# If $X\sim \operatorname{lognormal}$ then $Y:=(X-d\mid x\geq d)$ has approximately a Generalized Pareto distribution

Let $$X$$ be a random variable with lognormal distribution. Show that when sufficiently large then $$Y:=(X-d\mid x\geq d)$$ is approximately a random variable with generalized Pareto distribution.

Hint: Use the fact that $$\operatorname{erf}(x)\approx 1-\frac{1}{\sqrt{x}}e^{-\frac{x^2}{2}}$$ for large values of $$x$$.

My attempt: We recall that the density function for the lognormal distribution is given by $$f(x)=\frac{1}{x\sigma\sqrt{2\pi}}e^{\frac{-(\log x-\mu)^2}{\sigma}}\:\:\text{ for }x>0.$$ The comulative distribution function for a generalized Pareto random variable is given by $$G(x)=1-\left(1+\frac{\gamma x}{\theta}\right)^{\frac{-1}{\gamma}}.$$ The objective is to find parameters $$\gamma$$ and $$\theta$$ such that $$\mathbb{P}(Y\leq y)\approx G(y)$$, it is clear that $$\gamma$$ and $$\theta$$ will be expressed in terms of $$\sigma$$, $$\mu$$ and $$d$$. My attempt is: \begin{align} \mathbb{P}(Y\leq y) & =1- \frac{1-\int_0^{d+y}\frac{1}{x\sigma\sqrt{2\pi}} e^{\frac{-(\log x-\mu)^2}{\sigma}} \, dy }{1-\int_0^d\frac{1}{x\sigma\sqrt{2\pi}}e^{\frac{-(\log x-\mu)^2}{\sigma}} \, dy} \end{align}

We consider the changge of variable given by $$t=\frac{\log x -\mu}{\sqrt{2}\sigma}$$, then $$dt=\frac{1}{\sqrt{2}\sigma x} \, dx$$, so, $$dx=\sqrt{2}\sigma x\, dt$$. Therefores, we have

\begin{align} \mathbb{P}(Y\leq y) & =1- \frac{1-\frac{1}{\sqrt{\pi}}\int_{-\infty}^{\frac{\log(d+y) -\mu}{\sqrt{2}\sigma}}e^{-t^{2}} \, dy }{1-\frac{1}{\sqrt{\pi}}\int_{-\infty}^{\frac{\log(d) -\mu}{\sqrt{2}\sigma}}e^{-t^{2}} \, dy} \\ &= 1- \frac{1-\frac{1}{\sqrt{\pi}}\int_{-\infty}^0 e^{-t^2} dy- \frac{1}{\sqrt{\pi}}\int_0^{\frac{\log(d+y) -\mu}{\sqrt{2}\sigma}}e^{-t^2} \, dy }{1-\frac{1}{\sqrt{\pi}}\int_{-\infty}^0 e^{-t^2} \, dy-\frac{1}{\sqrt{\pi}}\int_0^{\frac{\log(d) -\mu}{\sqrt{2}\sigma}}e^{-t^2} \, dy} \\ &= 1- \frac{1-\frac{1}{2}- \frac{1}{\sqrt{\pi}}\int_0^{\frac{\log(d+y) -\mu}{\sqrt{2}\sigma}} e^{-t^{2}} \, dy }{1-\frac{1}{2}-\frac{1}{\sqrt{\pi}} \int_0^{\frac{\log(d) -\mu}{\sqrt{2}\sigma}}e^{-t^2} \, dy} \\ &= 1- \frac{\frac{1}{2}- \frac{1}{2} \operatorname{erf} \left(\frac{\log(d+y) -\mu}{\sqrt{2}\sigma}\right) }{\frac{1}{2}- \frac{1}{2} \operatorname{erf}\left(\frac{\log(d) -\mu}{\sqrt{2}\sigma}\right) } \\ &= 1- \frac{1- \operatorname{erf}\left(\frac{\log(d+y) -\mu}{\sqrt{2}\sigma}\right) }{1- \operatorname{erf} \left(\frac{\log(d) -\mu}{\sqrt{2}\sigma}\right) } \\ &\approx 1- \frac{\frac{1}{\sqrt{\frac{\log(d+y) -\mu}{\sqrt{2}\sigma}}}e^{-\frac{(\log(d+y)-\mu)^2}{4\sigma^2}} }{\frac{1}{\sqrt{\frac{\log(d) -\mu}{\sqrt{2}\sigma}}}e^{-\frac{(\log(d)-\mu)^2}{4\sigma^2}} } \leftarrow \text{by hint.}\\ &= 1- \sqrt{\frac{\log(d)-\mu}{\log(d+y)-\mu}}e^{-\frac{ (\log(d+y)-\mu)^2}{4\sigma^2}+\frac{(\log(d)-\mu)^2}{4\sigma^2}}\\ &=1- \sqrt{\frac{\log(d)-\mu}{\log(d+y)-\mu}}e^{\frac{1}{4\sigma^2}(\log(d+y)-\log(d))(\log(y+d)+\log(d)-\mu) }\\ &= 1- \sqrt{\frac{\log(d)-\mu}{\log(d+y)-\mu}}e^{-\frac{(\log(d+y)-\mu)^2}{4\sigma^2}+\frac{(\log(d)-\mu)^2}{4\sigma^2}}\\ &=1- \sqrt{\frac{\log(d)-\mu}{\log(d+y)-\mu}}e^{\frac{1}{4\sigma^2}\log\left(\frac{d+y}{d}\right)\left(\log(dy+d^2)-\mu\right) }\\ \end{align} I do not know how to continue, algebraically I have not been able calibrate the parameters to get what I need.

I ask for your help with this problem, any solution or suggestion will be well received.

• The statement is not true: it is never the case, for any $d,$ that a lognormal distribution truncated at $d$ has a generalized Pareto distribution. Does the exercise perhaps ask you to show that the distribution is approximately generalized Pareto?
– whuber
May 18, 2019 at 3:02
• If the "Generalized Pareto" distribution is the one described by Wikipedia, then no approximation of this sort will hold. This is a consequence of considerations of tail behavior like those discussed at stats.stackexchange.com/questions/86429: all Generalized Pareto distributions have heavier tails than all lognormal distributions. Could you explain the sense in which one of these distributions is intended to approximate the other?
– whuber
May 18, 2019 at 16:56
• @whuber I refer to it in the sense of Pickands-Balkema-de Haan Theorem, see en.wikipedia.org/wiki/… This theorem gives the existence of the constants that determine the Generalized Pareto Distribution, but does not give an explicit form. May 20, 2019 at 19:13
• That would be convergence in distribution, allowing for the scale parameter of the Generalized Pareto to vary with the cutoff $d.$ Because the result is not true, may I inquire about the origin of this statement?
– whuber
May 20, 2019 at 19:28
• Elaborating a little more on the last comment, the idea is that after changing variables, the problem amounts to understanding the scaling behavior of the tail of a normal random variable when viewed in a window $[n,n+xe^{-n}]$ for $x$ fixed and $n\to\infty$. When $n$ is large the density is roughly constant on such a small interval, but we can tease out the Pareto power-law behavior by being careful in our estimations. May 26, 2019 at 11:33

When $$X=e^{\mu+\sigma Z}$$ with $$Z$$ standard normal, I obtain that $$Y=(X-d\mid X\geq d)$$ is approximately generalized Pareto in the limit as $$d\to\infty$$, with $$\theta=\frac{d\sigma^2}{\log d-\mu},\qquad \gamma=\frac{\sigma^2}{\log d-\mu}.$$