The Lugannani–Rice Tail Approximation

Why approximate a tail by a saddlepoint?

A normal approximation can describe the center of a distribution reasonably well and still misstate a tail probability. This is especially consequential when the observed value is several standard deviations from the mean: the probability may be small, and a modest relative error can matter. The Lugannani–Rice approximation uses the cumulant-generating function to adapt the approximation to the particular tail being evaluated.

Its ingredients are the saddlepoint $t_x$, which makes the tilted distribution have mean $x$, a signed root $r$ measuring the distance to $x$ on a deviance scale, and a correction $q$ that accounts for local curvature at the saddlepoint. The formula combines these quantities with the standard normal distribution function $\Phi$ and density $\phi$. Throughout, the derivation below is for a continuous distribution with a density; nonlattice support alone is not enough for this density-based argument.

Definitions and conditions

Definition: Let $S$ have a moment-generating function finite on an open interval containing $0$, and define its cumulant-generating function by $K(t)=\log E[e^{tS}]$. For a value $x$ in the interior of the range of $K'$, a saddlepoint $t_x$ solves $K'(t_x)=x$. If $K''(t_x)>0$, define $$ r=\operatorname{sgn}(t_x)\sqrt{2\{t_xx-K(t_x)\}},\qquad q=t_x\sqrt{K''(t_x)}. $$ For $t_x\ne 0$, the Lugannani–Rice approximation to the lower tail is $$ P(S\le x)\ \approx\ \Phi(r)+\phi(r)\left(\frac{1}{r}-\frac{1}{q}\right). $$ The corresponding upper-tail approximation is $$ P(S\ge x)\ \approx\ 1-\Phi(r)+\phi(r)\left(\frac{1}{q}-\frac{1}{r}\right). $$

The upper-tail expression follows by subtracting the lower-tail approximation from one; the two correction terms therefore have opposite signs. When $S$ has a continuous distribution, $P(S\le x)=P(S\ge x)$ is not generally true, but $P(S\ge x)=1-P(S\le x)$ does hold because $P(S=x)=0$.

For sums of independent identically distributed continuous random variables, one commonly takes $S=S_n=\sum_{i=1}^nX_i$, with $K(t)=n\kappa(t)$, where $\kappa(t)=\log E[e^{tX_1}]$. The derivation requires a nondegenerate distribution, a saddlepoint in the interior of the CGF domain, and sufficient smoothness and Fourier regularity for a local saddlepoint density approximation under exponential tilting. For example, the tilted distributions must have densities and satisfy conditions that justify local Fourier inversion and the corresponding saddlepoint density approximation. The formula is an asymptotic approximation under such regularity conditions, not an exact identity for finite $n$.

From a saddlepoint density to the tail formula

The derivation has two stages. First, the density approximation is expressed in a coordinate that measures distance from the saddlepoint. Second, integrating that transformed density gives a normal term plus an endpoint correction. The integration step also displays the remainder that is being neglected in the approximation.

Suppose the continuous density of $S$ has the saddlepoint approximation $$ f_S(y)\ \approx\ \frac{\exp\{K(t_y)-t_yy\}}{\sqrt{2\pi K''(t_y)}}, \qquad K'(t_y)=y. $$ This is a density statement and needs density and regularity conditions; it does not follow merely from a nonlattice assumption. Define the signed deviance coordinate $$ w(y)=\operatorname{sgn}(t_y)\sqrt{2\{t_yy-K(t_y)\}}. $$ Since $K'(t_y)=y$, differentiating $w(y)^2/2=t_yy-K(t_y)$ gives $w\,dw/dy=t_y$. Thus $dy/dw=w/t_y$, with the positive Jacobian understood on either side of the mean. Also, $K(t_y)-t_yy=-w^2/2$. Transforming the approximate density therefore gives $$ f_W(w)\ \approx\ \phi(w)\frac{w}{q(w)},\qquad q(w)=t_y\sqrt{K''(t_y)}. $$ The factor $w/q(w)$ is positive and tends to one at the mean in the limiting sense; away from the mean it records the difference between the deviance coordinate and the local curvature scale.

At the threshold $x$, the transformed endpoint is $w(x)=r$ and $q(r)=q$. Integrating the transformed density from the left gives the following approximation to the lower tail: $$ P(S\le x)\ \approx\ \int_{-\infty}^{r}\phi(w)\frac{w}{q(w)}\,dw. $$ To expose the correction, write $h(w)=w/q(w)$ and $A(w)=(h(w)-1)/w$, with its continuous limiting value at zero where that value exists. Then $h(w)=1+wA(w)$, so integration by parts using $\phi'(w)=-w\phi(w)$ yields $$ \int_{-\infty}^{r}\phi(w)h(w)\,dw =\Phi(r)-\phi(r)A(r)+\int_{-\infty}^{r}\phi(w)A'(w)\,dw. $$ The boundary term at $-\infty$ vanishes under the stated regularity. At the endpoint, $$ -\phi(r)A(r) =-\phi(r)\frac{r/q-1}{r} =\phi(r)\left(\frac{1}{r}-\frac{1}{q}\right). $$ The Lugannani–Rice approximation retains the normal integral and this endpoint correction, while the remaining integral is of smaller order in the saddlepoint asymptotic regime under suitable regularity. This makes clear both the origin of the formula and why it is an approximation rather than an exact integration of the saddlepoint density.

Worked calculations

Worked Example: An Upper Tail for a Sum of Exponentials

Let $X_1,\ldots,X_{10}$ be independent exponential random variables with rate $1$, and let $S=\sum_{i=1}^{10}X_i$. Find the Lugannani–Rice approximation to $P(S\ge13)$.

For one exponential observation, $\log E[e^{tX_i}]=-\log(1-t)$ for $t<1$. Therefore $$ K(t)=-10\log(1-t),\qquad K'(t)=\frac{10}{1-t},\qquad K''(t)=\frac{10}{(1-t)^2}. $$ The saddlepoint equation $K'(t_x)=13$ gives $10/(1-t_x)=13$, hence $t_x=3/13$. At this value, $1-t_x=10/13$, so $$ K(t_x)=10\log(13/10),\qquad K''(t_x)=\frac{10}{(10/13)^2}=16.9. $$ The signed-root calculation is $$ t_xx-K(t_x)=3-10\log(1.3)\approx 0.37636,\qquad r=\sqrt{2(0.37636)}\approx 0.8676. $$ Since $t_x>0$, $r$ is positive. The curvature quantity is $$ q=\frac{3}{13}\sqrt{16.9}\approx 0.9487. $$ Using $\Phi(0.8676)\approx0.8072$ and $\phi(0.8676)\approx0.2739$, the upper-tail approximation is $$ P(S\ge13)\ \approx\ 1-0.8072+0.2739\left(\frac{1}{0.9487}-\frac{1}{0.8676}\right) \approx 0.1928-0.0270 \approx 0.1658. $$ The correction is negative here: $q>r>0$, so $1/q-1/r<0$. The sign follows directly from the upper-tail formula.

Worked Example: A Lower Tail for the Same Sum

For the same $S$, approximate $P(S\le8)$. The saddlepoint equation gives $10/(1-t_x)=8$, so $t_x=-1/4$, which is inside the CGF domain. We have $$ K(t_x)=-10\log(1.25)=10\log(0.8),\qquad K''(t_x)=\frac{10}{(1.25)^2}=6.4. $$ Then $$ t_xx-K(t_x)=-2-10\log(0.8)\approx0.23144,\qquad r=-\sqrt{2(0.23144)}\approx-0.6804, $$ where the negative sign is required because $t_x<0$. Also, $$ q=-\frac14\sqrt{6.4}\approx-0.6325. $$ With $\Phi(-0.6804)\approx0.2482$ and $\phi(-0.6804)\approx0.3163$, the lower-tail estimate is $$ P(S\le8)\ \approx\ 0.2482+0.3163\left(\frac{1}{-0.6804}-\frac{1}{-0.6325}\right) \approx 0.2482+0.0352 \approx0.2834. $$ Both $r$ and $q$ are negative, but their reciprocal difference is positive in this example. Keeping the signs through the calculation is essential.

Worked Example: The Normal Distribution Check

Suppose $S\sim N(\mu,\sigma^2)$, with $\sigma>0$, and take $x\ne\mu$. Its CGF is $K(t)=\mu t+\sigma^2t^2/2$, so the saddlepoint equation $\mu+\sigma^2t_x=x$ gives $t_x=(x-\mu)/\sigma^2$. It follows that $$ t_xx-K(t_x)=\frac{(x-\mu)^2}{2\sigma^2},\qquad r=\frac{x-\mu}{\sigma},\qquad q=t_x\sqrt{\sigma^2}=\frac{x-\mu}{\sigma}. $$ Thus $q=r$, the correction vanishes, and the Lugannani–Rice expressions reduce to the exact normal tails $P(S\le x)=\Phi((x-\mu)/\sigma)$ and $P(S\ge x)=1-\Phi((x-\mu)/\sigma)$. This check verifies the sign convention and shows that the formula preserves the normal case exactly.

Interpretation and a common pitfall

The root $r$ summarizes the CGF departure from its tangent at zero: its square is twice the CGF-based rate function $t_xx-K(t_x)$. The quantity $q$ instead combines the tilt $t_x$ with the standard deviation under that tilt, $\sqrt{K''(t_x)}$. For a normal distribution these scales agree. For a nonnormal distribution they need not, and their difference supplies the correction to the normal tail.

A common mistake is to treat “nonlattice” as if it meant “continuous,” then use a density approximation and integrate it. Those properties are different. For example, a random variable supported on $\{0,1,\sqrt{2}\}$ with positive probabilities is discrete and nonlattice, and its finite sums remain discrete; they have no density of the form used in the derivation. A lattice or discrete problem needs an appropriate point-mass or lattice-tail treatment. The continuous-density derivation here does not establish the formula for such a distribution.

A second practical caution is that the displayed formula contains $1/r$ and $1/q$, so it is not evaluated by direct substitution at $t_x=0$, the mean. The correction has a limiting interpretation under suitable conditions, but numerical evaluation near the mean requires care. Likewise, a well-defined saddlepoint alone does not guarantee a reliable approximation: density regularity and the asymptotic regime matter.

Key takeaway: For a continuous model satisfying suitable local regularity conditions, solve $K'(t_x)=x$, calculate $r$ and $q$ with their signs intact, and use the appropriate lower- or upper-tail formula. The correction comes from integrating the transformed saddlepoint density; it is not a license to use a density approximation for discrete nonlattice distributions.

Check Your Understanding

Work through these questions using the definitions and sign conventions above.

For a lower-tail threshold below the mean, what are the signs of $t_x$, $r$, and $q$, assuming $K''(t_x)>0$?
Explain why the transformed saddlepoint density contains the factor $w/q(w)$ in addition to $\phi(w)$.
For the exponential-sum example at $x=13$, verify that $1/q-1/r<0$. What does that imply for the upper-tail correction?
Why does a nonlattice distribution with finite support not meet the density requirement in the derivation?
For a normal distribution, show why the Lugannani–Rice correction term is zero at every $x\ne\mu$.