Log-normal returns Let $P_t$ denote a stock price distributed as  $\operatorname{lognormal}(\mu , \sigma^2 )$. Suppose we construct simple returns $R_t=\frac{P_t-P_{t-1}}{P_{t-1}}$. 
My question is:
What is the distribution of $R_t$, and what are the parameters of this distribution?
P.S. I have read somewhere that $R_t$ would have a shifted log-normal distribution, but I could not find the relation to the log-normal, and how the parameters would be related.
 A: I suppose you mean $P_t$ and $P_{t-1}$ are i.i.d. Note that we may express
$$ P_t = e^{\mu + \sigma Z_t}, P_{t-1} = e^{\mu + \sigma Z_{t-1}}$$
where $Z_t, Z_{t-1}$ are i.i.d. standard normal. Then
$$ R_t = \frac {P_t - P_{t-1}} {P_{t-1}} = \frac {P_t} {P_{t-1}} -1
= \frac {e^{\mu + \sigma Z_t}} {e^{\mu + \sigma Z_{t-1}}} -1 
= e^{\sigma (Z_t - Z_{t-1})} - 1$$
Since $Z_t - Z_{t-1} \sim \mathcal{N}(0, 2)$, we have $e^{\sigma (Z_t - Z_{t-1})} \sim \text{lognormal}(0,2\sigma^2)$ and thus the resulting $R_t$ is a shifted lognormal.
A: Just some quick thoughts which do not fit into the comment section. If you look at
$$
\log (1+R_t) = \log(P_t) - \log(P_{t-i}) \sim N(0,2\sigma^2)
$$
you find that it is normal distributed (as the difference of two normal distributions). 
So $R_t+1$ is lognormally distributed, and $R_t$ is a shifted lognormally distributed RV.
A: Since the returns involve changes in the stock price over consecutive time periods, the answer to your question depends on the joint distribution of the stock price over time.  Since you have only specified the marginal distribution of the stock price, you have not given enough information in your question to determine the distribution of interest.  Nevertheless, I will give you a useful result here that holds in a simple case, and which can be extended more broadly to other models with some additional algebraic work.
Stationary AR(1) model: Suppose the log-price follows the stationary model of the form:
$$\ln P_t = (1-\phi) \mu + \phi \ln P_{t-1} + \varepsilon_t \quad \quad \quad \varepsilon_t \sim \text{N}( 0, (1-\phi^2) \sigma^2),$$
where $-1 < \phi < 1$ is the auto-correlation parameter of the model.  In this case the stationary marginal distribution of the log-price is as you specified in your question (i.e., $\ln P_t \sim \text{N} (\mu, \sigma^2)$).  Now, since $\ln P_t = \ln P_{t-1} + \ln (1+R_t)$ the density function of the latter term is:
$$\begin{equation} \begin{aligned}
p(\ln (1+R_t) = r) 
&= \int \limits_{-\infty}^\infty p(\ln (1+R_t) = r | \ln P_{t-1} = s) \cdot \text{N}(s|\mu, \sigma^2) ds \\[6pt]
&= \int \limits_{-\infty}^\infty p(\ln P_t = r+s | \ln P_{t-1} = s) \cdot \text{N}(s|\mu, \sigma^2) ds \\[6pt]
&= \int \limits_{-\infty}^\infty \text{N}(r+s | \mu + \phi (s-\mu), (1-\phi^2) \sigma^2) \cdot \text{N}(s|\mu, \sigma^2) ds. \\[6pt]
\end{aligned} \end{equation}$$
To solve this integral equation we need to complete the square in the exponential terms of the normal density functions in the integrand.  This would lead to a normal distribution, so that $1+R \sim \text{LogN}$.  Obtaining the parameters of the distribution would require you to complete the square in the above integral, and perform the subsequent algebra.  I leave this as an exercise.
