Variance of a Stock price and relationship with volatility

Question

A bit of background. I know that the forward price of a stock (or its expected price) is given by $\mathbb{E}[S_T]=S_te^{(r-q)(T-t)}$. Here, $r$ and $q$ are not constant, but follow a curve. I was wondering whether the following is true: $\mathbb{Var}[S_T]=S_t^2e^{\sigma^2(T-t)}$, where $\sigma^2$ is the Black-Scholes volatility. I believe this to be true, but I cannot convince myself.

Could anyone help me out on this?

Edit. Thanks for the help guys. I was also wondering whether it was possible to determine this value. $\mathbb{Var}[e^{(r-q)(T-t)}]$. Just that value without the stock price?

Answer 1

This is not true. In the Black-Scholes setting, \begin{align*} S_T = S_t e^{(r-q-\frac{1}{2}\sigma^2)(T-t)+\sigma (W_T-W_t)}. \end{align*} Then $$E_t(S_T) = S_te^{(r-q)(T-t)}, $$ and \begin{align*} Var_t(S_T) &= E_t(S_T^2) - (E_t(S_T))^2\\ &=S_t^2e^{(2(r-q)+\sigma^2)(T-t)}-S_t^2e^{2(r-q)(T-t)}\\ &=S_t^2e^{2(r-q)(T-t)}\big[e^{\sigma^2(T-t)} -1\big]. \end{align*}

Answer 2

It is hard to digest how you arrive at your result. @Gordon has already provided the answer in risk neutral framework. So Here is answer in real world probabilities because you are interested in expected price and variance of stock price, not of any derivative contract.

Assume $S_t$ is stochastic process and follow geometric Brownian motion with following SDE: $$dS_t=\mu S_t dt + \sigma S_t dW_t$$ then $S_T$ follows lognormal distribution, such that: $$S_T|S_t \sim logN\left(lnS_t+ (\mu - \frac{\sigma^2}{2})(T-t), \quad \sigma^2(T-t)\right)$$ So, $$\mathbb{E}[S_T|S_t]=S_te^{\mu (T-t)},$$ and $$Var[S_T|S_t]=S_t^2 e^{2\mu (T-t)}[e^{\sigma^2(T-t)} -1]$$

In case of dividend, just subtract dividend rate ($q$) from $\mu$.

Here we used the fact that for log normally distributed random variable $X$, such that $X \sim logN(\mu, \sigma^2)$, the mean and variance is given by:

\begin{align} \mathbb{E}[X] &= e^{\mu + \tfrac{1}{2}\sigma^2}, \\ \operatorname{Var}[X] &= (e^{\sigma^2} - 1) e^{2\mu + \sigma^2}\\ &= (e^{\sigma^2} - 1)(\mathbb{E}[X])^2 \end{align}