CDF as a result of a Cauchy problem: how to solve it?

Question

I'm studying a particular class of random variables.

In order to find the CDF $F(x)$ of my variable, I should solve the following Cauchy problem:

$$ \begin{cases} F(x)=e^{-\lambda F'(x)} \\ F(0)=0 \end{cases} $$ where $\lambda>0$.

I solved the ODE, using the equivalent form

$$logF(x)=-\lambda F'(x)$$

The solution is

$$li^{-1}\Bigl(c-\frac{x}{\lambda}\Bigr)$$

where $c$ is the constant and $li$ is the logarithmic integral function.

In order to find the value of the constant $c$, I followed the steps below:

$$ F(0)=li^{-1}(c)=\frac{1}{li(c)}=0 \Leftrightarrow li(c)=\infty \Leftrightarrow c=1$$

So the resulting CDF should be

$$F(x)=li^{-1}\Bigl(1-\frac{x}{\lambda}\Bigr)$$

If I plot the aforementioned $F(x)$ it doesn't look like a valid CDF.

Are my reasonings right? Is there another way to solve the ODE $F(x)=e^{-\lambda F'(x)}$?

Answer 1

I think the steps you want are the following:

$$F(0)=li^{-1}(c)$$ $$li(F(0))=c$$ $$li(0)=c$$ $$0=c$$

$F(x)$ for various values can be plotted with Mathematica in the following manner:

plotCDF[λ_, color_] := ParametricPlot[{-λ LogIntegral[p], p}, {p, 0, 1},       
  PlotStyle -> color, Frame -> True, PlotLegends -> {"λ = " <> ToString[λ]}];
Show[{plotCDF[3, Red], plotCDF[2, Green], plotCDF[1, Blue]},
 PlotRangePadding -> None, AspectRatio -> 1/2, PlotLegends -> Automatic]

Using R one can evaluate the cdf and pdf using the pracma package:

# Load library
  library(pracma)

# Set lambda
  lambda <- 0.5

# Generate x values based on the cdf
  cdf <- (1:9999)/10000
  x <- -as.numeric(li(p)) * lambda

# Plot of CDF
  plot(x, cdf, type="l", las=1, xlab="x", ylab="", lwd=3, main=paste("lambda =", lambda))

# Add in pdf
  lines(x,-log(p)/lambda)
  
# Legend
  legend(min(x)+0.75*(max(x)-min(x)),0.8, c("PDF", "CDF"), lwd=c(1,3))

Both the Mathematica and R code avoid the direct calculation of the inverse of the logarithmic integral (which can be problematic).