is this a truncated normal?

Hi there,

After some algebra on a problem I have I derived the following distribution concering let's say the r.v. X.

s^(-1)*g*φ((x-g^(-1)*a)/(s*g^(-1))) *(1/ Φ(a/s)) (1).

where Φ is the cumulative normal and 1/k*φ((x-m)/k)=1/sqrt(2*π)*exp(-(x-m)^2/2*k^2), i.e. the normal density.

My question is :

Is the density in (1) some known distribution? I thought that it was a truncated normal distribution, but I am having second thoughts. I checked the wiki and the pdf of the truncated normal does not seem to match, is it? What would the mean and variance be?

Thanx in advance for any answers!!

I also have this form if it makes things easier..:

1/(s/b) φ( (x- (a/b^2)) /(s/b1) ) * ( 1/ Φ(a/(b*s) ) )
Last edited:


TS Contributor
Sorry the use of inverse ^(-1) is quite confusing here.
Do you mean f(x) = 1/(s/g)φ[x/(s/g) - (a/s)]/Φ(a/s)
Then it reduce to a 2-paremeter pdf of a truncated normal.
Let μ = a/s, σ = s/g.
Then f(x) = (1/σ)φ((x/σ) - μ)/Φ(μ) = (1/σ)φ((x - σμ)/σ)/Φ(μ)
Note (1/σ)φ((x - σμ)/σ) is the pdf of N(σμ, σ^2)

The support still need to be determined.
I assume you are taking the left-truncated normal
Let Z ~ N(0, 1)
Then Φ(μ) = Pr{Z ≤ μ} = Pr{Z ≥ -μ} = Pr{σZ + σμ ≥ -σμ + σμ = 0}
So the support should be the [0, ∞)