How to find normal and lognormal moments, given partial information?

Question

$Y=\ln(X)$. $X$ 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 $Y$?

Answer 1

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 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

$$m = \exp(\mu)\sqrt{x}.$$

Answer 2

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*}