MGF Domains and Valid Chernoff Parameters

Why the MGF Domain Matters

An exponential tail bound uses an expectation of the form $E[e^{tX}]$. That expectation is not automatically finite just because $t$ is a real number. Nor does every sign of $t$ give the right comparison for a chosen tail event. A parameter is useful only when its exponential moment exists and its sign makes the event imply a lower bound on $e^{tX}$.

The domain question therefore comes before any attempt to use or choose a Chernoff parameter. In particular, a formula for an MGF on part of the real line must not be extended past that region by algebra alone. At a boundary, finiteness must be checked rather than presumed.

MGF and Its Domain

Definition: For a real-valued random variable $X$, its moment-generating function at $t\in\mathbb{R}$ is the possibly infinite expectation $$ M_X(t)=E[e^{tX}]. $$ Its MGF domain is $$ \mathcal{D}_X=\{t\in\mathbb{R}:M_X(t)<\infty\}. $$ Since $e^{0X}=1$, $0\in\mathcal{D}_X$ and $M_X(0)=1$.

The expectation is well-defined as an extended nonnegative expectation, so at a parameter outside $\mathcal{D}_X$ we have $M_X(t)=\infty$. The MGF domain always contains zero, but it need not contain an interval around zero. It can have a finite endpoint, and that endpoint may or may not be included.

Theorem: The MGF domain $\mathcal{D}_X$ is an interval, possibly with one or both finite endpoints included. For any $s,t\in\mathcal{D}_X$ and $0<\alpha<1$, $\alpha s+(1-\alpha)t\in\mathcal{D}_X$.

Proof. Let $u=\alpha s+(1-\alpha)t$. Pointwise, $e^{uX}=(e^{sX})^\alpha(e^{tX})^{1-\alpha}$. Hölder's inequality, applied with exponents $1/\alpha$ and $1/(1-\alpha)$, gives

M_X(u) =E\!\left[(e^{sX})^\alpha(e^{tX})^{1-\alpha}\right] \leq \bigl(E[e^{sX}]\bigr)^\alpha \bigl(E[e^{tX}]\bigr)^{1-\alpha}<\infty.

Thus every point strictly between two points of the domain is also in the domain, which is exactly the interval property. The argument does not decide whether a finite endpoint is included; that requires evaluating the expectation there. $\square$

Which Parameters Give a Tail Bound?

The exponential Markov inequality applies Markov's inequality to the nonnegative random variable $e^{tX}$. The event direction depends on the sign of $t$: when $t>0$, exponentiation preserves order, while when $t<0$, it reverses order.

Theorem: If $a\in\mathbb{R}$ and $t>0$ with $t\in\mathcal{D}_X$, then $$ P(X\geq a)\leq e^{-ta}M_X(t). $$ If $t<0$ and $t\in\mathcal{D}_X$, then $$ P(X\leq a)\leq e^{-ta}M_X(t). $$ At $t=0$, either probability is at most $1=e^{-ta}M_X(0)$. The same bounds hold for strict tail events.

Proof. If $t>0$, then $X\geq a$ implies $e^{tX}\geq e^{ta}$. Markov's inequality for the nonnegative random variable $e^{tX}$ therefore gives

P(X\geq a) \leq P(e^{tX}\geq e^{ta}) \leq \frac{E[e^{tX}]}{e^{ta}} =e^{-ta}M_X(t).

This is a finite bound because $t\in\mathcal{D}_X$. For $t<0$, $X\leq a$ implies $e^{tX}\geq e^{ta}$, so the same application of Markov's inequality gives the lower-tail result. At $t=0$, the right-hand side is $1$, and every probability is at most $1$. Replacing a weak event by a strict event only makes its probability no larger. $\square$

For a finite upper-tail bound, the eligible parameters are $t\in\mathcal{D}_X$ with $t>0$; for a finite lower-tail bound, they are $t\in\mathcal{D}_X$ with $t<0$. The zero parameter is technically valid for either event but supplies only the trivial bound $1$. If $M_X(t)=\infty$, the extended inequality has an infinite right-hand side and provides no finite bound.

Worked Example: A Normal MGF and the Sign of the Parameter

Let $X\sim N(0,1)$. Its MGF is $M_X(t)=e^{t^2/2}$, finite for every real $t$. To bound $P(X\geq 2)$, choose $t=1$, which is positive and in the MGF domain. The exponential Markov bound is

P(X\geq 2) \leq e^{-1\cdot 2}e^{1^2/2} =e^{-3/2} \approx 0.2231.

The domain check alone is not enough: $t=-1$ is also in the domain, but it has the wrong sign for this upper-tail argument. In fact, it can give a false inequality if used in the upper-tail formula. Since $X\geq 0$ with probability $1/2$ by symmetry and has no atom at zero, $P(X\geq 0)=1/2$. The invalid substitution $a=0,\ t=-1$ would claim

P(X\geq 0)\leq e^{-(-1)\cdot 0}M_X(-1)=e^{1/2}\approx 1.6487.

This particular number does not contradict the probability, but the direction has no Markov justification and is not a valid upper-tail rule. A wrong-sign substitution can even make a false claim: taking $a=1,\ t=-1$ yields the proposed right-hand side $e\,e^{1/2}=e^{3/2}>1$, still uninformative; for meaningful use, the event inclusion, not merely an algebraic expression, is what establishes validity. For an actual failure, take $X\sim\operatorname{Exp}(1)$, $a=0$, and $t=-1$: $P(X\geq0)=1$, whereas the wrongly used expression is $M_X(-1)=1/2$. The claimed upper-tail inequality would be false.

Endpoints and Limits Require Separate Checks

An interval domain can have an excluded endpoint where the MGF diverges, or an included endpoint where it remains finite. A Chernoff parameter at a finite endpoint is valid exactly when it is in the domain; being a limit of valid interior parameters does not by itself establish that.

Worked Example: An Exponential MGF Has an Excluded Endpoint

Let $X\sim\operatorname{Exp}(2)$, with density $f(x)=2e^{-2x}$ for $x\geq0$. Direct integration gives

M_X(t)=\int_0^\infty e^{tx}2e^{-2x}\,dx =2\int_0^\infty e^{-(2-t)x}\,dx =\frac{2}{2-t},\qquad t<2.

If $t=2$, the integrand is the constant $2$, so the integral is infinite. If $t>2$, the integrand grows exponentially and the integral is also infinite. For $t<2$, it is integrable, including every negative $t$. Thus $\mathcal{D}_X=(-\infty,2)$; the endpoint $2$ is excluded.

For $a=3$, the parameter $t=1$ is positive and belongs to the domain. It gives

P(X\geq3)\leq e^{-3}M_X(1)=2e^{-3}\approx0.0996.

At $t=2$, the formal expression $e^{-6}M_X(2)$ is infinite, not a finite bound. The exact tail is $P(X\geq3)=e^{-6}\approx0.00248$, obtained by integrating the density. This example distinguishes a valid finite parameter from a boundary value at which the MGF diverges.

A limit toward an endpoint can sometimes be justified, but the justification depends on the random variable. For example, if $X\geq0$ almost surely and $t$ increases to a finite endpoint $b$, then $e^{tX}$ increases pointwise to $e^{bX}$. The monotone convergence theorem gives

\lim_{t\uparrow b}M_X(t)=E[e^{bX}],

where the limit and expectation may both be infinite. For a variable that can have either sign, $e^{tX}$ need not increase as $t$ increases, so this monotone-convergence argument does not apply. Evaluate the endpoint expectation directly or supply a suitable domination argument.

Worked Example: A Finite Endpoint Can Be a Valid Parameter

Let $\lambda=\log 2$, and define a positive-integer-valued random variable by

P(X=k)=c\,\frac{e^{-\lambda k}}{k^2},\qquad k=1,2,\ldots, \quad c=\left(\sum_{k=1}^{\infty}\frac{e^{-\lambda k}}{k^2}\right)^{-1}.

The normalizing sum is positive and finite, since it is bounded above by $\sum_{k=1}^{\infty}1/k^2<\infty$. Its MGF is

M_X(t)=c\sum_{k=1}^{\infty}\frac{e^{(t-\lambda)k}}{k^2}.

For $t<\lambda$, the series converges by comparison with a geometric series. At $t=\lambda$, it equals $c\sum_{k=1}^{\infty}1/k^2<\infty$, so the endpoint is included. For $t>\lambda$, its terms do not tend to zero, so the series diverges. Consequently, $\mathcal{D}_X=(-\infty,\lambda]$.

For the upper-tail event $X\geq4$, $t=\lambda$ is a valid positive endpoint parameter. The bound is

P(X\geq4)\leq e^{-4\lambda}M_X(\lambda) =\frac{c}{16}\sum_{k=1}^{\infty}\frac{1}{k^2}.

Indeed, the event consists of integers $k\geq4$, and for each such $k$, $e^{\lambda k}\geq e^{4\lambda}$; Markov's inequality applies because the endpoint expectation is finite. The corresponding expression at any $t>\lambda$ would have an infinite MGF and would not give a finite bound.

The zero parameter also deserves care. It always belongs to the domain and makes the tail bound equal to $1$, but it need not be approached by finite-MGF parameters from both sides. A one-sided limit at zero can be justified when there is a finite exponential moment in that direction. For example, if $M_X(\delta)<\infty$ for some $\delta>0$, then for $0\leq t\leq\delta$,

e^{tX}\leq 1+e^{\delta X}.

The right side is integrable, so dominated convergence gives $M_X(t)\to1$ as $t\downarrow0$. The analogous left-hand limit follows if $M_X(-\delta)<\infty$ for some $\delta>0$. Without a finite moment in the relevant direction, do not assume continuity there.

Common Mistakes and Practical Checks

Using the MGF formula outside its convergence region. A closed-form expression derived for some $t$-values is not evidence that the expectation is finite elsewhere. Return to the defining expectation or series.
Using the wrong sign for the tail. An upper tail requires $t>0$; a lower tail requires $t<0$. Check the event implication before applying Markov's inequality.
Including an endpoint by assumption. Convergence inside an interval does not settle convergence at its boundary. Test the endpoint itself.
Treating an infinite right-hand side as an informative bound. The extended inequality may remain formally true, but an infinite bound cannot control a probability more usefully than the bound $1$.
Assuming the limit at zero is two-sided. Zero is always in the domain, but nearby parameters may be available on only one side or not at all.

Key takeaway: Before using an exponential tail bound, check both conditions: the parameter must belong to the MGF domain so the expectation is finite, and its sign must match the tail direction. Treat zero and every domain endpoint explicitly.

Check Your Understanding

For each question, distinguish domain membership from the sign condition for the tail.

If $M_X(t)<\infty$ for $t=-2$ and $t=3$, what can you conclude about the MGF domain between those parameter values?
For an upper-tail bound on $P(X\geq a)$, why does a finite MGF at a negative parameter not justify the usual exponential Markov bound?
An MGF is finite for every $t<b$ and infinite for every $t>b$. What additional check determines whether $b$ is an eligible Chernoff parameter?
What bound does the parameter $t=0$ provide, and why is it usually called trivial?
If $X\geq0$ almost surely and $t\uparrow b$, which convergence theorem can relate $M_X(t)$ to $E[e^{bX}]$, and what may happen to that expectation?