#############################################################
## the response y has a Poisson disitribution
## but a Poisson regression is not appropriate
## a plain linear regression with Gaussian residual is fine
n=1000
x = rpois(n=n,lambda=5)
eps = rnorm(n,sd=.1)
y = x+eps

par(mfrow=c(2,2))
hist(y,breaks=seq(-.5,15.5,1),prob=TRUE)
title(sub='Red: Poisson mass distrib.')
h=0:15
lines(h,dpois(h,lambda=5),col=2,type='h')
points(h,dpois(h,lambda=5),col=2,pch=16)
hist(x,breaks=seq(-.5,15.5,1),prob=TRUE)
lines(h,dpois(h,lambda=5),col=2,type='h')
points(h,dpois(h,lambda=5),col=2,pch=16)
plot(x,y)
hist(eps,prob=TRUE)
h=seq(-.5,.5,.01)
lines(h,dnorm(h,sd=0.1),col=2,sub='Red: normal desnity')

plot(lm(y~x))

#######################################################
## now the opposite situation (a bit more artificial)
n=1000
x = rnorm(n=n,mean=3,sd=.001)
y = rpois(n=n,lambda=exp(x))

par(mfrow=c(2,2))
hist(y,breaks=40,prob=TRUE)
# title(sub='Red: Poisson mass distrib.')
h=(0:400)/10
lines(h,dnorm(h,mean=mean(y),sd=sd(y)),col=2)
# points(h,dpois(h,lambda=5),col=2,pch=16)
hist(x)
# lines(h,dpois(h,lambda=5),col=2,type='h')
# points(h,dpois(h,lambda=5),col=2,pch=16)
plot(x,y)
hist(eps,prob=TRUE)