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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.

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  • $\begingroup$ 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. $\endgroup$
    – Pierre PC
    Nov 24 at 9:47
  • $\begingroup$ @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. $\endgroup$
    – MikeG
    Nov 24 at 11:12
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    $\begingroup$ 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. $\endgroup$
    – Pierre PC
    Nov 24 at 11:41

1 Answer 1

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$\newcommand\la\lambda\newcommand\de\delta\newcommand\si\sigma\newcommand\Th\Theta$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):

enter image description here

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  • $\begingroup$ I see, that's interesting! Do you think we'll get better concentration if we sum over these? Like, consider $\sum_{n\le N}nPoisson(n^{-1})\cos(\theta_n)$. $\endgroup$
    – MikeG
    Nov 24 at 20:05

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