# Moments of Limited lognormal distribution

How do you calculate the moments of a lognormal distribution for values contained between two boundaries, given the distribution is also displaced by an amount? In insurance context, if the losses conform to a lognormal distribution plus a fixed amount, and we want to calculate the mean and standard deviation between a deductible and limit?

## 1 Answer

If $$Y$$ is log-normal with parameters $$\mu,\sigma$$ truncated to the interval $$(e^a, e^b)$$, then $$X=\ln Y$$ is truncated normal on the interval $$(a,b)$$. The moment generating function of $$X$$ is then known to be $$M_X(r)=e^{\mu t + \sigma^2 t^2 / 2} \left[ \frac{ \Phi(\beta- \sigma t) - \Phi(\alpha - \sigma t) }{\Phi(\beta) - \Phi(\alpha) } \right] ,$$ where $$\alpha=(a-\mu)/\sigma$$ and $$\beta=(b-\mu)/\sigma$$ (see wikipedia). Hence, the $$r$$'th moment of $$Y$$ is then given by $$E(Y^r)=E((e^X)^r)=E(e^{rX})=M_X(r).$$

• Thanks for this. Would you be able to expand as to what the mean and variance are for a lognormal distribution with parameters (μ,σ), with a shift displacement of D for the interval (a,b)? Thanks.
– Joe
Apr 6 at 12:45
• @Joe A shift displacement of D only, affect $EY$ and not the standard deviation of $Y$. Apr 7 at 13:44
• I agree for unlimited losses, but if I run a simulation on a lognormal distribution, using the simulated result and in scenario 1 have no displacement and in scenario 2 add a displacement amount, then partition the resultant value between a deductible and limit (same applied to each), the standard deviation differs.
– Joe
Apr 7 at 14:24