# Sub-exponential tail bound for Poisson multiplied by cosine of an independent uniform random variable

I am looking for the tail bound of the following random variable, hopefully of sub-exponential form:

$$\lambda_n^{-1}X_n\cos(\theta_n)$$, where $$X\sim Poisson(\lambda_n)$$ with $$\lambda_n\to 0$$, and another independent random variable $$\theta_n\sim Unif[0,2\pi]$$.

So basically I'd like to know how much can that cosine factor "pull" the whole distribution back to its center, therefore improve the concentration result--since $$\lambda_n^{-1}X_n$$ as sub-exponential$$(2\lambda_n^{-1},\lambda_n^{-1})$$ random variable is poorly concentrated.

But this is so hard to compute: say I want to compute the MGF. If I first integrate over $$\theta_n$$, I get the modified Bessel function of the first kind: $$\mathbb{E}e^{\beta\lambda^{-1}X\cos(\theta)}=\mathbb{E}I_0(\frac{\beta}{\lambda}X),$$

which I don't know how to evaluate. Otherwise I get $$\mathbb{E}e^{...}=\mathbb{E}\exp(\lambda(e^{\beta\lambda^{-1}\cos(\theta)}-1)),$$

which is also bizarre.

• Let me call $Z_n$ your variable and $Y_n:=X_n/\lambda_n$. A quick computation shows that the expectation $\mathbb E[\exp(tY_n)]$ goes to infinity as $n\to\infty$ for all $t>0$. Since $\mathbb E[\exp(tZ_n)]\geq\varepsilon\mathbb E[\exp(t(1-\delta)Y_n)]$ for some $\varepsilon,\delta>0$ by just considering $\theta_n\simeq0$, there will be no uniform sub-exponential bound. Nov 24, 2022 at 9:47
• @PierrePC: Yes, sure. Therefore I'd like to know some good quantitative bound for large but fixed $n$(depending on $\lambda_n$, ofc) to see what is the typical fluctuation of $Z_n$. Eg., if I simply bound the absolute value by $Y_n$ I can compute explicitly that $P(|Z_n|>t)\le P(Y_n>t)=P(Y_n-1>t-1)\le \exp(-\frac 12\min\{\lambda_n (t-1)^2/2,\lambda_n (t-1))$, implying if we denote $t:=1+\lambda_n^{-1}s$ we will get some O(1) fluctuation in $s$, or say $Z_n$ fluctuates on the scale $\lambda_n^{-1}$. But I am not sure how that $\cos(\theta)$ term changes the game. Nov 24, 2022 at 11:12
• What I am saying is that $Z_n$ and $Y_n$ should have very similar tails, for instance for any $\varepsilon>0$ small enough one has $\varepsilon\mathbb P(Y_n\geq(1+\varepsilon)t)\leq\mathbb P(Z_n\geq t)\leq\mathbb P(Y_n\geq t)$. For this reason it seems unlikely that your cosine factor would pull the distribution anywhere in a sense that it meaningful to you. Nov 24, 2022 at 11:41

$$\newcommand\la\lambda\newcommand\de\delta\newcommand\si\sigma\newcommand\Th\Theta$$There is hardly a reason for having the parameter $$n$$. So, let $$\la:=\la_n\to0$$, $$X:=X_n$$, $$\Th:=\theta_n$$, $$Y:=X/\la$$, $$Z:=Y\cos\Th$$. Note that $$EY=1$$, $$\si_Y:=\sqrt{Var\,Y}=1/\sqrt\la\to\infty$$, $$EZ=EY\,E\cos\Th=0$$, and $$\si_Z:=\sqrt{Var\,Z}=\sqrt{EZ^2}=\sqrt{EY^2\,E\cos^2\Th}=\sqrt{EY^2/2}\sim\si_Y/\sqrt2=1/\sqrt{2\la}$$. We want to estimate $$P(Z-EZ\ge z)=P(Z\ge z)$$ for real $$z>0$$ (note that the distribution of $$Z$$ is symmetric about $$0$$). It makes sense to measure the deviation $$z$$ of $$Z$$ from its (zero) mean using the standard deviation $$\si_Z=1/\sqrt{2\la}$$ of $$Z$$ or, equivalently, $$\si_Y=1/\sqrt\la$$ as the unit.

For $$z>0$$ we have $$P(Z\ge z)\le P(Y\ge z)=P(X\ge\la z)\le P(X>0).$$ Also, for $$z\in(0,1/(2\la)]$$, $$P(Z\ge z)\ge P(Y\ge2z)P(\cos\Th\ge1/2)=\tfrac13\,P(Y\ge2z) \\ =\tfrac13\,P(X\ge2\la z)=\tfrac13\,P(X>0).$$

So, $$P(Z\ge z)\asymp P(X>0)\sim\la\to0$$ for $$z\in(0,1/(2\la)]$$. Note also that $$1/(2\la)$$ is much greater than the standard deviation $$\si_Z\sim1/\sqrt{2\la}$$ of $$Z$$.

Similarly, for each $$k\in\{0,1,\dots\}$$ and each $$\de\in(0,1)$$, $$P(Z\ge z)\asymp\frac{\la^{k+1}}{(k+1)!} \tag{1}\label{1}$$ for $$z\in(k/\la,(k+1-\de)/\la]$$. Note also that for all real $$z>0$$ $$P(Z\ge z)=2e^{-\la}\sum_{j\ge\la z}\frac{\la^j}{j!}\arccos\frac{\la z}j.$$

The latter paragraph is illustrated below by the plots $$\Big\{P(Z\ge z)\Big/\frac{\la^{k+1}}{(k+1)!}\colon z\in(k/\la,(k+1-\de)/\la]\Big\}$$ of the ratios of the left-hand side of \eqref{1} to its right-hand side for $$k=3$$, $$\de=0.1$$, $$\la=0.1$$ (red), $$\la=0.03$$ (green), and $$\la=0.01$$ (blue): • I see, that's interesting! Do you think we'll get better concentration if we sum over these? Like, consider $\sum_{n\le N}nPoisson(n^{-1})\cos(\theta_n)$. Nov 24, 2022 at 20:05