Let Y be a lognormally distributed random variable with mean $\mu_Y$ and standard deviation $\sigma_Y$. Assume $\ln(Y)$ is normally distributed with mean $\u$ and variance $\sigma^2$. According to my source, the mean of the lognormal distribution satisfies
$\mu_Y=\ln(\frac{\mu\sigma}{\sqrt{1+w}})$
and variance $\sigma^{2}_Y=\ln(1+w)$
where $w=(\sigma_Y/\mu_Y)^2$.
I am questioning the correctness of this result. I have attempted to derive this formula as follows.
We know that if Y is the logarithm of the normal distribution with mean $\mu$ and variance $\sigma^2$, then the mean of Y is given by
(1) $\mu_Y=e^{\mu + \frac{1}{2}\sigma^2}$, and
(2) $\sigma_Y^2=(e^{\sigma^2} -1)\mu_Y^2$
Setting $w$ the same value as above, the second equation implies that $w=e^{\sigma^2} -1$ so that $\sigma^2=\ln(w+1)$ just as claimed above.
On the other hand, now that we have $\sigma^2$, we substitute this value into equation (1) we get
$\mu_Y=e^{\mu + \frac{1}{2}\ln(w+1)}$, which, to my algebra, yields $\mu =\ln(\frac{\mu_Y}{\sqrt{w+1}})$.
Hoping that somebody here can verify my sanity, or explain why the formula for the mean given in my source is correct.
Best Answer
We have $$ \mu_Y=e^{\mu+\sigma^2/2} $$ and $$ \sigma_Y^2=(e^{\sigma^2}-1)e^{2\mu+\sigma^2} $$ (see e.g. Wikipedia). Hence $$ w=\frac{\sigma_Y^2}{\mu_Y^2}=e^{\sigma^2}-1 \quad \Longrightarrow \quad \sigma^2=\ln(w+1). $$ Now $$ \ln\mu_Y=\mu+\frac{\sigma^2}2=\mu+\frac{\ln(w+1)}2=\mu+\ln(\sqrt{w+1}) $$ yielding $$ \mu=\ln\frac{\mu_Y}{\sqrt{w+1}}. $$ So you are correct.