1
$\begingroup$

I posted the following question in a comment on CDF of a log-concave discrete random variable. Since it is not related to my main question, I thought of reposting it as separate post.

Question:

Let $X \sim \operatorname{Pois}(\lambda)$. I'm interested in a large deviation bound for $X$; more specifically a sub-exponential type bound for $P(X \geq x)$ for all $x\geq 0$. While it's known that such a bound exists for $x \geq \lambda$ (Chernoff bound), it's not clear whether there exists one for $x< \lambda$. Is it possible to obtain a bound (or extend the bound obtained for $x \geq \lambda$) for these $x$ values using standard large-deviation techniques? If not, why not?

$\endgroup$
1
  • 5
    $\begingroup$ When $x<\lambda$, the event $\{X \ge x\}$ holds with at least constant probability, so we are not in the large-deviations regime. $\endgroup$
    – Alf
    Commented Feb 4, 2021 at 12:48

1 Answer 1

3
$\begingroup$

$\newcommand\Ga\Gamma\newcommand\Z{\mathbb Z}\DeclareMathOperator\Pois{Pois}$Let $t\mathrel{:=}\lambda$ and $k\mathrel{:=}x\in\Z\cap[0,t)$. Then for $X_t\sim \Pois(t)$ we have $$P(X_t\ge k)=1-\frac{\Ga(k,t)}{\Ga(k)};$$ this can be shown e.g. by integrating $$\Ga(k,t)=\int_t^\infty u^{k-1}e^{-u}\,du$$ $k-1$ times by parts.

It follows now by Corollary 3 of this paper or its arXiv preprint version that $P(X_t\ge k)>1/2$. Therefore (as was noted by user J.), standard large-deviation techniques are not applicable here.

However, using Theorem 1.1 and Proposition 2.7 of this paper or its arXiv preprint version, we have $$P(X_t\ge k)\le B_1(k,t)\mathrel{:=}1-e^{-t} \Big(1+\frac{(t+2)^k-t^k-2^k}{2 k!}\Big)$$ for $k\ge3$ and $$P(X_t\ge k)\le B_2(k,t)\mathrel{:=}1-\frac{e^{-t} t^{k-1}}{(k-1)!}$$ for $k\ge1$.

Both upper bounds, $B_1(k,t)$ and $B_2(k,t)$, on $P(X_t\ge k)$ are nontrivial, in the sense that they both are strictly less than $1$. Both bounds are rather accurate. The bound $B_2(k,t)$ on $P(X_t\ge k)$ is simpler but less accurate than $B_1(k,t)$. What has been said in this paragraph is illustrated by the following graphs $\Big\{\Big(k,\dfrac{P(X_{2k}\ge k)}{B_1(k,2k)}\Big)\colon k\in\{3,\dots,20\}\Big\}$ (red) and $\Big\{\Big(k,\dfrac{P(X_{2k}\ge k)}{B_2(k,2k)}\Big)\colon k\in\{3,\dots,20\}\Big\}$ (blue):

graphs of relative errors for B_1 and B_2

Here the maximum relative error of the bound $B_1(k,2k)$ on $P(X_{2k}\ge k)$ is $\approx0.001$ and the maximum relative error of the bound $B_2(k,2k)$ on $P(X_{2k}\ge k)$ is $\approx0.050$.

$\endgroup$
4
  • $\begingroup$ Thank you very much. I'll check out these papers. $\endgroup$
    – SL_MathGuy
    Commented Feb 4, 2021 at 19:15
  • $\begingroup$ I'm sorry for the delayed response but just wondering if the u.b when $k\geq 1$ should be $1- \frac{e^{-t} t^{k-1}}{(k-1)!}$ (NOT k!)? $\endgroup$
    – SL_MathGuy
    Commented Feb 14, 2021 at 12:51
  • 1
    $\begingroup$ @SL_MathGuy : That's right, it should have $(k-1)!$ in the denominator; this is now fixed. So, to have a closure, are you now satisfied with this answer? $\endgroup$ Commented Feb 14, 2021 at 16:28
  • $\begingroup$ I would say yes! I've accepted the answer. Thank you very much! $\endgroup$
    – SL_MathGuy
    Commented Feb 14, 2021 at 16:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .