# What is the formula for the Mode of the Log-Normal distribution followed by the Geometric Brownian Motion?

What is the formula for the Mode of the Log-Normal distribution followed by the Geometric Brownian Motion?

Intro__________________________

I was studying in Youtube this interesting MIT course of math in finance, where I learned about stochastic processes and the geometric Brownian motion (GBM), and it is stated the GBM follows a Log-Normal Distribution as it is also stated in the Wikipedia page. Here is where I have doubts, because for almost every demonstration are based on the mean value instead of the Mode, which formula at least it is not even shown in Wikipedia.

Questions______________________

A) It is right the Mode formula of the geometric Brownian motion I have found $$\nu[S_t] = S_0\ e^{\left(\mu -\frac{3}{2}\sigma^2\right) t}$$?

B) Does someone that is pricing their instruments using the Mode $$\nu[S_T]$$ will be more assertive in the short-term than someone else that it is pricing using the mean value $$M[S_T]$$ instead?

Notation is in the motivation section, but it is not required for giving the answers - but I will appreciate if the same notation is hold.

Motivation for question (A)

From the Wikipedia page for the Log-normal distribution I know the following: if $$Z$$ follows a standard normal distribution, then for some values $$m$$ and $$a>0$$ the variable $$Y = m+aZ$$ follows a Gaussian distribution with mean value $$m$$ and variance $$a^2$$, and the variable $$X = e^{m+aZ}$$ will follow a Log-Normal distribution with:

• The mean value is given by $$M = \exp\left(m+\frac{a^2}{2}\right)$$
• The Mode is given by $$\nu = \exp\left(m-a^2\right)$$
• The variance is given by $$V = \exp\left(2m+a^2\right)\cdot\left[\exp\left(a^2\right)-1\right]$$
• The probability density function (PDF) is given by $$f(x) = \dfrac{1}{xa\sqrt{2\pi}}\exp\left(-\dfrac{(\ln(x)-m)^2}{2a^2}\right)$$

I am using the same description than Wikipedia in order to reduce explanations but I change the characters to avoid confusion with the following section, and I am also avoiding the use of expected value term in purpose, since for a skewed distribution the "mean" value don't coincide with the value you expect to see more frequently which is given by the "Mode".

Now, for a Geometric Brownian motion that follows the stochastic differential equation (SDE): $$dS_t = \mu S_t\ dt+\sigma S_t\ dW_t$$ with $$W_t$$ is a Brownian motion, and the drift $$\mu$$ and the volatility $$\sigma$$ are constants. Then, it is stated that the solution of the SDE will be giving by (Ito's interpretation): $$S_t = S_0 \exp\left(\left(\mu-\frac{\sigma^2}{2}\right)t+\sigma W_t\right)$$

And it is shown without demonstration that the solution will follow:

1. The mean value of the process is given by $$M[S_t] = S_0\ e^{\mu t}$$
2. The variance of the process is given by $$V[S_t] = S_0^2\ e^{2\mu t}\left(e^{\sigma^2 t}-1\right)$$
3. The probability density function (PDF) of the process is given by $$f(s) = \dfrac{1}{s\sigma\sqrt{2\pi t}}\exp\left(-\dfrac{\left(\ln(s)-\ln(S_0)-\left(\mu-\frac{\sigma^2}{2}\right)t\right)^2}{2\sigma^2 t}\right)$$

This PDF kind of match what it is shown in the minute 33:40 of the following MIT video for the Black-Scholes equation with its respective defined terms, except from first denominator where I think there is some typos mistakes: a variable $$S$$ that should be $$S_T$$, and time $$T$$ that should be $$(T-t)$$ in order to match with Wikipedia definition.

Since Wikipedia page don't show which is the formula for the Mode $$\nu[S_t]$$, by matching terms I will get the following: $$a = \sigma\sqrt{t} \quad \rightarrow \quad a^2 = \sigma^2 t$$ $$m = \ln(S_0)+\left(\mu-\frac{\sigma^2}{2}\right)t$$

with them one recover $$M[S_t]$$ and $$V[S_t]$$ as shown in Wikipedia, and also I will have that the Mode is given by: $$\nu[S_t] = \exp\left(\ln(S_0)+\left(\mu-\frac{\sigma^2}{2}\right)t - \sigma^2 t\right) = S_0\ e^{\left(\mu -\frac{3}{2}\sigma^2\right) t}$$ which is the formula I am asking if it is right or not.

Motivation for question (B)

The last question is more conceptual, based in the mentioned issue that the "Expected Valued" is not given by the "mean value", but instead by the "Mode".

From what I think I have understood, looks like for pricing instruments based in SDEs the current estimated price at some future time $$t$$ is given by the mean value $$M[S_t]$$, as it is also used for all the theory like Martingale pricing, Risk-neutral measures, and for the Fundamental theorem of asset pricing where the idea of Arbitrage-free market is demonstrated.

But for what I understand from skewed unimodal distributions, under the assumption you see the market at random times, someone that is pricing using instead the Mode value $$\nu[S_t]$$ should have an statistical advantage compared with someone that is pricing with the mean value $$M[S_t]$$, since prices near their selection should appear more frequently, at least in the short-term, before the probabilistic weight of outliers carry the price near the mean value $$M[S_t]$$, so he should arbitrage against the first mean-value guy, closing their moves in the short-term, so the market price never arrives to the medium-long term, and "milking" instead the mean-value guy with several short-term moves.

And also, don't knowing if its worse, those who are pricing with the mean value $$M[S_t]$$ will see that their portfolios are behaving always worst than what have been planed, since $$\nu[S_t] \leq M[S_t]$$, so they will be over-adjusting their portfolios without never reaching the time where the mean value $$M[S_t]$$ sets, and taking more risk that they should.

I believe it is like when you overexcite the differentiating term $$D$$ on PID controllers, leading to unwanted oscillations in your controlled system. This last part just as an intuition, since it is just speculation.

Since for unimodal positive-skewed distributions I will see values near the mode more frequently than near the mean value (I tried to quantified how much in this question unsuccessfully), I am inclined to think that someone pricing using the Mode will be more assertive than someone using the mean value in the short-term, at least in this scenario of a Geometric Brownian motion that it is Log-Normally distributed.

As example, if $$0<\mu< \frac32\sigma^2$$ then I will have the mean value $$M[S_t]$$ will be increasing exponentially with the time $$t$$, but the values I will see more frequently given by the Mode $$\nu[S_T]$$ will be decreasing exponentially!.

Even worst, if the calculations of part (A) are right, the median will be $$\text{Med}[S_t]=S_0\ e^{\left(\mu -\frac{\sigma^2}{2}\right)t}$$ so if happen that $$0<\mu < \frac{\sigma^2}{2}$$ the mean value will still be growing exponentially, when at least $$50\%$$ of the values one could see at random times will be decreasing exponentially, so someone could easily bet against me $$50\%$$ of the time. This is why using the mean value looks very counter-intuitive to me on skewed scenarios. Note that the formula for the Median it is the same deterministic component for the solution of the SDE, and also that it is not listed in Wikipedia neither.

I specifically asked this question in the math blog instead of the finances one, because I want to know why this line of thought is flawed from the mathematics point of view, demonstrating why is that so.

I tried the following example within the mentioned extreme interval: $$S_0=1$$, $$\mu=1/10$$, and $$\sigma=1/2$$. You could see in the following image the mean value $$M[S_t]$$, the mode $$\nu[S_t]$$,and the median $$\text{Med}[S_t]$$ in their plot in Wolfram-Alpha:

As expected, the mean grows exponentially while the mode and the median converges to zero. Surely this parameter selection is an extreme case, but are these kind of examples where slight details becomes obvious.

Also to figure out how would look the GBM I made a code in online Octave:

The GBM paths look incredible spread. Against the superimposed mean, mode, and median curves, the paths "looks" are filling the whole graph instead of being concentrated near the mode. To check it I plotted later the histogram of the last point in time:

Now, you can see that the previous path's plot is misleading since the histogram shows an evident concentration onto the Mode value, which is indeed importantly higher than where the mean value is located, kind of justifying my intuition that using as "expected value" the mean value instead of the Mode is wrong for positively skewed unimodal distributions.

Here I left you the code I used, but be carefull since it is the first time I model a GBM so I am not $$100\%$$ sure it have no mistakes:

length = 501;
N = 50;
dt=1/50;
white_noise = sqrt(dt)*wgn(length-1,N,0);
simple_brownian = zeros(length,N);
t=0:1:length-1;
t=dt*t;

for m=1:1:N
simple_brownian(2:1:length,m) = cumsum(white_noise(:,m));
end

S0 = 1;
mu = 1/10;
sigma = (1/2);
sigma2 = (sigma)^2;

mean_val = S0*exp(mu*t);
mode_val = S0*exp((mu-3/2*sigma2)*t);
median_v = S0*exp((mu-1/2*sigma2)*t);

GBM = ones(length,N);
for m=1:1:N
for k=1:1:length
GBM(k,m) = S0*exp((mu-1/2*sigma2)*t(k)+sigma*simple_brownian(k,m));
end
end

figure (1),
hold on,
plot(t, mean_val,'r','Linewidth',2,t,mode_val,'y','Linewidth',2,t,median_v,'g','Linewidth',2), legend('mean value','mode value','median value'),
plot(t,GBM),
plot(t,mean_val,'r','Linewidth',2,t,mode_val,'y','Linewidth',2,t,median_v,'g','Linewidth',2),
axis([0 t(length) 0 2*S0*exp(mu*(t(length)))]),
hold off;

figure(2),
hist(GBM(length,:),round(N/2));

• Can you shorten your post to the essentials? As it stands, it takes a while to find out what you're trying to ask. Commented Nov 27, 2023 at 3:50
• This question might better fit in quant.stackexchange.com. Commented Nov 27, 2023 at 17:02
• yes, for the proof of the mode of lognormal see statproofbook.github.io/P/lognorm-mode.html Commented Nov 27, 2023 at 17:04
• @ThomasKojar thanks for the comments. I am not worried about the Mode of the Log-Normal distribution per se, but instead on the Mode of the geometrical Brownian motion, and how representative it is as a sample of the process. Commented Nov 28, 2023 at 4:20
• @Joako Same thing because you compute mode for each fixed t. Commented Nov 28, 2023 at 4:22

QUESTION A

Your procedure is correct, but if you want to derive it from first principles you must consider the following.

For a unimodal distribution, the mode is defined as the max of the pdf. In our case we are interested in

$$max_s( f(s))$$

which will give as a function of $$t$$ (and the parameters). To compute the maximum, we have to set the derivative to 0.

$$f'(s) = 0$$

This is a bit tedius. A nice trick you can use, is considering the fact that the logarithm being a monotonic function, does not affect the extrema. | So we can compute first

$$\ln(f)(s) = - \ln(s) + \frac{((\ln(s) - \ln(S_0) - (\mu-\sigma^2)t)^2}{2\sigma^2t} + K$$

Where $$K=\ln(\sigma t \sqrt{2 \pi})$$ includes all the constant terms (not depending on x) that will not affect the derivative.

Then by setting

$$\frac{d}{ds}\ln(f)(s) = 0$$

and solving the system in $$s$$ we find your solution for the mode:

## $$\nu[s] = S_0 e^{(\mu-\frac{3}{2}\sigma^2)t}$$

QUESTION B

This question in really deep and interesting, and I feel like I don't know enough stochastic processes to give a complete answer. I also don't know anything about finance so I cannot discuss the implication of this fact in financial terms.

What a can tell you is that the problem lies in the fact that the system is not ergodic. This creates the problem that, for same values of the parameters, the average path (i.e., the expected values in s of f(s,t), which is a function of t) will grow exponentially, while most individual trajectories go to zero. This is what you are observing and it is sometimes referred to as the extintion problem.

A brief discussion is at the end of this page: The average trajectory follows $$s(t) =e^{\mu t}$$, which explodes when $$\mu >0$$. Yet, when $$0<\mu<\sigma^2/2$$ the probability of any trajectory of hitting 0 (extinction) is $$1$$!

This somehow paradoxical result is analogous to the fact that a browniam motion will reach any value as time passes, even when the mean trajectory (the expected value) is $$0$$.

As I said, I don't anything about finance, but I guess that if you try to use the mode instead of the mean you will end up being fairly close to the solution most of the time, but really wrong sometimes. And by really, I mean exponentially (consider that those few paths that go up are so big that are able to drag the average to an exponential growth even if most paths are descending).

I hope this helps a bit. But I think that your perplexity raises from the fact that you are assuming ergodicity of the process. Maybe this paper (which apparently discusses some economical implication of the non ergodicity of the GBM) might help!

Expanding on this, I think that a nice way to think about this fact is the following:

Consider a really basic discrete process in which the value of your stock varies according to

$$s_{t+1} = s_t + (0.1) b_t s_t$$

where $$b_t = \pm 1$$ with equal probability, i.e., each day your stock will increase of decrease its value by 10%. Starting at $$s_0=1$$, let's see what happens after $$4$$ days.

• You see that every realization with an even number of $$-1$$ and plus $$+1$$, like $$(+,-,+,-)$$ or $$(+,+,-,-)$$ will result is a value smaller than $$1$$ ($$s_4 \approx 0.98$$)
• But the path with the realization (+,+,+,+) leads to a value much higher than one ($$s_4 \approx 1.464$$)
• Of course, there is also (-,-,-,-) which leads to $$s_4 \approx 0.65$$ but if you compute the expected values of $$s_4$$ (i.e., the average of all the possible combination of + and -) you will notice that even if most trajectory will lead to a decrease in value, the average will increase.

And this is the problem: even if most realizations will leave you with a lower value than the initial one, those few leading to a higher value will "dominate" the average. But notice that at every single step, the probability of going up or down is the same.

• Thanks for the detailed answer, I have found it very useful. I didn't know the process wasn't ergodic, so I wonder how representative are really the statistical representation. In the first link you shared are using the the characteristics of the median to show the characteristics of the process, since $S_t = S_0 \text{Median}[S_t]e^{\sigma W_t}$, but I have mixed feelings about it, since you could have indeed a exponential increasing mean value and median but an exponential decreasing mode, so I believe that for every scenario where $\mu < 3/2 \cdot \sigma^2$,. (...) Commented Dec 6, 2023 at 22:49
• (...) you will have that these central moments don't going to fully represent how the paths are behaving. But also, thinking in something similar than your last section (which is similar to was I did here for the Kelly Criterion), if I try to take something similar to the geometric mean which exponential represent the process, I will have that $$S_t = \left[ e^{\mu-\sigma^2/2+\sigma\frac{W_t}{t}}\right]^t$$ so since $W_t/t$ should be decreasing as $1/\sqrt{t}$ so the median do looks like a representative path, (...) Commented Dec 6, 2023 at 22:59
• (...) even when it could be infinitely random at first, but I think the process itself will smooth it at this fake start since the process must be continuous. I am still trying to figure out how it will behave but your answer gives me a lot of inspiration. Commented Dec 6, 2023 at 23:00
• I just realized I messed up my last comments: it is $S_t = \text{Med}[S_t]e^{\sigma W_t}$ so it is also $S_t/S_0 = \left[e^{\mu-\sigma^2/2 +\sigma \frac{W_t}{t}}\right]^t$. My apollogies. Commented Dec 7, 2023 at 0:06