Saddlepoint Approximations Near the Mean

Why the Mean Needs Special Treatment

The Lugannani–Rice approximation is especially useful for estimating a tail probability near a target value. At the mean, however, its usual expression appears to involve division by zero: both terms in its correction diverge. The important point is that they diverge at the same rate, and their difference has a finite limit. Taking that limit gives a usable central probability approximation and shows how the correction depends on skewness.

We work with a continuous random variable $S$ whose cumulant generating function is finite in an open interval containing zero. This local existence condition ensures that the derivatives used below exist. We also assume $K''(0)>0$. For a continuous model, the Lugannani–Rice CDF expression is an approximation, not an exact identity; its central limit can nevertheless be derived directly from its defining quantities.

The Formula and Its Apparent Singularity

Definition: Let $K(t)=\log E[e^{tS}]$ be the cumulant generating function of $S$, for $t$ in an open interval containing zero. For a target $x$ near $E[S]=K'(0)$, let $t$ solve the saddlepoint equation $K'(t)=x$. Define $$ r=\operatorname{sgn}(t)\sqrt{2\{tx-K(t)\}}, \qquad q=t\sqrt{K''(t)}. $$ The Lugannani–Rice CDF approximation is $$ P(S\le x)\ \approx\ \Phi(r)+\phi(r)\left(\frac{1}{r}-\frac{1}{q}\right), $$ where $\Phi$ and $\phi$ are the standard normal CDF and density.

At $x=K'(0)$, the saddlepoint is $t=0$. Then $r=q=0$, so neither reciprocal in the correction is defined. The formula cannot be evaluated there by simply substituting zero. Instead, we evaluate the limit as $t$ tends to zero. Since $x=K'(t)$, the expression inside the braces in the definition of $r$ can be written $tK'(t)-K(t)$; the quantity under its square root is $2\{tK'(t)-K(t)\}$.

Deriving the Central Limit of the Correction

Write $\kappa_j=K^{(j)}(0)$ for the cumulants of $S$. In particular, $\kappa_1=E[S]$, $\kappa_2=\operatorname{Var}(S)>0$, and $\kappa_3$ is its third cumulant. Because the moment generating function is finite near zero, $K$ has a Taylor expansion there. Using $x=K'(t)$, expand the two quantities that determine $r$ and $q$:

\begin{aligned} 2\{tK'(t)-K(t)\} &=\kappa_2t^2+\frac{2\kappa_3}{3}t^3+O(t^4),\\ K''(t)&=\kappa_2+\kappa_3t+O(t^2). \end{aligned}

For sufficiently small $t$, $\kappa_2>0$ makes the square roots well defined. The sign convention for $r$ gives $r=t\sqrt{\kappa_2}$ times a factor tending to one from either side of zero. Expanding the square roots yields

\begin{aligned} r&=t\sqrt{\kappa_2} \left(1+\frac{\kappa_3}{3\kappa_2}t+O(t^2)\right),\\ q&=t\sqrt{\kappa_2} \left(1+\frac{\kappa_3}{2\kappa_2}t+O(t^2)\right). \end{aligned}

Taking reciprocals of these expansions gives

\begin{aligned} \frac{1}{r} &=\frac{1}{t\sqrt{\kappa_2}} -\frac{\kappa_3}{3\kappa_2^{3/2}}+O(t),\\ \frac{1}{q} &=\frac{1}{t\sqrt{\kappa_2}} -\frac{\kappa_3}{2\kappa_2^{3/2}}+O(t). \end{aligned}

The divergent terms cancel on subtraction. Since $r\to0$, $\Phi(r)\to\frac12$ and $\phi(r)\to\phi(0)=1/\sqrt{2\pi}$. This proves the central form.

Theorem: Under the conditions above, the continuous extension of the Lugannani–Rice approximation to $x=E[S]$ is $$ \frac12+\phi(0)\frac{\kappa_3}{6\kappa_2^{3/2}}. $$ The limit of the correction itself is $\kappa_3/(6\kappa_2^{3/2})$.

In particular, the adjustment is positive for positive skewness and negative for negative skewness. It vanishes for a symmetric model with $\kappa_3=0$, though that fact alone does not make every such model symmetric.

Worked Central Calculations

Worked Example: A Normal Model

Suppose $S\sim N(\mu,v)$, where $v>0$. Its cumulant generating function is $K(t)=\mu t+vt^2/2$, so $\kappa_2=v$ and $\kappa_3=0$. At the mean $x=\mu$, the central Lugannani–Rice limit is

$$ \frac12+\phi(0)\frac{0}{6v^{3/2}}=\frac12. $$

Here the result agrees with the exact probability: a normal distribution is continuous and symmetric about $\mu$, so $P(S\le\mu)=1/2$. This check also confirms that the central correction disappears when the third cumulant is zero.

Worked Example: The Sum of 25 Exponential Variables

Let $S=X_1+\cdots+X_{25}$, where the $X_i$ are independent exponential random variables with rate $1$. For one summand, $K_X(t)=-\log(1-t)$ for $t<1$. Thus $K(t)=-25\log(1-t)$, and its second and third derivatives at zero give $\kappa_2=25$ and $\kappa_3=50$.

The central correction is $$ \frac{\kappa_3}{6\kappa_2^{3/2}} =\frac{50}{6(25)^{3/2}} =\frac{50}{6\cdot125} =\frac{1}{15} \approx 0.06667. $$

Since $\phi(0)=1/\sqrt{2\pi}\approx0.3989423$, the central CDF approximation is $$ \frac12+\frac{0.3989423}{15} \approx \frac12+0.0265962 =0.5265962. $$

The sum has positive skewness, so a probability above one half is consistent with the sign of the correction. The calculation uses cumulants of the sum, not of a single summand.

Worked Example: A Sum With Two Exponential Rates

Let $S$ be the sum of 10 independent rate-$1$ exponential variables and 10 independent rate-$2$ exponential variables. For an exponential variable with rate $\lambda$, the second and third cumulants are $1/\lambda^2$ and $2/\lambda^3$. Independence makes the cumulants add. Therefore

$$ \kappa_2=10(1)+10\left(\frac14\right)=12.5, \qquad \kappa_3=10(2)+10\left(\frac{2}{8}\right)=22.5. $$

Its mean is $10(1)+10(1/2)=15$, and the central correction is $$ \frac{22.5}{6(12.5)^{3/2}} \approx \frac{22.5}{265.1650} \approx 0.08485. $$

Multiplying by $\phi(0)$ gives $0.3989423(0.08485)\approx0.03386$. Thus the central Lugannani–Rice approximation is $P(S\le15)\approx0.53386$. The independent summands need not have a common distribution for this cumulant calculation; the required conditions here are a finite cumulant generating function near zero, positive variance, and a continuous model.

Stable Evaluation Near the Mean

Close to the mean, calculating $1/r$ and $1/q$ separately can lose numerical precision: each may be large even though their difference stays finite. The expansion above shows what to use instead when $t$ is sufficiently small:

\frac{1}{r}-\frac{1}{q} =\frac{\kappa_3}{6\kappa_2^{3/2}}+O(t).

At the mean, use the limit exactly; do not attempt to evaluate either reciprocal. For a nearby target, a numerical implementation can use a local series for the difference, with further terms if the desired precision requires them. It should also compute $tK'(t)-K(t)$ carefully near zero: direct subtraction of nearly equal quantities may be inaccurate, so a Taylor expansion or a numerically stable evaluation of the cumulant generating function is preferable.

For a sum of $n$ independent, identically distributed continuous variables with variance $\sigma^2>0$ and third cumulant $\kappa_3(X_1)$, the sum has $\kappa_2=n\sigma^2$ and $\kappa_3=n\kappa_3(X_1)$. Its central correction is therefore $$ \phi(0)\frac{\kappa_3(X_1)}{6\sigma^3\sqrt{n}}. $$ This displays the familiar order $n^{-1/2}$: the skewness adjustment decreases as the sample size grows.

Scope and Common Pitfalls

Substituting $t=0$ into the unextended formula. Both reciprocals are undefined there. Take their difference’s limit instead.
Dropping the factor of two inside $r$. Substituting $x=K'(t)$ gives $2\{tK'(t)-K(t)\}$ under the square root, not just the expression inside the braces.
Using the wrong cumulants. For a sum, the central correction uses the variance and third cumulant of the entire sum, or equivalently the scaled single-summand cumulants shown above.
Applying the continuous CDF formula unchanged to a lattice distribution. Discrete probabilities can have lattice effects at the same order as the central correction. A lattice setting requires a lattice-adjusted approximation; the continuous formula alone need not give the correct central probability.

The central limit of the Lugannani–Rice expression is valuable because it replaces an apparent singularity with a finite, interpretable skewness adjustment. It also provides a practical diagnostic: if a computation near the mean changes sharply when evaluated by direct subtraction rather than by the limiting expansion, numerical cancellation may be responsible.

Check Your Understanding

Use the definitions and central-limit derivation above to answer these questions.

Why are $1/r$ and $1/q$ individually undefined at the mean, while their difference has a finite limit?
For a model with $\kappa_2=9$ and $\kappa_3=27$, what is the central Lugannani–Rice approximation?
What sign should the central adjustment have for a distribution with negative third cumulant?
For $n$ independent identical summands, how does the central correction scale with $n$?
Why should the continuous Lugannani–Rice CDF expression not be applied unchanged to a lattice distribution?