Lamperti Transforms for Approximate Transition Densities

Why transform before approximating a transition

For a discretely observed diffusion, a likelihood contribution depends on the transition density between successive observations. When the diffusion coefficient varies with the state, a Gaussian approximation made directly in the original state can misrepresent the local scale of the transition. For a scalar diffusion with a smooth, nonzero coefficient, the Lamperti transform changes coordinates so that the transformed process has unit diffusion. A Gaussian short-interval approximation is then natural in the transformed coordinate, but converting its density back requires a Jacobian.

The earlier material on Lamperti Coordinates and Boundary-Aware Simulation introduced the coordinate change in a simulation setting. Here the focus is the transition-density approximation used for discretely observed data. Throughout, suppose that on an open interval $I$, the process satisfies

dX_t=b(X_t)\,dt+\sigma(X_t)\,dW_t,

where $W_t$ is standard Brownian motion, $b$ is the drift, and $\sigma$ is the diffusion coefficient. Assume $\sigma(x)>0$ on $I$, and that $b$ and $\sigma$ are sufficiently smooth and locally Lipschitz for a non-explosive solution up to any boundary or exit time under consideration. The positivity assumption fixes an orientation; if $\sigma<0$ throughout the state interval, its absolute value can be used in the coordinate definition.

Definition: For a fixed reference point $x_0\in I$, the Lamperti coordinate is $$ \Phi(x)=\int_{x_0}^{x}\frac{du}{\sigma(u)}. $$ It is strictly increasing, with $\Phi'(x)=1/\sigma(x)$. Its image $J=\Phi(I)$ is an interval, and $Y_t=\Phi(X_t)$ is the transformed state while $X_t\in I$.

The transformed drift and its transition approximation

Because $\Phi'(x)=1/\sigma(x)$ and $\Phi''(x)=-\sigma'(x)/\sigma(x)^2$, Itô’s formula gives the transformed dynamics. The correction involving $\sigma'$ is essential: transforming the drift by simply dividing $b$ by $\sigma$ omits the Itô correction.

Result: If $\sigma$ is continuously differentiable and the Itô formula applies, then, up to exit from $I$, $$ dY_t=\mu(Y_t)\,dt+dW_t,\qquad \mu(\Phi(x))=\frac{b(x)}{\sigma(x)}-\frac{1}{2}\sigma'(x). $$

To verify the result, apply Itô’s formula to $Y_t=\Phi(X_t)$:

\begin{aligned} dY_t &=\Phi'(X_t)b(X_t)\,dt +\Phi'(X_t)\sigma(X_t)\,dW_t +\frac12\Phi''(X_t)\sigma(X_t)^2\,dt\\ &=\left(\frac{b(X_t)}{\sigma(X_t)}-\frac12\sigma'(X_t)\right)dt+dW_t. \end{aligned}

The stochastic coefficient is exactly one because $\Phi'(x)\sigma(x)=1$. Writing the drift as a function of $Y_t$ means evaluating the displayed expression at $x=\Phi^{-1}(Y_t)$. This proves the transformed SDE. $\square$

For an observation interval of length $h>0$, starting from $X_t=x$, the transformed Euler approximation freezes the transformed drift at the starting value:

Y_{t+h}\mid Y_t=\Phi(x) \ \approx\ N\!\left(\Phi(x)+\mu(\Phi(x))h,\ h\right).

This is an approximation, not an assertion that the transformed process has constant drift over the interval. Its Gaussian density in the transformed endpoint $y'=\Phi(x')$ is converted to a density in $x'$ by the change-of-variables rule. Since $dy'/dx'=1/\sigma(x')$, the endpoint coefficient appears in the denominator.

Approximate transition density: For $x,x'\in I$, the Lamperti short-interval approximation is $$ \widetilde p_h(x'\mid x)= \frac{1}{\sigma(x')\sqrt{2\pi h}} \exp\!\left\{-\frac{\left[\Phi(x')-\Phi(x)-\mu(\Phi(x))h\right]^2}{2h}\right\}. $$ It is obtained by approximating the transformed increment as Gaussian with frozen starting drift, then transforming the endpoint density back to the original state.

The Jacobian is not optional. For fixed $x$, if $q_h(y'\mid\Phi(x))$ denotes the approximating density in transformed coordinates, then $dy'=dx'/\sigma(x')$, so $\widetilde p_h(x'\mid x)=q_h(\Phi(x')\mid\Phi(x))/\sigma(x')$. This establishes the formula from the Gaussian approximation and the density change of variables. $\square$

Worked calculations

Worked Example: A Nonconstant Coefficient and the Endpoint Jacobian

Let $I=(0,\infty)$, $b(x)=0.4-0.2x$, and $\sigma(x)=0.5x$. Take $x=1.2$, $x'=1.3$, and $h=0.10$. Since $1/\sigma(x)=2/x$, one possible Lamperti coordinate is $\Phi(x)=2\log x$.

At the starting state, $\sigma(1.2)=0.6$, $b(1.2)=0.16$, and $\sigma'(1.2)=0.5$. Thus $\mu(\Phi(1.2))=0.16/0.6-0.5(0.5)=0.0166667$. The transformed endpoint increment is $2\log(1.3/1.2)=2\log(13/12)\approx0.160085$.

The centered increment in the exponent is therefore $0.160085-(0.0166667)(0.10)\approx0.158419$. Its squared value divided by $2h=0.20$ is approximately $0.1255$. The Gaussian standard-deviation factor is $\sqrt{2\pi(0.10)}\approx0.792665$, and the endpoint coefficient is $\sigma(1.3)=0.65$. Consequently,

$\widetilde p_h(1.3\mid1.2)\approx [0.65(0.792665)]^{-1}\exp(-0.1255) \approx1.941(0.8821)\approx1.712.$

The value is a density per unit of the original $x'$-scale. The factor $1/0.65$ converts from transformed scale; omitting it would give a density with respect to the wrong coordinate.

Worked Example: A Case Where the Approximation Is Exact

Consider geometric Brownian motion $dX_t=0.1X_t\,dt+0.4X_t\,dW_t$, with $x=2$, $x'=2.2$, and $h=0.25$. On $I=(0,\infty)$, $\Phi(x)=\log(x)/0.4$. The transformed drift is constant: $\mu=0.1/0.4-0.4/2=0.05$.

In this case the transformed SDE is exactly Brownian motion with constant drift $0.05$, so its Gaussian transition law is exact, not merely a frozen-drift approximation. The transformed increment is $\log(2.2/2)/0.4=\log(1.1)/0.4\approx0.238276$, and its mean over the interval is $0.05(0.25)=0.0125$. The endpoint Jacobian is $1/\sigma(2.2)=1/(0.4(2.2))=1/0.88$.

The Gaussian standard deviation is $\sqrt{0.25}=0.5$, or equivalently its density denominator is $\sqrt{2\pi(0.25)}\approx1.253314$. The exponent is $- (0.238276-0.0125)^2/(0.5)\approx-0.10195$. Therefore the exact original-scale density is approximately $[0.88(1.253314)]^{-1}\exp(-0.10195)\approx0.9067(0.9031)\approx0.819$. Here the Lamperti expression reproduces the exact density because the transformed drift is constant.

Worked Example: A Square-Root Coefficient Near a State Boundary

Take $b(x)=0.5(2-x)$, $\sigma(x)=\sqrt{x}$, and $I=(0,\infty)$. The Lamperti coordinate is $\Phi(x)=2\sqrt{x}$. The transformed drift is $\mu(\Phi(x))=b(x)/\sqrt{x}-1/(4\sqrt{x})$. At $x=4$, this is $[0.5(2-4)]/2-1/8=-0.5-0.125=-0.625$.

For $x'=4.4$ and $h=0.10$, the transformed increment is $2\sqrt{4.4}-2\sqrt{4}\approx4.195235-4=0.195235$. The centered increment is $0.195235-(-0.625)(0.10)=0.257735$, giving exponent $-0.257735^2/0.20\approx-0.3322$. Since $\sigma(4.4)=\sqrt{4.4}\approx2.097618$, the prefactor is $[2.097618(0.792665)]^{-1}\approx0.6016$. Thus the approximate density is $0.6016\exp(-0.3322)\approx0.432$.

The computation is local to positive states. A Gaussian approximation on the whole transformed line assigns positive probability to $Y_{t+h}<0$, even though the valid transformed state space here is $J=(0,\infty)$. If boundary behavior is relevant over the observation interval, an unrestricted Gaussian approximation may therefore need a boundary-aware alternative.

Using the approximation in a likelihood

For observations $x_0,\ldots,x_n$ at times with intervals $h_i=t_i-t_{i-1}>0$, the approximation supplies one transition factor for each adjacent pair:

\widetilde L(\theta)=\prod_{i=1}^{n} \widetilde p_{h_i}(x_i\mid x_{i-1};\theta).

The model quantities $b$, $\sigma$, and $\Phi$ in each factor depend on the parameter value $\theta$. Both the exponent and the Jacobian can therefore contribute to parameter estimation. In particular, dropping $1/\sigma(x_i)$ does not merely alter a constant: if the diffusion coefficient has unknown parameters, the omitted factor can change the likelihood’s dependence on $\theta$.

A direct Gaussian approximation in the original state instead freezes the original drift and diffusion, giving a normal increment with mean $b(x)h$ and variance $\sigma(x)^2h$. That approximation is simple but uses a local scale based only on the starting state. The Lamperti approximation accounts for the nonlinear coordinate and endpoint Jacobian. Neither approximation is universally accurate: the direct version can be useful for very short intervals with mild coefficient variation, while the transformed version can be more appropriate when state-dependent scale is prominent and the transform is well behaved.

Diagnostics, assumptions, and limitations

The approximation presumes that the path remains in the region where $\sigma$ is nonzero and the Lamperti map is smooth. If $\sigma$ vanishes, changes sign, or has a singular derivative within the relevant state range, the stated transform and drift formula may fail or require a restricted domain. Likewise, large observation intervals allow the transformed drift to vary substantially, so freezing it at the starting point may be inaccurate.

Useful checks include verifying that $\Phi'(x)\sigma(x)=1$, retaining the $-\sigma'(x)/2$ drift correction, and checking that the Jacobian uses the endpoint $x'$, not the starting point $x$. For fitted models, inspect transformed standardized increments \[ \frac{\Phi(x_i)-\Phi(x_{i-1})-\mu(\Phi(x_{i-1}))h_i}{\sqrt{h_i}}. \] Systematic trends in these residuals with state or interval length, or strong changes in estimates when long intervals are excluded, can signal inadequate local approximation. A Gaussian approximation may also place mass outside the transformed state space; apparent numerical stability does not make those transitions physically valid.

Key takeaway: Transform first, include the Itô correction in the transformed drift, and transform the endpoint density back with $1/\sigma(x')$. The result is a useful local approximation, not a substitute for checking interval length, coefficient regularity, and boundary behavior.

Check Your Understanding

Work each item, then compare with the brief solutions below.

For $\sigma(x)=3x$ on $x>0$, find a Lamperti coordinate.
For $b(x)=ax$ and $\sigma(x)=cx$, where $c>0$, derive the transformed drift.
Why does the approximate density contain $\sigma(x')$ rather than $\sigma(x)$ in its Jacobian?
Name two warning signs that the short-interval approximation may be unreliable.

Solutions. (1) Integrating $1/(3x)$ gives $\Phi(x)=\log(x)/3$, up to an additive constant. (2) Since $b/\sigma=a/c$ and $\sigma'=c$, the transformed drift is the constant $a/c-c/2$. (3) The density change of variables uses the derivative at the endpoint: $d\Phi(x')/dx'=1/\sigma(x')$. (4) Examples include transformed residuals that vary systematically with state, strong sensitivity to interval length, or appreciable Gaussian probability outside the transformed state space.