Suppose we have a random variable $S_t$ with a log normal distribution distribution, where $S_t$ represents the price of a stock at a time $t$. Suppose that we have the annual volatility $\sigma$, of $S_t$. I know that the expectation of $S_t$ will be $e^{\mu+1/2(\sigma)^2}$, and that the variance will be $(e^{(\sigma^{2})}-1)(e^{2\mu+\sigma^{2}})$.


  1. Is the sigma here the volatility of the stock?
  2. Why isn't the expectation of $S_t$ equal to $\mu$?
  3. Why isn't variance equal to $\sigma^{2}$?
  4. If we wanted to find something like $p(40\le S_{.4}\le55)$, how would we go about it?
  5. How do we find the values $\sigma$ and $\mu$?

1 Answer 1


If $S_t \sim LogNormal (\mu, \sigma^2)$. It is important to remember that a LogNormal distribution $\implies S_t > 0$.

Now by the definition of the Log Normal distribution: $\log S_t \sim Normal (\mu, \sigma^2)$; $\mu$ is the mean of $\log S_t$ and $\sigma^2$ is the volatility (variance) of $\log S_t$. $\log(x)$ here is the natural log; $\log (x) = log_e(x) = \ln (x)$. Notice how $\mu \in \mathbb{R}$ but $ S_t > 0$ so automatically $\mu \neq E (X)$.

If you want to find $P(40 \leq S_{0.4} \leq 55)$ then we just apply the definition of the LogNormal:

\begin{align} $P(40 \leq S_{0.4} \leq 55)$ & = P( \log 40 \leq \log S_{0.4} \leq \log 55) \\ &=P\left(\frac{\log 40 - \mu}{\sigma} \leq \frac{\log S_{0.4} - \mu}{\sigma} \leq \frac{\log 55 - \mu}{\sigma} \right)\\ &= P\left(Z \leq \frac{\log 55 - \mu}{\sigma}\right) - P\left(Z \leq \frac{\log 40 - \mu}{\sigma}\right) \end{align}

where $Z\sim Normal(0,1)$. This final probabilities can be found bey looking at e.g. z tables, provided you have $\mu$ and $\sigma$.

How to estimate $(\mu, \sigma)$? Well depends on your approach to statistics, but wikipedia explains quite well how to find the maximum likelihood estimates, provided you have obtained some data. You could take a Bayesian approach if you prefer.


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