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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.

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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.

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  • $\begingroup$ many thanks @BGM, that derivation is very helpful and clear! In terms of notation, would it be correct to write: $R_t \sim lognormal(-1,2 \sigma^2)$, is this the accurate relation between lognormal and shifted lognormal? $\endgroup$ – AlexMe Nov 21 '18 at 10:40
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    $\begingroup$ No. Do not mix it up with the normal case, in which $mu$ is a location parameter. In lognormal case it is no longer a location parameter. Also, the support of a shifted lognormal is no longer $(0, +\infty)$ while the support of an ordinary lognormal is always $(0, +\infty)$. One relationship you may want to exploit: Consider $cP_t = ce^{\mu + \sigma Z_t} = e^{\ln c + \mu + \sigma Z_t} \sim \text{lognormal}(\ln c + \mu, \sigma^2)$. So when you multiply it by a scalar, you will affect the parameter $\mu$ in this way. $\endgroup$ – BGM Nov 21 '18 at 15:35
  • $\begingroup$ that makes a lot of sense @BGM, thanks a lot! Only when $P_t$ is multiplied by a scalar $c$, the parameter $\mu$ is affected the way you have described (which I think would require $c>0$). $\endgroup$ – AlexMe Nov 21 '18 at 21:49
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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.

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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.

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  • $\begingroup$ Many thanks @Ben, that's an interesting and useful insight, I have never looked at log-prices from that perspective! $\endgroup$ – AlexMe Nov 24 '18 at 9:13
  • $\begingroup$ another minor question @Ben, how come in the second equality we are conditioning $P_{t}$ on the future value of $P_{t+1}$? Was it supposed to be $P_{t-1}$? $\endgroup$ – AlexMe Nov 24 '18 at 9:36
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    $\begingroup$ Yeah, sorry, that was a typo - fixed. $\endgroup$ – Ben Nov 24 '18 at 22:22

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