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