The Saddlepoint Equation and Its Root

Why the Saddlepoint Equation Has a Root Question

A saddlepoint approximation is indexed by a parameter chosen to match a specified target value. That parameter is not available for every target: it must lie in the domain of the cumulant-generating function (CGF), and the target must be achievable as a derivative of the CGF there. Determining whether that parameter exists, and whether it is unique, is therefore part of setting up the approximation—not a detail to postpone until after the calculation.

We derive the equation for an independent, identically distributed sum and its sample mean. The key fact is that the derivative of the CGF is strictly increasing on its interior domain when the distribution is nondegenerate. Its range then gives an exact criterion for the existence of a finite interior root.

From the CGF to the Saddlepoint Equation

Definition: For a random variable \(X\), its moment-generating function is \(M(t)=E[\exp(tX)]\) wherever this expectation is finite. The cumulant-generating function is \(K(t)=\log M(t)\) on the effective domain \(D=\{t\in\mathbb{R}:M(t)<\infty\}\). Write \(D^\circ=(a,b)\) for the interior of this domain.

The domain \(D\) is an interval: if the exponential moments are finite at two arguments, they are finite at every argument between them, by Hölder's inequality. The saddlepoint is taken in \(D^\circ\), where the CGF is differentiable. Assume throughout this tutorial that \(X\) is nondegenerate. Then \(K''(t)>0\) throughout \(D^\circ\).

To see why, define the exponentially tilted distribution at \(t\in D^\circ\) by weighting the original distribution in proportion to \(\exp(tX)\). Under this distribution, denoted here by \(P_t\), differentiation of \(K\) gives

K'(t)=E_t[X], \qquad K''(t)=\operatorname{Var}_t(X)>0.

These derivatives are valid at interior points: a small interval around \(t\) remains in \(D\), so exponential moments in that interval dominate the factors involving \(X\) needed to differentiate under the expectation. The tilted distribution assigns positive weight wherever the original distribution does. Thus it remains nondegenerate, and its variance is strictly positive.

For independent and identically distributed \(X_1,\ldots,X_n\), let \(S_n=\sum_{i=1}^n X_i\) and \(\bar X_n=S_n/n\). Independence gives the CGFs \(nK(t)\) for \(S_n\) and \(nK(t)\) for the scaled sum \(n\bar X_n\). To target a sum value \(s\), the saddlepoint equation is the stationary-point equation for \(ts-nK(t)\):

s-nK'(t)=0, \qquad\text{or equivalently}\qquad K'(t)=\frac{s}{n}.

If the target is written as a sample-mean value \(x=s/n\), the same equation is \(K'(t)=x\). The factor \(n\) belongs with the sum target; it cancels when the target is expressed as a sample mean.

Definition: For a target sample mean \(x\), an interior saddlepoint is a real number \(\hat t(x)\in D^\circ\) satisfying \(K'(\hat t(x))=x\). For a target sum \(s\), it satisfies \(nK'(\hat t(s))=s\). A target with no such value has no finite interior saddlepoint.

Existence, Uniqueness, and Dependence on the Target

Because \(K''\) is positive, \(K'\) is strictly increasing on \(D^\circ\). Define its endpoint limits, allowing infinite values, by

L=\lim_{t\downarrow a}K'(t), \qquad U=\lim_{t\uparrow b}K'(t).

These limits describe precisely which sample-mean targets have an interior saddlepoint. In particular, a target equal to a finite endpoint limit is not included: it may be approached as \(t\) approaches the boundary, but it is not attained at any \(t\in D^\circ\).

Theorem: Suppose \(X\) is nondegenerate and \(K\) is finite on the interior \(D^\circ=(a,b)\) of its CGF domain. An interior saddlepoint \(\hat t(x)\) exists if and only if \(L<x<U\). When it exists, it is unique. As a function of \(x\) on \((L,U)\), it is continuously differentiable and

\frac{d\hat t(x)}{dx}=\frac{1}{K''(\hat t(x))}>0.

For the sum target \(s\), the same conditions apply to \(x=s/n\). Equivalently, an interior root exists exactly when \(nL<s<nU\), and its derivative with respect to \(s\) is \(1/\{nK''(\hat t(s))\}\).

Proof. The function \(K'\) is continuous and strictly increasing on \((a,b)\), since \(K''(t)>0\). By its endpoint limits and the intermediate value theorem, its image is exactly \((L,U)\). Thus every \(x\) in that interval is attained once, and no \(x\) outside it is attained. Strict increase gives uniqueness.

At any root \(t=\hat t(x)\), \(K''(t)>0\). The inverse function theorem applied to \(K'\) gives a continuously differentiable local inverse with derivative \(1/K''(t)\). Uniqueness makes these local inverses agree wherever they overlap, so they define the claimed differentiable function throughout \((L,U)\). Differentiating the sum equation \(nK'(\hat t(s))=s\) gives \(nK''(\hat t(s))\,d\hat t/ds=1\), which proves the sum-target derivative. \(\square\)

The equation also identifies a maximizer. For an interior root, the derivative of \(g_x(t)=tx-K(t)\) is \(x-K'(t)\), which is positive for \(t<\hat t(x)\) and negative for \(t>\hat t(x)\). Hence \(\hat t(x)\) is the unique global maximizer of \(g_x\) over \(D^\circ\). For a sum target, the corresponding function is \(ts-nK(t)\), with the same root.

Worked Examples

Worked Example: A Normal Sample Mean

Suppose \(X_i\sim N(\mu,\sigma^2)\), with \(\sigma^2>0\), independently. Find the saddlepoint for target sample mean \(x\).

The normal CGF is \(K(t)=\mu t+\sigma^2t^2/2\), finite for every real \(t\). Therefore \(K'(t)=\mu+\sigma^2t\), whose range is all of \(\mathbb{R}\). Solving \(K'(t)=x\) gives

\mu+\sigma^2\hat t=x \quad\Longrightarrow\quad \hat t(x)=\frac{x-\mu}{\sigma^2}.

Every real target has one finite root. For example, if \(\mu=2\), \(\sigma^2=4\), and \(x=5\), then \(\hat t=(5-2)/4=3/4\). For a sum target \(s\) from \(n\) observations, substitute \(x=s/n\), giving \(\hat t=(s/n-\mu)/\sigma^2\).

Worked Example: A Bernoulli Sample Mean

Let \(X_i\sim\operatorname{Bernoulli}(p)\), where \(0<p<1\). Determine which targets have an interior saddlepoint and solve for it.

The CGF and its derivative are

K(t)=\log(1-p+pe^t), \qquad K'(t)=\frac{pe^t}{1-p+pe^t}.

The domain is all of \(\mathbb{R}\), and \(K'(t)\) increases from \(0\) to \(1\). Thus a finite root exists exactly for \(0<x<1\). Solving the equation,

x=\frac{pe^{\hat t}}{1-p+pe^{\hat t}} \quad\Longrightarrow\quad x(1-p)=pe^{\hat t}(1-x) \quad\Longrightarrow\quad \hat t(x)=\log\!\left(\frac{x(1-p)}{p(1-x)}\right).

For \(p=1/4\) and \(x=1/2\), this gives \(\hat t=\log(3)\): substituting \(e^{\hat t}=3\) into \(K'\) gives \((3/4)/(3/4+3/4)=1/2\). At \(x=0\) or \(x=1\), the equation has no finite root; those targets are limits of \(K'(t)\) as \(t\) tends to minus or plus infinity.

Worked Example: An Exponential Sample Mean

Let \(X_i\) have an exponential distribution with rate \(\lambda>0\). Find the root and its target range.

For \(t<\lambda\), direct integration gives

M(t)=\int_0^\infty \lambda e^{-(\lambda-t)y}\,dy =\frac{\lambda}{\lambda-t}, \qquad K(t)=-\log(1-t/\lambda), \qquad K'(t)=\frac{1}{\lambda-t}.

The interior domain is \((-\infty,\lambda)\), and \(K'(t)\) ranges over \((0,\infty)\). Therefore every positive target mean has one root. Solving \(K'(t)=x\) yields \(\hat t=\lambda-1/x\), which is less than \(\lambda\) for every \(x>0\). With rate \(\lambda=2\) and target \(x=1\), the root is \(t=1\), and \(K'(1)=1/(2-1)=1\). A zero or negative target has no interior root.

Why the Domain Matters

The target range is determined by the image of \(K'\), not simply by whether the target seems plausible from the support of \(X\). In the Bernoulli example the support restricts sample means to \([0,1]\), but only targets strictly between 0 and 1 have finite interior roots. In the exponential example the support is positive, and the derivative range likewise excludes zero. A boundary target can sometimes correspond to a limiting value of the saddlepoint, but that is not an interior solution of the equation.

For a sum, keep the scale consistent. If \(K\) is the CGF of one observation and the target is \(s\) for \(S_n\), solve \(nK'(t)=s\). If instead the target is \(x\) for \(\bar X_n\), solve \(K'(t)=x\). Using \(K'(t)=s\) for a sum, or inserting an extra \(n\) into the sample-mean equation, changes the target and generally gives the wrong root.

Common Mistake

Wrong reasoning: “The CGF is defined at \(t=0\), so every target must have a saddlepoint.” The existence of one point in the CGF domain does not imply that \(K'\) takes every real value there.
Correction: Find the interior domain and the range of \(K'\). A finite interior root exists only when the target lies strictly between the endpoint limits \(L\) and \(U\). For Bernoulli data, for instance, the CGF exists for all real \(t\), but its derivative has range only \((0,1)\).

Key takeaway: For a nondegenerate iid model, the saddlepoint for a target mean \(x\) is the unique interior solution of \(K'(t)=x\), provided \(x\) lies in the range of \(K'\). The root increases smoothly with the target, at rate \(1/K''(\hat t(x))\).

Check Your Understanding

Use the CGF domain and derivative range to answer each question.

For an iid sum with one-observation CGF \(K\), write the saddlepoint equation for target sum \(s\), and then rewrite it for target sample mean \(x=s/n\).
Why does nondegeneracy imply that \(K'\) is strictly increasing on the interior CGF domain?
For a Bernoulli distribution with \(0<p<1\), does a finite interior root exist for target mean \(x=0.8\)? What about \(x=1\)? Explain.
For the exponential model with rate \(\lambda\), state the CGF domain and the range of \(K'\). Which target means have a finite root?
If \(K'(\hat t(x))=x\), what is \(d\hat t(x)/dx\), and why is it positive?