Poisson Tail Bounds from Its MGF

Why the Poisson MGF gives useful tail bounds

A Poisson model is a natural starting point for counts: \(X\sim\operatorname{Pois}(\lambda)\), with \(\lambda>0\), takes values \(0,1,2,\ldots\) and has mean \(\lambda\). Its exact tail probabilities can be computed by summing its probability mass function. For large means or many different thresholds, however, a compact bound can be easier to evaluate and interpret. The moment-generating function (MGF) gives such bounds through the exponential Markov inequality.

The central idea is to replace a count-tail event by an event involving \(e^{tX}\). Choosing \(t>0\) emphasizes large values and bounds an upper tail; choosing \(t<0\) emphasizes small values and bounds a lower tail. Optimizing \(t\) produces an explicit exponential bound. The resulting expression is useful, but it is generally not the exact tail probability.

The Poisson MGF and its exponential form

For a Poisson variable, the MGF is finite for every real \(t\). Directly summing over the probability mass function gives

The series converges for every real \(t\), because it is the exponential series evaluated at the finite number \(\lambda e^t\). Thus any real Chernoff parameter used below is in the MGF domain. The exponential Markov inequality says that if \(Y\geq0\) and \(c>0\), then \(P(Y\geq c)\leq E[Y]/c\). Applying it to \(Y=e^{tX}\) gives the upper-tail bound for \(t>0\). For \(t<0\), the function \(x\mapsto e^{tx}\) decreases, so a lower-tail event instead implies a large value of \(e^{tX}\).

Optimizing the upper and lower tails

For \(u>0\), define the Poisson rate function

and set \(h(0)=1\), its continuous limiting value. For \(u\geq0\), \(h(u)\geq0\): for \(u>0\), \(h'(u)=\log u\), so \(h\) decreases up to \(u=1\), increases after \(u=1\), and attains its minimum \(h(1)=0\).

Thresholds \(a\) need not be integers. The event notation has its usual meaning for the integer-valued variable \(X\); for instance, \(X\geq a\) is \(X\geq\lceil a\rceil\). The theorem remains valid for real \(a\), because the exponential Markov argument uses the threshold \(a\) itself.

Proof. First suppose \(a>\lambda\). For every \(t>0\), if \(X\geq a\), then \(e^{tX}\geq e^{ta}\). Exponential Markov therefore gives

The exponent \(g(t)=\lambda(e^t-1)-ta\) has derivative \(g'(t)=\lambda e^t-a\) and second derivative \(g''(t)=\lambda e^t>0\). Its unique minimum over \(t>0\) is at \(t^\ast=\log(a/\lambda)>0\). Substitution yields

which proves the upper-tail claim.

Now let \(0<a<\lambda\). For \(t<0\), the event \(X\leq a\) implies \(e^{tX}\geq e^{ta}\). Thus

The same strictly convex function \(g(t)\) has its unique minimum at \(t^\ast=\log(a/\lambda)<0\), which is now an admissible lower-tail parameter. Substituting this value again gives \(g(t^\ast)=-\lambda h(a/\lambda)\). Finally, because \(X\) is nonnegative and integer-valued, \(X\leq0\) means \(X=0\), whose probability is \(e^{-\lambda}\). This establishes the endpoint directly. \(\square\)

For a lower threshold \(a=0\), one can also see the limiting expression in the rate function: \(\exp\{-\lambda h(0)\}=e^{-\lambda}\), which matches the exact probability. At \(a=\lambda\), the optimizing parameter is \(t=0\), and the exponential bound is \(1\); there is no nontrivial tail bound at the mean from this optimization.

Worked comparisons with exact probabilities

How to read the bound and avoid common errors

The same rate-function expression describes both tails, but the parameter constraints differ: an upper-tail deviation uses \(t>0\), while a lower-tail deviation uses \(t<0\). The optimized exponent is nonpositive because \(h(u)\geq0\), and it is strictly negative whenever \(u\ne1\). Consequently, the bound decreases exponentially as the rate-function cost of the threshold grows.

A common mistake is to use \(t>0\) for a lower-tail event and claim that \(X\leq a\) implies \(e^{tX}\geq e^{ta}\). For positive \(t\), the exponential is increasing, so the implication goes in the opposite direction: \(X\leq a\) gives \(e^{tX}\leq e^{ta}\). The correction is to choose \(t<0\), making the exponential decreasing; then \(X\leq a\) does imply \(e^{tX}\geq e^{ta}\), and Markov's inequality applies.

Another important distinction is between a bound and an exact probability. Optimizing the exponential Markov expression finds the best bound within this one-parameter family of exponential Markov bounds. It does not make the resulting number equal to the Poisson tail in general. Exact probabilities still come from summing the mass function, and the examples show that the bound can be conservative by different amounts in different settings.