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<blockquote data-quote="Kevin Lauder" data-source="post: 26469" data-attributes="member: 2187">Thank you for responding so quickly!I see, so you&#039;re saying that it would be better to calculate the extreme value from the maximum/minimum itself, and not even use the standard deviation?Its probably not standard, but the log-normal distribution I am using is from Sortino/Satchell p 57:(\tag{1}\text{lognormal}(x) = \left\{ \begin{array}{l} \frac{\alpha}{\tau-x}\exp(\beta(\ln(\tau-x) - \mu)) \; \mathrm{if}\, \tau &gt; \overline{x} \\&nbsp; \frac{\alpha}{x-\tau}\exp(\beta(\ln(x-\tau) - \mu))\; \mathrm{if}\, \tau &lt; \overline{x}\end{array}\right.)where,(\tag{1a}\alpha=\frac{1}{\sigma\sqrt{2\pi}})(\tag{1b}\beta=-\frac{1}{2\sigma^2})If I understand correctly are you saying that the log-normal is not defined wherever the ((x-\tau)) or the ((\tau-x)) argument to ln in (1), above is less than zero?</blockquote>

[QUOTE="Kevin Lauder, post: 26469, member: 2187"] Thank you for responding so quickly! I see, so you're saying that it would be better to calculate the extreme value from the maximum/minimum itself, and not even use the standard deviation? Its probably not standard, but the log-normal distribution I am using is from Sortino/Satchell p 57: ( \tag{1} \text{lognormal}(x) = \left\{ \begin{array}{l} \frac{\alpha}{\tau-x}\exp(\beta(\ln(\tau-x) - \mu)) \; \mathrm{if}\, \tau > \overline{x} \\ \frac{\alpha}{x-\tau}\exp(\beta(\ln(x-\tau) - \mu))\; \mathrm{if}\, \tau < \overline{x} \end{array}\right. ) where, ( \tag{1a} \alpha=\frac{1}{\sigma\sqrt{2\pi}} ) ( \tag{1b} \beta=-\frac{1}{2\sigma^2} ) If I understand correctly are you saying that the log-normal is not defined wherever the ((x-\tau)) or the ((\tau-x)) argument to [I]ln[/I] in (1), above is less than zero? [/QUOTE]

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