# Is it possible that under Black-Scholes: $\ln S_{T} \sim N \left ( \ln S_t - \frac{1}{2}\sigma^2(T-t), \sigma^2(T-t) \right )$

I have a slide on which there is written that under Black-Scholes model:

$$\ln S_{T} \sim N \left ( \ln S_t - \frac{1}{2}\sigma^2(T-t), \sigma^2(T-t) \right )$$

Now, here there is a good explanation on why:

$$\ln{\frac{S_{T}}{S_t}} \sim N\left ((\mu - \frac{1}{2}\sigma^2)(T-t), \sigma^2 (T-t) \right )$$

but this did not solve my problem. In fact I cannot see how one can get from the second equation to the first one. I think that there is something wrong. Am I right?

• If the drift is zero both representations are the same. Check your slides if there is the assumption of a zero interest rate somewhere or if $S_T$ is the discounted stock price.
– Phun
Jun 4, 2016 at 13:10
• @Phun Thank you for your answer. There is in fact an assumption that the interest rate is set to zero. I did not know that under risk neutral probability pricing the drift term can be replaced with the interest rate. This explains why I can write that: $$\ln{\frac{S_{T}}{S_t}} \sim N\left ((- \frac{1}{2}\sigma^2)(T-t), \sigma^2 (T-t) \right )$$. But how can I explain that $\ln S_t$ is inside $N(\cdot)$? Jun 4, 2016 at 13:50
• Just use $a+b N(0,1) \sim N(a,b^2)$ and $\log(A)-\log(B) = \log(A/B)$
– Phun
Jun 4, 2016 at 14:17
• Got it. Since $S_t$ is known, I have added $\ln S_t$ to $\ln S_T / \ln S_t$, this leaves me with $\ln S_T$ which is distributed as a normal with mean $\ln S_t - \frac{1}{2}\sigma^2(T-t)$ and variance $\sigma^2 (T-t)$. Now everything is clear. Thanks! Jun 4, 2016 at 14:58
• @BehrouzMaleki hmm, not really. I actually mentioned that question in my query. There is in fact an hyperlink to it under "here", and it did not (totally) solve my doubt. Jun 6, 2016 at 8:35

Starting from the Black-Scholes model that $$\dfrac{dS}{S} = \mu \:dt + \sigma\:dW_t$$ where $W_t$ is a standard Brownian motion, and $\sigma$ and $\mu$ are constant where $\sigma > 0$. Here $W_t$ is a Brownian motion under the physical measure $\mathbb{P}$. We can then use Girsanov's theorem to change the measure to risk neutral measure $\mathbb{Q}$ where we can now have $\mu \to r$, but this requires $\sigma \neq 0$.

Using stochastic calculus we we can write the right hand side of the above as a stochastic process $dX_t$ and then we can solve this trivially by using the Dolean stochastic exponential, where if we take our limits of integration to be be the domain $[t,T]$ then we recover $$S_T = S_t \exp \left( \left(r - \dfrac{\sigma^2}{2}\right)(T-t) + \sigma(W_T - W_t)\right).$$ Now we observe that $S_T$ has a log-normal distribution with

$$\log(S_T) \sim N\left(\log(S_t) + \left(r - \dfrac{\sigma^2}{2}\right)(T-t), \sigma^2(T-t)\right)$$ We can then de-mean both sides by $\log(S_t)$ where $$\log(S_T) - \log(S_t) \sim N\left(\log(S_t) + \left(r - \dfrac{\sigma^2}{2}\right)(T-t), \sigma^2(T-t)\right) - \log(S_t)$$ $$\log\left(\dfrac{S_T}{S_t}\right)\sim N\left(\left(r - \dfrac{\sigma^2}{2}\right)(T-t), \sigma^2(T-t)\right)$$ If we then consider the case where $r=0$ then we recover the equations stated in the question. The subtle aspects of the above are:

• Knowing how to integrate a stochastic process properly.
• Knowing how to apply Girsanov's theorem to change the measure from $\mathbb{P} \to \mathbb{Q}$ such that in our equations $\mu \to r$.

I hope this helps.

Thanks to @Phun and @oliversm I solved the problem. So I'm posting here the solution in case someone will need it.

Under Black-Scholes assets dynamics are determined by a Geometric Brownian Motion, and we can define the price of a security at time $t+\Delta t$ as:

$$S_{t+\Delta t}=S_{t}\exp\left(\left(r-\frac{1}{2}\sigma^{2}\right)\Delta t+\sigma\sqrt{\Delta t}\varepsilon\right)\qquad\varepsilon\sim N(0,1)$$

defining $T=t+\Delta t$ and substituting above leads to:

$$S_{T}=S_{t}\exp\left(\left(r-\frac{1}{2}\sigma^{2}\right)\left(T-t\right)+\sigma\sqrt{T-t}\varepsilon\right)\qquad\varepsilon\sim N(0,1)$$

Now, under risk neutral probability pricing the drift term $\mu$ can be replaced with the interest rate, and setting $r=0$ leads to:

$$S_{T}=S_{t}\exp\left(-\frac{1}{2}\sigma^{2}\left(T-t\right)+\sigma\sqrt{T-t}\varepsilon\right)\qquad\varepsilon\sim N(0,1)$$

Following the procedure illustrated here, it is easy to show that:

$$\frac{S_{T}}{S_{t}}\sim\ln N\left(-\frac{1}{2}\sigma^{2}\left(T-t\right),\sigma^{2}\left(T-t\right)\right)$$

or equivalently:

$$\ln S_{T}\sim N\left(-\frac{1}{2}\sigma^{2}\left(T-t\right),\sigma^{2}\left(T-t\right)\right)$$

At this point let's define $S = \ln (S_T / S_t) = \ln(S_T) - \ln(S_t)$. $S_t$ is known at time $t$, so we can add $\ln S_t$ to $S$. $S+\ln S_t$ will be normally distributed with mean:

$$\ln S_t-\frac{1}{2}\sigma^{2}\left(T-t\right)$$

and variance:

$$\sigma^{2}\left(T-t\right)$$

So:

$$S+\ln S_{t}\sim N\left(\ln S_{t}-\frac{1}{2}\sigma^{2}\left(T-t\right),\sigma^{2}\left(T-t\right)\right)$$

But since $S+\ln(S_t)=\ln(S_T)$ it follows that:

$$\ln S_{T}\sim N\left(\ln S_{t}-\frac{1}{2}\sigma^{2}\left(T-t\right),\sigma^{2}\left(T-t\right)\right)$$