CDF as a result of a Cauchy problem: how to solve it? 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 $F(x)=e^{-\lambda F'(x)}$?
 A: 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).
