Right-truncating a lognormal distribution

Hello,

I would like to compute the probability that P(z<a) where

z=exp(x*beta+e) and e is distributed iid N(0,sigma^2).

I would like to evaluate the CDF of z, i.e.,

P(e< log(a)-x*beta),

for various values of x*beta. However, I know that z cannot take a value greater than a certain number, b. How can I obtain the truncated distribution to evaluate the above probability?

Many thanks in advance.

[Matt J]

However, I know that z cannot take a value greater than a certain number, b.

This needs to be explained. If z is log-normal, how can it be bounded by b?

[george pepper]

From an underlying theory, I know that observations where z is greater than a constant, b, cannot exist in my data.

[Matt J]

But that contradicts your original claim that z is lognormal. Clearly z is not lognormal if it is bounded by b. You must say what the distribution actually is before we can talk about how to compute the CDF.

[Matt J]

Possibly you want a conditional CDF?

prob(z<a | z<b)

[george pepper]

I think this is right. How can I code this?

Answer 1

[Matt J]

I think this is right. How can I code this?

function out=conditionalCDF(a,b,x,beta,sigma)

    pa=normcdf(log(a)-x*beta,0,sigma);
    pb=normcdf(log(b)-x*beta,0,sigma);

    out=min(1, pa./pb);

end

[george pepper]

Thanks so much for your time, Matt! Have a great day.