How to find normal and lognormal moments, given partial information? $Y=\ln(X)$. $X$ is lognormal and $Y$ is normal. If all I know is the arithmetic mean of $Y$ and the standard deviation of $X$. What is the formula to calculate the arithmetic mean of $X$ and the standard deviation of $Y$?
 A: To say that $Y$ is Normal with mean $\mu$ and standard deviation $\sigma$ is equivalent to $Y = \mu + \sigma \eta$ where $\eta$ is standard normal, having PDF proportional to $\exp(-\eta^2/2)$.  We need formulae for the moments of the lognormal variate $X = \exp(Y) = \exp(\mu + \sigma\eta)$, so for natural numbers $k$ we compute
$$\mathbb{E}(X^k) = \mathbb{E}(\exp((\mu + \sigma\eta)k))= C\int_\mathbb{R}\exp(-\eta^2/2 + \mu k + \sigma \eta k)\,d\eta = \\C\int_\mathbb{R}\exp\left(-\left(\eta-k\sigma\right)^2/2\right) \exp\left( k\mu + k^2\sigma^2/2\right)\,d\eta = \exp(k\mu + k^2\sigma^2/2).$$
Setting $k=1, 2$ gives the absolute first and second moments from which we find the mean and variance of $X$ are
$$m = \exp(\mu + \sigma^2/2);\quad s^2 = \exp(2\mu + \sigma^2)\left(\exp(\sigma^2)-1\right).$$
The question asks how to find $m$ and $\sigma^2$ when $\mu$ and $s^2$ are known.  The second equation can be rewritten
$$s^2 = \exp(2\mu)x(x-1)$$
where $x = \exp(\sigma^2).$  The only positive root of this quadratic is
$$x = \frac{1}{2} \left(1+\sqrt{1+4 s^2\exp(-2\mu)}\right)$$
whence
$$\sigma^2 = \log(x) = \log\left( \frac{1}{2} \left(1+\sqrt{1+4 s^2\exp(-2\mu)}\right)\right).$$
Now that $\mu$ and $\sigma^2$ are available, the formula for the first moment gives $m$ immediately. For computational purposes it can be conveniently rewritten
$$m = \exp(\mu)\sqrt{x}.$$
A: If $X$ is lognormally distributed than the variance of this random variable is:
\begin{align*}
Var(X)=\exp(2\mu+\sigma^2)[\exp(\sigma^2)-1]
\end{align*}
Since you know the standard deviation of $X$ -- and hence its variance -- in addition to $\mu$ (the mean of Y) you can solve the formula for $\sigma$. With sigma at hand you can solve for the expected value of $X$:
\begin{align*}
E(X)=\exp(\mu+0.5\sigma^2)
\end{align*}
