Why is the Geometric Brownian Motion defined as $S(t) = e^{\alpha + \beta t + \sigma W(t)}$ used as a model for stock prices? $S(t)$ has a lognormal distribution which is right skewed. Another problem is that while this cannot go negative, it also cannot change trend - always goes up to infinity or down to zero. Even a simple random walk will generate a plot more similar (to a human eye) of an actual price action graph.
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4$\begingroup$ Nobody really assumes stock prices follow a geometric Brownian motion nowadays. Yes, it’s part of the Black Scholes model, but nobody uses this in reality for pricing or heading. It’s just the simplest time continuous model and is used for teaching purposes (It allows for closed form solutions, is always positive, a Markov process, has jagged paths, etc.). The last 50 years of research have produced many more realistic models. $\endgroup$– KevinApr 15, 2020 at 6:58
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$\begingroup$ Might also be interesting: papers.ssrn.com/sol3/papers.cfm?abstract_id=2956257 $\endgroup$– vonjdApr 15, 2020 at 9:12
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2 Answers
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We don't model the prices, we model the returns.
The stock prices aren't explicitly modelled as log-normal, but rather this is a consequence of the actual model used to describe the returns. The core of the model used in the Black-Scholes model is to assume a geometric Brownian motion for the change in the price $S_t$ where over some small time increment $\mathrm{d}t$ the returns are modelled as \begin{equation} \frac{\mathrm{d}S}{S} = \mu \mathrm{d}t + \sigma \mathrm{d}W_t \end{equation} The key idea here is that we assume no history or memory in the markets which isn't already accounted for in the current price (and hence forms a Markov process). Assuming this is a geometric Brownian motion is just a nice mathematical assumption which to leading order is not that bad. Most of the time it works okay, but isn't great when it tries to capture a good description of more extreme behaviour, as it underestimates the tails when compared to market data.
Skew and positivity aren't a problem
The fact the price process is strictly positive and never hits zero is actually a desirable thing when describing stocks, as this is very physical. The fact there is a positive skew is just a result of this, and this is not really a problem, especially if this matches the market data.
Likewise, the process is always finite, and does not shoot off to infinity, which again is desirable.
I think you have some confusion here. Discounting the process (aka correcting for inflation) then the process is driven by the Wiener process $W_t$, which is finite and a Martingale process, and just as likely to go up as it is down (cf. the reflection principle). I think if you do a bit of reading around Brownian motions you will see this.
Don't judge models by eye!
Even a simple random walk will generate a plot more similar (to a human eye) of an actual price action graph
I completely disagree with this, primarily based on the fact that a quantitative model should never assessed or compared to data by eye. If you want to scientifically assess two models you want to compare the quality of their predictions, such as:
- How well do the processes match in a statistical sense? Do they have the same mean, variance, skew, kurtosis, etc.?
- If I use my model to price options on the process, how well do the prices match those seen in the market. (Beware, most models aren't actually used for pricing, but rather for hedging).
- What are the corner cases of my models? Are they mean reverting, do they predict negative values, do they have reflecting boundaries, etc.? Some of these are mathematically desirable, and others physically/economically desirable.
- Is my model causal, adapted, stationary, invertible, etc. (These crop up with several discrete time series models).
To assess the model statistically a very good starting point is to inspect the residuals, and if these look like white noise. If so, it's a good start, and if not, then there is likely room to improve the model. These sorts of model assessments are impossible to do by eye. A trivial model which looks great by eye is $S_{t+1} = S_t$, and would likely be indistinguishable by eye from a slightly more sophisticated $\text{ARMA}(p,q)$ model, but the former is pretty useless for most things, whereas the latter isn't.
If you want a model to adapt, use a more sophisticated model
The geometric Brownian motion model for the price process is very simplistic. The reason for this is because it is the simplest model that was first thought up that produced some interesting financial insight. It allowed for hedging, and the pricing of derivatives and all sorts of options. However, it is very simple and doesn't adapt. Some other models with varying degrees of adaptive behaviour might include:
- An Ornstein-Uhlenbeck (OU) model is mean reverting, and popular for interest rates.
- The Cox-Ingersol-Ross (CIR) model is again mean reverting, but has a variable volatility (perhaps you might call this as changing the trend). It can also have varying behaviour at the boundary based on the Feller condition.
- The Heston model models the volatility process.
- The SABR model can even predict negative interest rates (many thought this couldn't happen, but in recent years it has occurred multiple times).
I have named just a few models here, but there is a tradeoff between how "realistic" a model is, and its analytic tractability. We will often favour something simpler and because we can do useful stuff with a simple model. It is of no use having a complicated model which can't be simulated from nor be used to make predictions.
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$\begingroup$ Great answer, +1. Would you agree that Ito processes can’t model skewed/fat tailed returns? $dS/S=r_tdt+\sigma_tdW_t$, where $\sigma_t$ is any (suitable) volatility process, perhaps even multi-dimensional (cf. Heston). We can even add jumps (cf. Merton). We always have $\mathbb{E}[dS/S]=r_tdt$ (+ jump corrections) and $\mathbb{E}[(dS/S)^2]=\sigma_t^2dt$ (+ correlation terms in higher dimensions). However, higher moments are zero (of higher order than $dt$), i.e. instantaneous returns ($dS/S$) are conditionally normally distributed (due to $dW$) and thus symmetric and have no fat tails? $\endgroup$– AlexMay 29, 2021 at 20:46
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1$\begingroup$ The usual (semi-hand waving) argument with Ito processes is that as the increments are normal random variables, then the sum of them is normal too, and so should have finite variance. This only half works for discrete summations, but for continuous processes and integrals this argument readily fails, where of course the solutions of most SDEs are most definitely not normal. Much of this hinges on making assumptions about $L^2$ integrability about the drift and diffusion functions. Dropping this, and I think you can easily have fat tailed returns. (I expect you can get this without dropping it) $\endgroup$– oliversmJun 4, 2021 at 15:45
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1$\begingroup$ I think the fat tails can quickly appear with certain stochastic volatility models, but I'm not too familiar with the wide range of models and SDEs and their models. (As always, see Kloeden and Platen). I think adding skew could be fairly easy. $\endgroup$– oliversmJun 4, 2021 at 15:48
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1$\begingroup$ However, there is a sharp distinction between instantaneous returns $dS/S$ and macroscopic returns $\Delta S/S$. We can only observe macroscopic effects (at least ignoring the HFT relm for now), and these are the integrated processes. Thus I think to make the jump/inference that what is not present in the infinitesimal description doesn't emerge in the macroscopic scale is erroneous. @Alex $\endgroup$– oliversmJun 4, 2021 at 15:51
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1$\begingroup$ I understand the theory to suggest: $dS/S$ seems to be conditionally normally distributed, symmetric and light-tailed? Conditional on information up to time $t$, we have $\mathbb{E}[dS/S]=\mu_t dt$ and $\mathbb{V} [dS/S]=\sigma_t^2dt$. Higher moments of instantaneous returns are zero (because they include terms like $dt^2$). This aligns with the normality from the Brownian increment $dW_t$. $\endgroup$– AlexJun 6, 2021 at 23:29
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1) Prices are assumed to be lognormal and skewed to accommodate the observation that under normal circumstances returns are normal (which they’re not, but it’s the simplest model we can fit). That’s why the “geometric” part
2) Stock prices by definition can only go down till zero(in case of bankruptcy) so we don’t need a model that can accommodate negative values.
3) Why cannot it Change trend? The W(t) part is random and can take negative values and the volatility multiple will drive it in that direction
4) Brownian motion is synonymous to random walk in continuous time, while random walk is discrete.
answered Apr 15, 2020 at 6:00