Why a Point Probability Needs a Lattice Adjustment
For a continuous random variable, a saddlepoint approximation estimates a density: probability is spread over intervals, and a density is probability per unit of the variable. A discrete random variable instead assigns probability directly to individual support points. For a sum on a lattice with spacing \(h\), the point mass is approximately the local density multiplied by \(h\). Omitting this factor gives the wrong scale whenever the lattice spacing is not one.
The adjustment follows naturally from exponential tilting. Tilting makes the point of interest the mean under a new distribution; a local limit approximation then estimates the probability of that point. The exponential change of measure supplies the large-deviation term, while the local limit supplies both the Gaussian curvature term and the lattice span.
Setup and the Point-Probability Formula
Let \(X_1,\ldots,X_n\) be independent and identically distributed copies of a nondegenerate lattice random variable \(X\), and write \(S_n=\sum_{i=1}^n X_i\). Suppose its cumulant generating function \(K(t)=\log E[e^{tX}]\) is finite on an open interval containing the value of \(t\) under consideration. For a target point \(s=nx\) in the support of \(S_n\), the saddlepoint equation is \(K'(t)=x\). When this equation has a solution, that value of \(t\) makes the tilted mean equal to the target per summand.
The finite-support assumption gives a simple set of sufficient conditions and a direct proof below. Under suitable moment and characteristic-function regularity conditions, the same approximation holds for broader lattice distributions; the formula should not be taken as an unconditional guarantee for arbitrary distributions.
Derivation by Exponential Tilting
For \(t\) with \(K(t)<\infty\), define the tilted distribution of one summand by $$ P_t(X=y)=e^{ty-K(t)}P(X=y). $$ Its probabilities sum to one because \(E[e^{tX-K(t)}]=1\). Differentiating the finite moment generating function gives \(E_t[X]=K'(t)\) and \(\operatorname{Var}_t(X)=K''(t)\). For a sequence of independent summands, each tilted in this way, the likelihood ratio between the original and tilted joint distributions at a sample with sum \(s\) is \(e^{-ts+nK(t)}\). Therefore the following identity is exact: $$ P(S_n=s)=e^{nK(t)-ts}P_t(S_n=s). $$
Choose \(t\) so that \(K'(t)=x=s/n\). Under the tilted distribution, \(S_n\) has mean \(nx=s\) and variance \(nK''(t)\). A local limit approximation at this mean gives $$ P_t(S_n=s)\ \sim\ \frac{h}{\sqrt{2\pi nK''(t)}}. $$ Combining it with the exact change-of-measure identity gives the theorem’s formula. In particular, the exponential term describes the cost of reaching the target under the original distribution, and the prefactor describes the local concentration around that target under the tilted distribution.
Why the Local Approximation Contains \(h\)
The span factor can be justified directly by Fourier inversion. Under the tilted law, let \(\phi_t(u)=E_t[e^{iuX}]\). For a lattice variable of maximal span \(h\), inversion over one frequency period gives, at the tilted mean \(s=nx\), $$ P_t(S_n=s)=\frac{h}{2\pi}\int_{-\pi/h}^{\pi/h}\bigl(e^{-iux}\phi_t(u)\bigr)^n\,du. $$ The factor \(h/(2\pi)\) is part of lattice Fourier inversion; it is not a correction added after the calculation.
Here are the details of the asymptotic evaluation under the theorem’s finite-support assumptions. The centered characteristic function \(e^{-iux}\phi_t(u)\) has expansion $$ e^{-iux}\phi_t(u)=1-\frac{K''(t)u^2}{2}+O(|u|^3) $$ as \(u\) tends to zero. Consequently, after setting \(u=v/\sqrt{nK''(t)}\), its \(n\)th power tends to \(e^{-v^2/2}\) for each fixed \(v\). In a sufficiently small neighbourhood of zero, its modulus is bounded by \(e^{-c u^2}\) for some \(c>0\). Away from that neighbourhood in \([-\pi/h,\pi/h]\), its modulus is strictly less than one: equality at a nonzero frequency would imply that the support lies on a lattice with a larger span, contrary to maximality of \(h\). Compactness then bounds the modulus there by some \(\rho<1\).
Thus the part of the inversion integral away from zero is exponentially small in \(n\). In the neighbourhood of zero, the change of variable above and the Gaussian bound permit dominated convergence, and the limiting integral is \(\int_{-\infty}^{\infty}e^{-v^2/2}\,dv=\sqrt{2\pi}\). Including the inversion factor gives $$ P_t(S_n=s)\ \sim\ \frac{h}{2\pi\sqrt{nK''(t)}}\sqrt{2\pi} =\frac{h}{\sqrt{2\pi nK''(t)}}. $$ This proves the local approximation used in the derivation. Together with the exact tilting identity, it proves the stated point-probability approximation. \(\square\)
Worked Applications
Worked Example: A Binomial Point Probability
Let \(S_n\sim\operatorname{Binomial}(n,p)\), the sum of \(n\) independent Bernoulli variables with success probability \(p\). Its summands have maximal span \(h=1\), and their cumulant generating function is $$ K(t)=\log(1-p+pe^t). $$ For a target \(k\), put \(x=k/n\), with \(0<k<n\). The saddlepoint equation is $$ K'(t)=\frac{pe^t}{1-p+pe^t}=x, \qquad e^t=\frac{x(1-p)}{p(1-x)}. $$ At this solution, \(K''(t)=x(1-x)\). Substitution in the exponent gives \(n[K(t)-tx]=-nD(x\|p)\), where $$ D(x\|p)=x\log\!\left(\frac{x}{p}\right)+(1-x)\log\!\left(\frac{1-x}{1-p}\right). $$ Therefore $$ P(S_n=k)\ \approx\ \frac{\exp\{-nD(x\|p)\}}{\sqrt{2\pi n x(1-x)}}. $$
For \(n=20\), \(p=0.3\), and \(k=10\), \(x=0.5\), and $$ D(0.5\|0.3)=\tfrac12\log(5/3)+\tfrac12\log(5/7) =\tfrac12\log(25/21)\approx 0.08718. $$ The approximation is consequently $$ \frac{\exp\{-10\log(25/21)\}}{\sqrt{2\pi(20)(0.5)(0.5)}} =\frac{(21/25)^{10}}{\sqrt{10\pi}} \approx \frac{0.17490}{5.60499} \approx 0.03121. $$ The exact mass is \(\binom{20}{10}(0.3)^{10}(0.7)^{10}\approx 0.03082\). The approximation is about \(1.3\%\) above the exact value in this example.
Worked Example: A Poisson Sum
Suppose \(X_1,\ldots,X_n\) are independent with \(X_i\sim\operatorname{Poisson}(\lambda)\). Their sum has lattice span \(1\), and for one summand $$ K(t)=\lambda(e^t-1),\qquad K'(t)=\lambda e^t,\qquad K''(t)=\lambda e^t. $$ For a target \(s=nx\), the saddlepoint is \(t=\log(x/\lambda)\), provided \(x>0\). Then \(K''(t)=x\), and $$ n[K(t)-tx] =n[x-\lambda-x\log(x/\lambda)] =-n\{x\log(x/\lambda)-x+\lambda\}. $$ The point-probability approximation is $$ P(S_n=nx)\ \approx\ \frac{\exp[-n\{x\log(x/\lambda)-x+\lambda\}]}{\sqrt{2\pi nx}}. $$
For example, take \(n=12\), \(\lambda=2\), and target \(s=30\). Then \(x=30/12=2.5\), and \(t=\log(2.5/2)=\log(1.25)\). The exponent is $$ -12\{2.5\log(1.25)-2.5+2\} =-12\{2.5\log(1.25)-0.5\}\approx -0.69431. $$ The prefactor is \(1/\sqrt{2\pi(30)}\approx 0.07284\), so the approximation is \(e^{-0.69431}/\sqrt{60\pi}\approx 0.03637\). The exact mass, available because \(S_{12}\sim\operatorname{Poisson}(24)\), is \(e^{-24}24^{30}/30!\). The approximation is constructed for the mass at \(30\), not for the probability of an interval around \(30\).
Worked Example: A Lattice with Span Two
Let \(X_i=2B_i\), where \(B_i\) is Bernoulli with success probability \(p\). Each \(X_i\) takes values \(0\) and \(2\), so its maximal lattice span is \(h=2\). Write \(S_n=2\sum_i B_i\), and consider the admissible target \(s=2k\), where \(x=s/n=2k/n\). The cumulant generating function and its second derivative are $$ K(t)=\log(1-p+pe^{2t}),\qquad K''(t)=4q(1-q), $$ where \(q=pe^{2t}/(1-p+pe^{2t})\). The saddlepoint equation \(K'(t)=2q=x\) sets \(q=k/n\). The exponential term is the same binomial deviance as for \(\sum_i B_i\), but the prefactor is $$ \frac{2}{\sqrt{2\pi n\,4q(1-q)}} =\frac{1}{\sqrt{2\pi n q(1-q)}}. $$
This agrees with the fact that \(P(S_n=2k)=P(\sum_i B_i=k)\): changing the units of each summand doubles the span and multiplies the variance by four, so the two changes cancel in the prefactor. Using span \(1\) for \(X_i\) would instead produce half the correct leading approximation.
Common Pitfall and Practical Interpretation
A common mistake is to write the continuous saddlepoint density prefactor \(1/\sqrt{2\pi nK''(t)}\) as the approximation to a lattice point mass, without checking the support spacing. That quantity has units of inverse \(X\)-units; a point mass is dimensionless. Multiplying by \(h\) converts the local density scale to the probability assigned to one lattice point. In particular, the span is the maximal spacing of the support, not a convenient spacing chosen for the calculation.
The formula estimates an individual admissible point probability. It is not a continuity-corrected tail approximation, and it does not by itself approximate a cumulative probability. Near the mean, the saddlepoint is \(t=0\), so the exponential factor is one and the formula reduces to the local normal approximation \(h/\sqrt{2\pi nK''(0)}\). Away from the mean, the saddlepoint changes the exponential rate as well as the variance used in the prefactor. As with other asymptotic approximations, accuracy at a particular finite \(n\) should be assessed in context rather than presumed.
Check Your Understanding
Check that you can connect the saddlepoint, the tilted local probability, and the lattice spacing.
- For a sum of Bernoulli variables, what is the maximal lattice span, and why?
- In the point-probability formula, which factor comes from exponential tilting, and which factor comes from the tilted local limit approximation?
- Why does a variable supported on the even integers have a different leading local approximation from an otherwise comparable variable supported on every integer?
- For a Poisson sum with target \(s=nx>0\), what equation determines the saddlepoint \(t\)?
- Explain why the point-mass formula is not itself an approximation to a tail probability.