgroup$ – zoidberg Dec 12 '18 at 2:04
  • $\begingroup$ Actually, take a look at this: math.stackexchange.com/questions/1652781/… $\endgroup$ – zoidberg Dec 12 '18 at 2:13
  • $\begingroup$ So I think the bound you'll get will be $e^{-\Omega(\sqrt{t})}$ $\endgroup$ – zoidberg Dec 12 '18 at 2:56
  • $\begingroup$ I think you are right. It seems that $\frac{1}{b}\sum_{i=1}^kX_i^2Y_i^2>(\frac{1}{b}\sum_{i=1}^kX_i^2)^2$ close to 50% of the time. So the values seems very similar. Using the later value we get $\exp\left(\frac{-k}{4} \left(2 \sqrt{t}-2-\log (t)\right)\right)$, which would be great. $\endgroup$ – Thomas Ahle Dec 12 '18 at 15:56
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I found that a good way to handle this sum is using Bernstein's inequality:

Let $X_1, \ldots, X_n$ be independent zero-mean random variables. Suppose that $|X_i|\leq M$ almost surely, for all $i$. Then, for all positive $t$,

$$\Pr \left (\sum_{i=1}^n X_i \ge t \right ) \leq \exp \left( -\frac{t^2/2}{\sum \mathbb{E} X^2+ Mt/3} \right).$$

We have $$ \Pr[X^2Y^2\ge t] \le E[(XY)^{2p}]t^{-p} \le 2(2p/e)^{2p}t^{-p} \le 2 e^{-\sqrt t}, $$ taking $p=\sqrt{t}/2$. Hence we can take a union bound over all $XY$ to get: $$ \begin{align} \Pr\left( \sum_k X_k^2Y_k^2\ge (1+\epsilon)k \right) &\le \exp \left( -\frac{k\epsilon^2/2}{9+ M\epsilon/3} \right) +2k \exp(-\sqrt M)\le2\delta \end{align} $$ When taking $M=\log^2((2 k)/\delta)$ and $$\begin{align}t &= \frac{18 \log \left(\frac{1}{\delta}\right)}{\epsilon^2}+\frac{2 \log \left(\frac{1}{\delta}\right) \log^2 \left(\frac{2 k}{d}\right)}{3 \epsilon} =O(\epsilon^{-2}\log1/\delta+\epsilon^{-1}\log^32k/\delta). \end{align}$$

This matches exactly what we would expect from the central limit theory in the first term, and nearly bound from a "single large term" in the second term.

The only remaining question is whether we can get rid of the third power, to just have $\epsilon^{-1}\log^2(k/\delta)$ in the second term. I don't know how to do that though.

Update

The previous approach lost a factor $\log1/\delta$ in $t$. This can be avoided by the following neat trick: Instead of using Bernstein's inequality, just use $(XY)^2\le M^{1/2}|XY|\le \sqrt M (X^2+Y^2)/2$.

That gives us $$\begin{align} \Pr[\sum_iX_i^2Y_i^2 \ge (1+\epsilon)k \mid X^2Y^2\le M] &\le \exp\left(\lambda k X^2Y^2k\right)/\exp\left(\lambda(1+\epsilon)\right) \\&\le \exp\left(\lambda k \sqrt M (X^2+Y^2)/2\right)/\exp\left(\lambda(1+\epsilon)k\right) \\&= \frac{1}{1-\lambda k\sqrt M}/\exp\left(\lambda(1+\epsilon)k\right) \\&=\frac{(1+\epsilon)k}{\sqrt M}\exp(1 - \frac{(1+\epsilon)k}{\sqrt M}) \\&=\frac{(1+\epsilon)k}{\log(2k/\delta)}\exp(1 - \frac{(1+\epsilon)k}{\log(2k/\delta)}) %\\&=\sqrt{(1+\epsilon)k}\exp(1 - \sqrt{(1+\epsilon)k}). \end{align}$$

Taking $\lambda=((1+\epsilon)k-M)/(M (1+\epsilon)k)$ and $M=(1+\epsilon)k$. The union bound over from the single elements is then $+2k\exp(-\sqrt{(1+\epsilon)k})$.

We see that taking $k\approx\epsilon^{-2}\log1/\delta + \epsilon^{-1}(\log1/\delta)^2$ now suffices.

The method is nearly as versatile as the Bernstein approach, and it gives the optimal values.

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