I have two random variables $s\sim \mathcal{N}(\nu,\sigma)$ and $a\sim \mathcal{U}(0,A)$, calculate a third r.v. $t=(1-s)/a$ and want to find its distribution $p(t|\nu,\sigma,A)$.

My reasoning is, that for any value of $a$, there is exactly one $s=(1-a)/t$ such that the $(a,s)$ pair will produce $t$ and I will therefore have to integrate over the probabilities for these values:

$$p(t|\nu,\sigma,A) = \int_0^A p_a(x)p_s((1-x)/t)\,dx\\$$

which can be expressed as the sum of two error functions.

But already when checking this step numerically, I get a discrepancy between estimated (by sampling) and calculated distribution:

![](http://i45.tinypic.com/2da0ew5.png)

*Why do the histogram and the calculated distribution do not match?*



This is the code I used:

    n=10000000

    A=0.7
    nu=.7
    sigma=.1

    # sampling from the target distribution
    s=rnorm( n, mean=nu, sd=sigma )
    a=runif( n, min=0, max=A )
    t=(b-a)/s
    hist(t,200, freq=F, xlim=c(0,5), ylim=c(0,2.5))


    # plot the analytical result
    analytic <- function(t, A, nu, sigma ){
      tmp <- function(x, A, t, nu, sigma){
         return (1/A*dnorm( (1-x)/t, mean=nu, sd=sigma) )
      }
      return( integrate( tmp, 0, A, A=A, t=t, nu=nu, sigma=sigma)$value)
    }

    x=seq(0,5,by=.01)
    y=rep(0,length(x))
    for( i in seq(1,length(x)) ){
      y[i]=analytic(x[i], A, nu, sigma)
    }
    lines( x, y, col="red", type="l")