Let $X_1, X_2, \dots, X_n$ i.i.d. $N(0,\sigma^2)$ and let $Y_1, Y_2, \dots, Y_n$ be independent and identical Bernoulli random variables (where $Y_i$ may depend on $X_i$).

I am searching for a tailbound / concentration inequality of the form $$P(|\frac{1}{n}\sum_{i=1}^n X_iY_i - \mathbb{E}(X_iY_i)|>z)\leq 2\exp(-c z^2)$$ for some specific value of $c$ (which should obviously depend on the variance of $X_iY_i$).

If the tailbound holds I am very interested in a concrete (and "sharp") value of $c$!).

Possible way to go:

For me it looks like as if $X_iY_i-\mathbb{E}(X_iY_i)$ will still be sub-Gaussian (with which parameter!?). If this could be checked, then one would have to apply a Hoeffding bound to bound the sum of independent subgaussian random variables (which are sub-Gaussian again) and would be done.

However I am having trouble showing that $X_iY_i-\mathbb{E}(X_iY_i)$ is sub Gaussian and finding the correct parameter.

Any help is greatly appreciated.

Edit: as Whuber pointed out with reference to wikipedia it is easy to see that $Z_i:=X_iY_i-\mathbb{E}(X_iY_i)$ is sub-Gaussian by checking the $\Psi_2$ condition. (Here done in more detail than needed, see Whuber's comment: it would have been enough to check the condition directly for $X_iY_i$)

Indeed: since $Z^2 \leq 2(X_iY_i)^2 + 2 \mathbb{E}(X_iY_i)^2$ and $|E(X_iY_i)|<d$ for some $d>0$ we have for all $a>0:$ \begin{align*} \mathbb{E}(\exp(aZ^2)) & \leq \mathbb{E}(\exp(2a (X_iY_i)^2 + 2a\mathbb{E}(X_iY_i)^2))\\ & =\exp(2a\mathbb{E}(X_iY_i)^2)\mathbb{E}(\exp(2a (X_iY_i)^2))\\ & \leq \exp(2ad^2) \mathbb{E}(\exp(2a (X_iY_i)^2))\\ & \leq \exp(2ad^2) \mathbb{E}(\exp(2a X_i^2)) < \infty, \end{align*} since $X_i$ itself is sub-Gaussian and hence follows the $\Psi_2$ condition. Hence $Z_i:=X_iY_i -\mathbb{E}(X_iY_i)$ is sub-Gaussian with some parameter $b$ and $\sum_{i=1}^n Z_i$ is sub-Gaussian with parameter $nb$

However: I am still in need of a concrete value of the constant $c$ (or equivalently: $b$) (as sharp as possible)

The true dependency between $X_i$ and $Y_i$ is too complicated to be given here, hence a more general bound on $c$ would be sufficient (which I feel should be possible, since |Y_i| is bounded by 1). However, if it is of any help/as a starter one could think that for each $i$, the relationship of $(X_i,Y_i)$ could be described by a logistic regression model.

  • $\begingroup$ Since $|X_iY_i|\le |X_i|$ and $|\mathbb{E}(X_iY_i)|\le |\sigma|$, isn't the sub-Gaussianity immediate? And when it comes down to finding a "sharp" value of $c$, isn't it equally evident its value must depend on precisely how each $Y_i$ depends on $X_i$? So, aside from these trivialities, there doesn't seem to be much more one could say without more specific information about the $Y_i$. $\endgroup$
    – whuber
    Jul 25, 2017 at 15:43
  • $\begingroup$ While I suspect too that the parameter of the sub-Gaussian of $Z:= X_iY_i-\mathbb{E}(X_iY_i)$ is the same as of $X_i$ (i.e. $\sigma$), I don't think it is enough to bound only the first moment of $X_iY_i$ to check for the sub-Gaussian. I think one must at least bound all absolute moments of $Z$? But by doing this I get $$\mathbb{E}(|Z|^p)\leq 2^p E(|X_iY_i|^p) \leq 2^p \mathbb{E}(|X_i|^p)$$ which will at least induce an additional factor of $2$. It's not what I expected + I don't know if the moments are sufficient. $(X_i,Y_i)$ could for example be related through a logistic regression model. $\endgroup$
    – chRrr
    Jul 25, 2017 at 16:21
  • $\begingroup$ The bounds on the moments are irrelevant: for any constant $\mu$, $Z$ and $Z-\mu$ are either both subgaussian or both not. Provided, then, that $Z$ has finite expectation, it makes no difference whether you study $Z$ or $Z-\mathbb{E}(Z)$. $\endgroup$
    – whuber
    Jul 25, 2017 at 16:26
  • 1
    $\begingroup$ See en.wikipedia.org/wiki/Sub-Gaussian_distribution. One easy characterization of a subgaussian variable $Z$ is that for all $a$, $\mathbb{E}(e^{aZ^2})\lt \infty$. Obviously the finiteness of this expectation is unchanged when $Z$ is shifted to $Z-\mu$, QED. $\endgroup$
    – whuber
    Jul 25, 2017 at 16:34
  • 1
    $\begingroup$ The strict inequalities in probabilities make no difference at all, because my examples can be approached arbitrarily closely while still respecting them. The point is that the $Y_i$ can suppress the chances of large $|X_iY_i|\in\{|X_i|, 0\}$ appearing or they can allow large $|X_iY_i|$ to appear at essentially the same rate they would in the underlying Normal distribution for $X_i$--and any behavior in between is possible. To make any progress, you must supply information about the rates at which the $Y_i$ reduce the chances of large $|X_iY_i|$ appearing. $\endgroup$
    – whuber
    Jul 25, 2017 at 18:10

2 Answers 2


For explicit absolute constants in the bound, you can use symmetrization and contraction. Indeed, if $\epsilon_1,...,\epsilon_n$ are independent random variables taking values $\pm1$ with probability $1/2$ (and independent of the $X_i$ and the $Y_i$, then by symmetrization (noting that $t\to\exp(\lambda t)$ is convex) we get

$E[ \exp(\lambda \sum_{i=1}^n (X_iY_i - E[X_iY_i]) \le E\exp(2\lambda \sum_{i=1}^n \epsilon_i X_i Y_i).$

Next, by contraction you get rid of the Bernoulli (even if $X_i,Y_i$ are dependent):

$E\exp(2\lambda \sum_{i=1}^n \epsilon_i X_i Y_i) \le E\exp(2\lambda \sum_{i=1}^n X_i)$.

But now $\sum_{i=1}^n X_i$ is $N(0, n\sigma^2)$ so the right hand side equals $\exp(2 \lambda^2 n \sigma^2)$. This implies by Markov inequality with $\lambda =t/(2\sigma\sqrt n)$

\[ P\left( \frac{1}{\sqrt n}\sum_{i=1}^n (X_i Y_i-E[X_i Y_i]) > t \sigma \right) \le e^{-\lambda t \sigma \sqrt n + 2n\lambda^2\sigma^2 } = e^{-t^2/4}. \] We may use the same argument for the other direction.

What is symmetrization?

If $\epsilon_1,...,\epsilon_n$ are iid $\pm 1$ as above, independent of $Z_1,...,Z_n$ which are iid with $E[Z_i]=0$, then for any convex positive function $F$,

$E[ F(\sum_{i=1}^nZ_i) ]\le E[ F(\sum_{i=1}^n\epsilon_i Z_i) ]$.

Why is this true? of $Z_1',...,Z_n'$ are independent copies of $Z_1,...,Z_n$ then by Jensen's inequality \begin{align} E[ F(\sum_{i=1}^nZ_i) ] & \le E[ F(\sum_{i=1}^n (Z_i-Z_i') ] \\ &= E[ F(\sum_{i=1}^n \epsilon_i(Z_i-Z_i') ] \\ &\le \frac 1 2 E[ F(2\sum_{i=1}^n \epsilon_iZ_i) ] + \frac 1 2 E[ F(2\sum_{i=1}^n -\epsilon_i Z_i') ] \\ &= E[ F(2\sum_{i=1}^n \epsilon_i Z_i) ]. \end{align} Here, the first inequality is Jensen's and the $Z_i'$ have mean zero, the first equality is because $\epsilon_i(Z_i-Z_i')$ has the same distribution as $Z_i-Z_i'$ (because its distribution is symmetric), the second inequality is Jensen's again, and the last equality is because $\epsilon_iZ_i$ and $-\epsilon_iZ_i'$ are equal in distribution.

What is contraction?

If you have deterministic $x_1,...,x_n$ and random signs $\epsilon_1,...,\epsilon_n$ as above, then \[ \sup_{\boldsymbol y\in [-1,1]^n} E_\epsilon [F(\sum_{i=1}^n \epsilon_i y_i x_i) = E_\epsilon[F(\sum_{i=1}^n \epsilon_i x_i). \] (Here the only random variables are the $\epsilon_i$s). Why is this true? Because the function $G:\boldsymbol y \to E[F(\sum_{i=1}^n \epsilon_i y_i x_i)$ from $[-1,1]^n\to R$ is convex, and convex functions over polytopes such as $[-1,1]^n$ attains their maximum at a vertex. Here, the vertices of $[-1,1]^n$ are of the form $y_i=\pm 1$ for each $i=1,...,n$. But for any $\boldsymbol y$ of such form, $G(\boldsymbol y) = E_\epsilon[F(\sum_{i=1}^n \epsilon_i x_i)$ because $(\epsilon_1,...,\epsilon_n)$ has the same distribution as $(y_1\epsilon_1,...,y_n\epsilon_n)$ when $y_i=\pm 1$.

Contraction for random $X_1,...,X_n$ and $Y_1,...,Y_n$...

Now if $X_1,...,X_n$ and $Y_1,...,Y_n$ are random with $Y_i\in[-1,1]$ (as in your example), then by conditioning on the $X_i,Y_i$'s, \begin{align} E[ F( 2 \sum_{i=1}^n \epsilon_i X_i Y_i) ] & = E\Big[ E[ F( 2 \sum_{i=1}^n \epsilon_i X_i Y_i) \Big|X_1,...,X_n,Y_1,...,Y_n] \Big] \\ &\le E\Big[ \max_{\boldsymbol y\in[-1,1]^n} E[ F( 2 \sum_{i=1}^n \epsilon_i X_i y_i) \Big|X_1,...,X_n] \Big] \\ & = E\Big[ E[ F( 2 \sum_{i=1}^n \epsilon_i X_i) \Big|X_1,...,X_n] \Big] = E[ F(2 \sum_{i=1}^n \epsilon_i X_i ) ]. \end{align}


I think I got one answer. It doubt it to be the sharpest bound, but at least it is a usable bound.

I am using Theorem 2.1 from this rough draft:

It is easy to see that (III) in Theorem 2.1 holds: $$\mathbb{E}((X_iY_i-\mathbb{E}(X_iY_i))^{2k})\leq \frac{(2k)!}{2^k k!} (2\sigma)^{2k}=\frac{(2k)!}{2^k k!} \theta^{2k},$$ where $\theta = 2\sigma$ From the proof $ (III) \rightarrow (I)$ given in Appendix A of the rough draft (p.39) we then may immediately conclude that for all $\lambda>0$ we have

$$\mathbb{E}(\exp(\lambda (X_iY_i-\mathbb{E}(X_iY_i))) \leq \exp(\frac{(\lambda \sqrt{2}\theta)^2}{2}),$$

i.e., by definition, $Z_i:=(X_iY_i-\mathbb{E}(X_iY_i))$ is sub-Gaussian with parameter $\sqrt{2}\theta = 2^{\frac{3}{2}}\sigma$.

Since $Z_1,\dots Z_n$ are independent sub-Gaussian r.v. with parameter $2^{\frac{3}{2}}\sigma$. It follows from the Hoeffding bound that also $$P(\frac{1}{n}|\sum_{i=1}^n (X_iY_i-\mathbb{E}(X_iY_i))| \geq t) \leq 2 \exp(-\frac{t^2 n}{2^4\sigma^2})= 2\exp(-ct^2)$$ where $c \equiv c_n = \frac{n}{2^4 \sigma^2}$ (hence the value $c$ is not a pure constant, but depends also on $n$).

Moreover I think that one might still improve the factor $\frac{1}{2^4}$ appearing in $c$.

  • $\begingroup$ If by "parameter" you mean the value "$c$" in your question, then this result cannot be correct. (Test it in the case $n=1$ and $Y_1$ is an independent Bernoulli$(1/2)$ variable, making $X_1Y_1$ a Normal$(0,\sigma^2)$ variable.) You might be switching back and forth between understanding $\sigma$ as the precision and $\sigma$ as the standard deviation of the underlying Normal distribution, or perhaps there's a typo in the question and you meant to write $1/c^2$ instead of $c$. $\endgroup$
    – whuber
    Jul 26, 2017 at 13:54
  • $\begingroup$ for $n=1$ it follows from $$\mathbb{E}(\exp(\lambda (X_iY_i-\mathbb{E}(X_iY_i))) \leq \exp(\frac{(\lambda \sqrt{2}\theta)^2}{2})$$ that $$P(|X_iY_i-\mathbb{E}(X_iY_i))|\geq z) \leq 2 \exp(-\frac{z^2}{2 (2^{\frac{3}{2}\sigma})^2}),$$ hence the corresponding value of $c$ would be $$c = \frac{1}{16\sigma^2}.$$ If $X_iY_i$ would be $N(0,\sigma^2)$ then this bound is, as expected, rather rough ($c=\frac{1}{16\sigma^2}$ vs $c_{normal} = \frac{1}{2\sigma^2}$), but I can't see anything wrong in it. (a sharper bound would be, of course, still better) $\endgroup$
    – chRrr
    Jul 26, 2017 at 14:22
  • $\begingroup$ Could you please, once and for all, explain what you mean by the "parameter" of a sub-Gaussian? Your edit to this answer indicates it's not $c$, but $c$ explicitly is what you refer to in the question. $\endgroup$
    – whuber
    Jul 26, 2017 at 15:17
  • $\begingroup$ Definition: A r.v. $X$ with $\mathbb{E}(X)=\mu$ is called sub-Gaussian if $\exists \sigma>0$ s.t. for all $\lambda>0$ $$\mathbb{E}(\exp(\lambda)(X-\mu))\leq \exp(\frac{\sigma^2\lambda^2}{2})$$ $\sigma$ is called the sub-Gaussian parameter. Short: $X \sim sG(\sigma)$. If $X_1,\dots, X_n$ ind. $sG(\sigma)$ one gets the Hoeffding bound $$ P(\frac{1}{n}|\sum_{i=1}^n(X_i-\mu_i))| \geq t) \leq 2 \exp(-\frac{t^2n}{2 \sigma^2}) \leq 2\exp(-ct^2)$$ for some "constant" $c$.$c$ is not the sG parameter but a "constant" (depending on n) bounding the $exp$-fct.I search a tight $c$. $\endgroup$
    – chRrr
    Jul 26, 2017 at 15:48

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