Q&A Forum

Can someone share the class code of "Change of Varbile" in the example of phi(z) we did in the lecture? I got my result from code kind of inappropriate from the theretical value.

For now I get code like this:
g_2 <- function(x,z){
  return(z*exp(-((z*y)^2)/2))
}

y <- runif(M)
theta.2 <- mean(g_2(y,z))
Phi.2 <- 0.5 + theta.2/sqrt(2*pi)

Also could you also explain how this g_2 function come from?
Thank you    

I'm sorry Darth Knight, but I still cannot find the correct inverse function here.
I currently get the cdf to be F(x) = -exp[(-x^2)/16] + 1
And I always end up with the inverse function
f(x) = sqrt(-16*ln(1-x)) which does not make sense.
I guesss there exists some error on my integral but I do not know where it is...    

Just make a correction, the function should be
 inv.fun <- function(x){
  result <- -log(1-x)
  return <- 4*sqrt(result)
}
but the domain is still confusing

Hi Cat,

Your observation is correct. c has to be greater than 1.

My question is whether g(x)=1 is really true.
I recommend you plot the g(x) first.

Hi Matt,

Yes, the discrete case of the inverse transformation method can be confusing.
The reason is that while the idea uses the inverse CDF,
we actually don't need to do the actual inversion for the implementation.

See the generation scheme on page 18 of Generating Random Variable I.
The algorithm does not use inverse CDF anywhere.

What we need to do is:

• Find out the CDF values at the discontinuity points (in this case 0,1,2,3,...), and
  chop the interval (0,1) with those CDF values;
• Generate u from uniform(0,1);
• Track which sub-interval u falls into so that we can assign new random variables accordingly.

This procedure is actually what we did in our lab 9.
The sample code for logarithm distribution (we saw this on Tuesday) would be helpful
since the Poisson distribution is somewhat similar to the "logarithm" distribution (Both have infinite discontinuities.)

Please let me know if you need further clarification.