I'm writing an economic overview and I need to get an explanation in the context of log-normal distribution being derived from the idea of multiplicative influence of factors and in order to explain an example below.

The model where factors are considered being accumulated by multiplication is usually selected when the growth of the variable is proportional to its value. When the factors are accumulated additively, central limit theorems tell us that the distribution of sums will tend to normal. In the case of multiplication, we can take take the logarithm of the product and apply CLT to the sum of logarithms, thus obtaining log-normal distribution.

EDIT: Simplifed example of so-called "volatility drag":

Imagine, we have a product series, e.g.

$$S(t)=u_1 \cdot u_2 \cdot...\cdot u_T$$

$$ln(S(t)) = ln(u_1) + ln(u_2) + ... + ln(u_T), \ \ ln(S(t)) \sim N(\cdot,\cdot)$$

where $u_i$ are compounded factors with some distribution. If $u_t$ makes the process $S(t)$ to jump down, multiplying it by 0.5, then іn order to return $u_{t+1}$ needs to equal $2$ on the next step which has less probability (perhaps unless if the distribution is not significantly positively skewed).

The log-normal distribution has positive skewness that depends on its variance, which means that right tail is larger. The expectation also equals $\exp(\mu + \sigma^2/2)$, which means that log-normal variable tends to be dragged into bigger values as variance grows. However, it should inherit the above property that if one factor makes series to jump down, it makes hard for them to resurrect, i.e. negative skewness.

Where I am wrong?

EDIT2: This is my previous example to make sure I haven't misunderstood something:

The economic process consists of log-normally distributed returns $S_t/S_{t-1}$. If I understand correctly, in the case of the compounded return of two return series with the same arithmetic mean, but different variance, the series with a lower variance has a higher compounded return (and geometric mean). The reason is that if an investment looses 50% of its value, it has to make a return of +100% (very different from +50%) to come back to the initial value. In other words, if the process jumps down by an amount 0.5, it needs to achieve 2 on the next step which has less probability (perhaps if the distribution is not significantly positively skewed).


1 Answer 1


I see a couple things that may be contributing to your confusion:

  1. Confusing the limiting distribution with the conditional distribution.
  2. Confusing skewness with probability of high values.

For the first one, I'll give an example:

Lets say that a stock starts at value $S_0$ and is subject to an iid sequence of multiplicative returns $r_i$ to get a future value $S_N=S_0\prod_1^N r_i$. You correctly pointed out that if we take logs, we can show that $S_N\dot{\sim} \textrm{Lognormal}$ to a reasonable degree of accuracy for sufficiently large N.

However, let's say that we want the one-step forecast. There is no reason to believe $S_1$ will be lognormally distributed, as I would be skeptical that convergence is adequate here. So, the short-term behavior of a stock need not have the same skew, just like the iid sum of even highly skewed RVs can approach a symmetric Gaussian.

For the second case, look at the shape of a particular lognormal:

enter image description here

You can see that the median is less than the mean, which means that the typical return is less than the mean and the values are actually clustered around low values!

Therefore, skewness does not mean "heavy tailed", and the "heavy tailed-ness" of a distribution has less to do with the amount of probability in the tails (which is a vague concept) than with how the tails decay towards $0$.

  • $\begingroup$ >>However, let's say that we want the one-step forecast. There is no reason to believe $S_1$ will be lognormally distributed >> No, I haven't stated that, I just meant that it has some own distribution that should be positively skewed to compensate low values affecting $S(t)$. As far as I understand this effect is empirically observable, e.g. if $r_i$ are, say, normally distributed, our series really show volatility drag. $\endgroup$
    – Slowpoke
    Commented May 7, 2016 at 22:17
  • $\begingroup$ @hcl14 since the lower bound for a multiplicative factor is 0 and it can theoretically go to $\infty$ we will naturally have positive skewness. I'm sure you are correct that the short-term distributions are positively skewed. Note that the answer in the link you had on "volatility drag" essentially makes the point that expected value is not a good proxy for typical compounded return. So the fact that the expected value of the stock increases as volativlity increases may not have the typical intuitive interpretation. $\endgroup$
    – user237392
    Commented May 7, 2016 at 23:05
  • $\begingroup$ Thank you. I'm sorry for this, but I am taking this entire probabilistic mechanism quite slowly. Can you please explain me more clear: the median of log-normal doesn't depend on variance. Can we really say that volatility drag has explanation in the properties of log-normal distribution? The answer in the question I referred seems to do so, but in the way very unclear to me, as the stochastic equation places $\mu - \sigma^2/2$ in its linear trend. The author meant that SDE corrects the trend by a volatility to account this effect which I'm trying to describe. $\endgroup$
    – Slowpoke
    Commented May 7, 2016 at 23:16
  • $\begingroup$ @hcl14 the link is describing Geometric Brownian Motion, which has two parts, a stochastic part that is lognormally distributed, and a leading factor that is basically the compounded return. The "volatility drag" appears in the leading coefficient, not in the lognormal part, so they are not directly related. $\endgroup$
    – user237392
    Commented May 7, 2016 at 23:20
  • $\begingroup$ But the drag that appears in the compound return is connected with the variance coefficient of the log-normal. This is because that our series are returns $u_i = S_i/S_{i-1}$, and $S_t = u_0...u_t$, $S_t \sim LN(\cdot,\cdot)$, so the leading coefficient is the expectation of log-normal. $\endgroup$
    – Slowpoke
    Commented May 7, 2016 at 23:37

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