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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?

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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 $$ E(Y^r)=E((e^X)^r)=E(e^{rX})=M_X(r). $$

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  • $\begingroup$ 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. $\endgroup$
    – Joe
    Apr 6 at 12:45
  • $\begingroup$ @Joe A shift displacement of D only, affect $EY$ and not the standard deviation of $Y$. $\endgroup$ Apr 7 at 13:44
  • $\begingroup$ 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. $\endgroup$
    – Joe
    Apr 7 at 14:24

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