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I need to generate some random data from lognormal distribution, where I set the mode and standart deviation of that lognormal distribution. For this purpose I choose to use random numbers generator from lognormal distribution. This generator takes two numbers, that are mean and sd of underlying normal distribution.

So far its clear I need to derive mean and sd of normal distribution, which is underlaying for lognormal distribution where I know mode and sd. I know the equations for derivation of mean and sd:

NOTATION:

n(x) = mean of normal distribution

sd(x) = sd of normal distribution

n(y) = mean of lognormal distribution

sd(y) = sd of lognormal distribution

mode(y) = mode of lognormal distribution

EQUATIONS:

$$ n(x) = 2*ln(n(y)) - (1/2)*ln(sd(y)^2 + n(y)^2) $$ $$ sd(x) = -2*ln(n(y)) + ln(sd(y)^2 + n(y)^2) $$ $$ mode(y) = exp(n(y) - sd(y)^2) $$

Here I stuck because I cant get the equation for $n(y)$ from these equations, that I need to compute $n(x)$. So far I ended:

$$ mode(y) = exp(4*ln(n(y))-3/2*ln(n(y)^2 - sd(y)^2)) $$ $$ mode(y)^{2/3}*sd(y)^2 = n(y)^2 * (n(y)^{2/3} - mode(y)^{2/3}) $$

Can anybody help me to complete this derivation? Or there are any other ways to anallytically derive mean of lognormal distribution from mode and sd of lognormal distribution?

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1 Answer 1

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If the underlying normal distribution has mean $\mu$ and variance $\sigma^2$ then the lognormal has mean $\exp(\mu+\sigma^2/2)$, mode $\exp(\mu - \sigma^2)$, and variance $(\exp(\sigma^2) -1)\exp(2\mu+\sigma^2)$. You say you set the mode and standard deviation of the lognormal, let's say as $m$ and $s$

Then $m=\exp(\mu - \sigma^2)$ so $\log_e(m) = \mu - \sigma^2$ and $$\mu = \log_e(m) + \sigma^2$$ leaving us to find $\sigma$. Meanwhile $s^2 = (\exp(\sigma^2) -1)\exp(2\mu+\sigma^2)$ which with some substitution and manipulation gives $$ (\exp(\sigma^2))^4 - (\exp(\sigma^2))^3- \dfrac{s^2}{m^2} =0 $$ which is a quartic equation in $\exp(\sigma^2)$ with what looks like one root greater than $1$ when $\frac{s^2}{m^2} \gt 0$ and so it may be possible to solve for $\sigma$ even if unattractive in a closed form

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