has a global maximum at 0. is shown on top for M = 0.5, and at the bottom for M = 3 (both in blue). As M grows the approximation of this function by a Gaussian function (shown in red) improves. This observation underlies Laplace's method.
Suppose the function has a unique global maximum at x0. Let M be a constant and consider the following two functions:
Note that x0 will be the global maximum of and as well. Now observe:
As M increases the ratio for will grow exponentially while the ratio for does not change. Thus, significant contributions to the integral of this function will come only from points x in a neighbourhood of x0, which can then be estimated.
To state and motivate the method, we need several assumptions. We will assume that x0 is not an endpoint of the interval of integration, that the values cannot be very close to unless x is close to x0, and that
Since has a global maximum at x0, and since x0 is not an endpoint, it is a stationary point, so the derivative of vanishes at x0. Therefore, the function may be approximated to quadratic order
for x close to x0 (recall ). The assumptions ensure the accuracy of the approximation
(see the picture on the right). This latter integral is a Gaussian integral if the limits of integration go from −∞ to +∞ (which can be assumed because the exponential decays very fast away from x0), and thus it can be calculated. We find
A generalization of this method and extension to arbitrary precision is provided by Fog (2008).
Suppose is a twice continuously differentiable function on and there exists a unique point such that:
Lower bound: Let . Since is continuous there exists such that if then By Taylor's Theorem, for any
Then we have the following lower bound:
where the last equality was obtained by a change of variables
Remember so we can take the square root of its negation.
If we divide both sides of the above inequality by
and take the limit we get:
since this is true for arbitrary we get the lower bound:
Note that this proof works also when or (or both).
Upper bound: The proof is similar to that of the lower bound but there are a few inconveniences. Again we start by picking an but in order for the proof to work we need small enough so that Then, as above, by continuity of and Taylor's Theorem we can find so that if , then
Lastly, by our assumptions (assuming are finite) there exists an such that if , then .
Then we can calculate the following upper bound:
If we divide both sides of the above inequality by
and take the limit we get:
Since is arbitrary we get the upper bound:
And combining this with the lower bound gives the result.
Note that the above proof obviously fails when or (or both). To deal with these cases, we need some extra assumptions. A sufficient (not necessary) assumption is that for
and that the number as above exists (note that this must be an assumption in the case when the interval is infinite). The proof proceeds otherwise as above, but with a s lightly different approximation of integrals:
When we divide by
we get for this term
whose limit as is . The rest of the proof (the analysis of the interesting term) proceeds as above.
The given condition in the infinite interval case is, as said above, sufficient but not necessary. However, the condition is fulfilled in many, if not in most, applications: the condition simply says that the integral we are studying must be well-defined (not infinite) and that the maximum of the function at must be a "true" maximum (the number must exist). There is no need to demand that the integral is finite for but it is enough to demand that the integral is finite for some
Let’s look at the Taylor expansion of around x0 and translate x to y because we do the comparison in y-space, we will get
Note that because is a stationary point. From this equation you will find that the terms higher than second derivative in this Taylor expansion is suppressed as the order of so that will get closer to the Gaussian function as shown in figure. Besides,
The figure of with equals 1, 2 and 3, and the red line is the curve of function .
3. The larger is, the smaller range of is related
Because we do the comparison in y-space, is fixed in which will cause ; however, is inversely proportional to , the chosen region of will be smaller when is increased.
4. If the integral in Laplace’s method converges, the contribution of the region which is not around the stationary point of the integration of its relative error will tend to zero as grows.
Relying on the 3rd concept, even if we choose a very large Dy, sDy will finally be a very small number when is increased to a huge number. Then, how can we guarantee the integral of the rest will tend to 0 when is large enough?
The basic idea is to find a function such that and the integral of will tend to zero when grows. Because the exponential function of will be always larger than zero as long as is a real number, and this exponential function is proportional to the integral of will tend to zero. For simplicity, choose as a tangent through the point as shown in the figure:
is denoted by the two tangent lines passing through . When gets smaller, the cover region will be larger.
If the interval of the integration of this method is finite, we will find that no matter is continue in the rest region, it will be always smaller than shown above when is large enough. By the way, it will be proved later that the integral of will tend to zero when is large enough.
If the interval of the integration of this method is infinite, and might always cross to each other. If so, we cannot guarantee that the integral of will tend to zero finally. For example, in the case of will always diverge. Therefore, we need to require that can converge for the infinite interval case. If so, this integral will tend to zero when is large enough and we can choose this as the cross of and
You might ask that why not choose as a convergent integral? Let me use an example to show you the reason. Suppose the rest part of is then and its integral will diverge; however, when the integral of converges. So, the integral of some functions will diverge when is not a large number, but they will converge when is large enough.
Based on these four concepts, we can derive the relative error of this Laplace's method.
Importantly, the accuracy of the approximation depends on the variable of integration, that is, on what stays in and what goes into .
The derivation of its relative error
First of all, let me set the global maximum is located at which can simplify the derivation and does not lost any important information; therefore, all the derivation inside this sub-section is under this assumption. Besides, what we want is the relative error as shown below
So, if we let
and , we can get
since . Now, let us find its upper bound.
Owing to we can separate this integration into 5 parts with 3 different types (a), (b) and (c), respectively. Therefore,
where and are similar, let us just calculate and and are similar, too, I’ll just calculate .
For , after the translation of , we can get
This means that as long as is large enough, it will tend to zero.
For , we can get
and should have the same sign of during this region. Let us choose as the tangent across the point at , i.e. which is shown in the figure
is the tangent lines across the point at .
From this figure you can find that when or gets smaller, the region satisfies the above inequality will get larger. Therefore, if we want to find a suitable to cover the whole during the interval of , will have an upper limit. Besides, because the integration of is simple, let me use it to estimate the relative error contributed by this .
Based on Taylor expansion, we can get
and then substitute them back into the calculation of ; however, you can find that the remainders of these two expansions are both inversely proportional to the square root of , let me drop them out to beautify the calculation. Keeping them is better, but it will make the formula uglier.
Therefore, it will tend to zero when gets larger, but don't forget that the upper bound of should be considered during this calculation.
About the integration near , we can also use Taylor's Theorem to calculate it. When
and you can find that it is inversely proportional to the square root of . In fact, will have the same behave when is a constant.
Conclusively, the integral near the stationary point will get smaller as gets larger, and the rest parts will tend to zero as long as is large enough; however, we need to remember that has an upper limit which is decided by whether the function is always larger than in the rest region. However, as long as we can find one satisfying this condition, the upper bound of can be chosen as directly proportional to since is a tangent across the point of at . So, the bigger is, the bigger can be.
In the multivariate case where is a -dimensional vector and is a scalar function of , Laplace's approximation is usually written as:
In extensions of Laplace's method, complex analysis, and in particular Cauchy's integral formula, is used to find a contour of steepest descent for an (asymptotically with large M) equivalent integral, expressed as a line integral. In particular, if no point x0 where the derivative of vanishes exists on the real line, it may be necessary to deform the integration contour to an optimal one, where the above analysis will be possible. Again the main idea is to reduce, at least asymptotically, the calculation of the given integral to that of a simpler integral that can be explicitly evaluated. See the book of Erdelyi (1956) for a simple discussion (where the method is termed steepest descents).
The appropriate formulation for the complex z-plane is
for a path passing through the saddle point at z0. Note the explicit appearance of a minus sign to indicate the direction of the second derivative: one must not take the modulus. Also note that if the integrand is meromorphic, one may have to add residues corresponding to poles traversed while deforming the contour (see for example section 3 of Okounkov's paper Symmetric functions and random partitions).
An extension of the steepest descent method is the so-called nonlinear stationary phase/steepest descent method. Here, instead of integrals, one needs to evaluate asymptotically solutions of Riemann–Hilbert factorization problems.
Given a contour C in the complex sphere, a function defined on that contour and a special point, say infinity, one seeks a function M holomorphic away from the contour C, with prescribed jump across C, and with a given normalization at infinity. If and hence M are matrices rather than scalars this is a problem that in general does not admit an explicit solution.
An asymptotic evaluation is then possible along the lines of the linear stationary phase/steepest descent method. The idea is to reduce asymptotically the solution of the given Riemann–Hilbert problem to that of a simpler, explicitly solvable, Riemann–Hilbert problem. Cauchy's theorem is used to justify deformations of the jump contour.
The nonlinear stationary phase was introduced by Deift and Zhou in 1993, based on earlier work of Its. A (properly speaking) nonlinear steepest descent method was introduced by Kamvissis, K. McLaughlin and P. Miller in 2003, based on previous work of Lax, Levermore, Deift, Venakides and Zhou. As in the linear case, "steepest descent contours" solve a min-max problem. In the nonlinear case they turn out to be "S-curves" (defined in a different context back in the 80s by Stahl, Gonchar and Rakhmanov).
Azevedo-Filho, A.; Shachter, R. (1994), "Laplace's Method Approximations for Probabilistic Inference in Belief Networks with Continuous Variables", in Mantaras, R.; Poole, D. (eds.), Uncertainty in Artificial Intelligence, San Francisco, CA: Morgan Kaufmann, CiteSeerX10.1.1.91.2064.