logisticNormal

R functions for the evaluation of the logistic-normal integral using the methods proposed in Pirjol (2013)

In this paper the logistic-normal integral is defined as $$\varphi(z,t)=\int_{-\infty}^\infty \frac{1}{1+e^x} e^{-\frac{1}{2t}(x-z)^2} \frac{dx}{\sqrt{2\pi t}}$$

The function $\varphi(z,t)$ has the properties:

• Its values for positive and negative argument $z$ are related as $\varphi(z,t) + \varphi(-z,t)=1$

• For any $z\in \mathbb{R}$, we have $\varphi(z+t,t)= e^{-z-\frac12 t} (1 - \varphi(z,t))$

Using $\varphi(0,t)=\frac12$ (which follows by taking $z=0$ in the first relation), repeated application of the recursion relation gives exact evaluations at $z=kt$ with $k\in \mathbb{Z}$. Remarkably, the recursion relation implies that knowledge of $\varphi(z,t)$ in an interval in $z$ of length $t$ is sufficient to determine it anywhere else. The recursion relation can be used to derive also a Poisson summation formula for $\varphi(z,t)$.

$\varphi(z,t)$ is related to the Mordell integral. This relation is discussed further in:

D. Pirjol, Addendum: The logistic-normal integral and its generalizations, Journal of Computational and Applied Mathematics 237, 460-469 (2013)

The function phiLogisticNormal(z,t,Nmax) evaluates $\varphi(z,t)$ by Poisson summation truncating the sum to $Nmax$ terms.

The function phiLNexact(t,k) with $k\in \mathbb{Z}$ gives the exact value of $\varphi(kt, t)$

The function phiLinInterp(x,t) computes $\varphi(x,t)$ by linear interpolation from the exact values at the points $(kt,(k+1)t)$ bracketing $x$

Generalized logistic-normal integrals

Define the generalized logistic-normal integrals $$\varphi_j(z,t) = \int_{-\infty}^\infty \frac{x^j}{1+e^x} e^{-\frac{1}{2t}(x-z)^2} \frac{dx}{\sqrt{2\pi t}} , \quad j=1,2,\cdots$$

The functions $\varphi_j(z,t)$ can be evaluated exactly on certain grids of points $z$ with step $t$:

• Even index $j=0,2,4,\cdots$ can be evaluated exactly at $z_k=kt$ with $k\in \mathbb{Z}$. For example $\varphi(0,t)=\frac12,\varphi_2(0,t)=\frac12 t$

• Odd index $j=1,3,5,\cdots$ can be evaluated exactly at $z_k=(k+1/2)t$ with $k\in \mathbb{Z}$. For example $\varphi_1(\frac12,t)=0$

The derivatives of the logistic-normal integral can be expressed in terms of these functions. For example, the first few derivatives are given below.

$$\partial_z \varphi(z,t) = -\frac{z}{t} \varphi(z,t) + \frac{1}{t} \varphi_1(z,t) \qquad (1)$$

$$\partial_z^2 \varphi(z,t) = (\frac{z^2}{t^2} - \frac{1}{t} ) \varphi(z,t) - \frac{2z}{t^2} \varphi_1(z,t) + \frac{1}{t^2} \varphi_2(z,t)\qquad (2)$$

$$t \partial_t \varphi(z,t) = \varphi_1(z,t) - (z+\frac12) \varphi(z,t)\qquad (3)$$

Relation to moments of logistic-normal random variables

The logistic-normal random variable $X \sim logitnorm(\mu,\sigma)$ is defined as $X=\frac{1}{1+e^{-Z}}$ with $Z\sim N(\mu,\sigma)$. The expectation and higher moments of $X$ are expressed in terms of the logistic-normal integral and its generalizations. The first two moments are

$$\mathbb{E}[X]=\varphi(-\mu,\sigma^2)$$

$$\mathbb{E}[X^2] = e^{\mu+\frac12\sigma^2} \Big( -\frac{1}{\sigma^2} \varphi_1(-\mu-\sigma^2,\sigma^2) - (1 + \frac{\mu}{\sigma^2}) \varphi(-\mu-\sigma^2,\sigma^2)\Big) $$

Sample usage

# plot the logistic-normal integral 
x <- seq(-5,5, 0.1)
n <- length(x)

d <- c()
for (i in 1:n) d <- c(d,phiLogisticNormal(x[i], 1.0, 10))
plot(d~x, type ="l",main="Logistic-normal integral (t=1)")

y <- seq(-5,5,1)

ex <- c()
for (i in 1:11) ex <- c(ex, phiLNexact(1.0,y[i]))
                        
points(ex~y, type = "p", col="blue")

Numerical evaluation comparing with exact results (blue dots)

A more detailed comparison

# compare the Poisson summation formula with the exact evaluation (t=1.0 - 50.0)
t <- 1.0
v1 <- c()
v2 <- c()
for (k in 1:50) v1 <- c(v1,phiLNexact(t, k))
for (k in 1:50) v2 <- c(v2,phiLogisticNormal(k*t, t, 10))
v3 <- v1 - v2
vd <- data.frame("exact"=v1, "Poisson" = v2, "diff"=v3)
vd

Keeping N=10 terms in the Poisson sum, the error is reasonably small for $t$ sufficiently large ($t > 0.5$), and all $x$.

For small t, the exact evaluation gives the values of $\varphi(z,t)$ on a fine grid with step $t$. Its values at other points can be obtained by interpolation. In the example below the blue dots show the exact values for $\varphi(k t, t)$ with $t=0.1$.

The function varphi1(z,t,Nmax,eps) computes $\varphi_1(z,t)$ using (1) by central finite differences in $z$ with step $eps$. Default values $Nmax=12,eps=10^{-6}$

The function momentsLogisticNorm(mu,sig,Nmax) computes the mean and variance of a logistic-normal random variable $Z \sim logitnorm(\mu,\sigma)$

> momentsLogisticNorm(0.6,0.5)
[1] 0.6381209 0.0120817