Why Deterministic Models Fail¶

Recall (see § Stylized Facts of Financial Returns): the five empirical constraints — heavy tails, volatility clustering, the leverage effect, absence of return autocorrelation, and aggregational Gaussianity — have been documented. Given these properties, we now test whether a deterministic ordinary differential equation can reproduce them. The answer is no, and the reasons are structural, not a matter of calibration. This is the logical hinge of the chapter: deterministic dynamics fail, so randomness is unavoidable.

Concept Definition¶

A deterministic model for an asset price $S(t)$ is an ordinary differential equation (ODE) of the form

\[ \frac{dS}{dt} = f(S, t) \]

whose solution is uniquely determined by the initial condition $S(0) = S_0$. Given $S_0$ and $f$, the price at every future time is exactly known.

The canonical example is exponential growth:

\[ \frac{dS}{dt} = \mu S, \qquad S(0) = S_0 \]

with unique solution

\[ S(t) = S_0 e^{\mu t} \]

More generally, any ODE satisfying standard Lipschitz conditions produces a solution that is continuously differentiable: the derivative $S'(t)$ exists at every $t$.

Why Deterministic Models Fail: Five Structural Failures¶

Failure 1 — Zero Variance¶

The log-return over a period $\Delta t$ under the exponential-growth model is

\[ r_t = \log \frac{S(t + \Delta t)}{S(t)} = \mu \Delta t \]

This is a constant, not a random variable:

\[ \operatorname{Var}(r_t) = 0 \]

Real log-returns satisfy $\operatorname{Var}(r_t) = \sigma^2 \Delta t > 0$ with annualised volatility $\sigma$ typically in the range 15–30 % for equities. A model with zero variance cannot reproduce any observed dispersion of returns, regardless of how $\mu$ is chosen.

Failure 2 — Paths Are Too Smooth¶

The Picard–Lindelöf theorem guarantees that, under mild regularity on $f$, the solution of $dS/dt = f(S,t)$ is continuously differentiable. In particular, the limit

\[ \lim_{\Delta t \to 0} \frac{S(t + \Delta t) - S(t)}{\Delta t} = f(S(t), t) \]

exists and is a deterministic finite value at every $t$.

Real price paths behave differently. As $\Delta t \to 0$, the normalised empirical increment $r_t^{\text{real}} / \sqrt{\Delta t}$ — where $r_t^{\text{real}}$ is the actual observed log-return — converges in distribution to a non-degenerate random variable, not a constant. This is the hallmark of Brownian motion, whose paths are continuous but nowhere differentiable. The derivative $S'(t)$ does not exist for a typical realisation of a stochastic price process.

Quadratic variation distinguishes the two cases

For a differentiable path, the quadratic variation over $[0,T]$ is zero: $\sum_i (S_{t_{i+1}} - S_{t_i})^2 \to 0$ as the partition is refined. For Brownian motion paths the same sum converges to $\sigma^2 T > 0$. Real return data exhibit non-zero, measurable quadratic variation, decisively ruling out differentiable (ODE) paths.

Failure 3 — No Volatility Clustering¶

Recall (see § Stylized Facts of Financial Returns — Stylized Fact 2): squared returns are strongly autocorrelated. Under any deterministic ODE, volatility is identically zero at all times — there is nothing to cluster. A model that cannot generate variability cannot generate time-varying variability.

Failure 4 — Cannot Produce Heavy Tails¶

Under the exponential-growth ODE the log-return distribution is a point mass:

\[ P(r_t = \mu \Delta t) = 1 \]

This distribution has no tails whatsoever. Real returns have heavy tails (see § Stylized Facts of Financial Returns — Stylized Fact 1). No reparametrisation of $\mu$ can resolve this; the issue is structural.

Failure 5 — No Leverage Effect¶

Recall (see § Stylized Facts of Financial Returns — Stylized Fact 3): $\operatorname{Corr}(r_t, \sigma_{t+1}^2) < 0$. Under a deterministic ODE, $r_t = \mu\Delta t$ is a constant and $\sigma_{t+1}^2 = 0$ identically, so the correlation is undefined. There is no mechanism by which a price decline can feed back into increased volatility.

The Naive Fix and Why It Fails¶

A natural first attempt is to bolt random shocks onto the ODE:

\[ S(t + \Delta t) = S(t)\,e^{\mu \Delta t} + \varepsilon_t, \qquad \varepsilon_t \sim \mathcal{N}(0,\, \sigma^2 \Delta t) \]

Problem 1 — Prices can go negative. Because $\varepsilon_t$ is unbounded below, there is positive probability that $S(t + \Delta t) < 0$, which is economically inadmissible.

Problem 2 — Noise does not scale with price. The additive term $\varepsilon_t$ has constant variance $\sigma^2 \Delta t$ regardless of the current price level. Empirically, a $100 stock and a $1000 stock with the same percentage volatility have very different dollar volatilities. Additive noise conflates the two.

Problem 3 — The continuous-time limit diverges. Dividing the increment by $\Delta t$:

\[ \frac{S(t + \Delta t) - S(t)}{\Delta t} = \mu S(t) + \frac{\varepsilon_t}{\Delta t} \]

As $\Delta t \to 0$, the term $\varepsilon_t / \Delta t \sim \mathcal{N}(0,\, \sigma^2/\Delta t)$ has standard deviation growing like $1/\sqrt{\Delta t} \to \infty$. The limit does not exist in $L^2$; additive noise has no consistent continuous-time interpretation.

The Right Fix: Multiplicative Noise¶

All three problems above are resolved by making the noise proportional to the current price, so that the discrete update takes the multiplicative form

\[ S(t + \Delta t) = S(t)\exp\!\left(\mu \Delta t + \sigma\sqrt{\Delta t}\cdot Z\right), \qquad Z \sim \mathcal{N}(0,1) \]

This ensures positivity ($e^x > 0$), correct scaling (drift and diffusion both proportional to $S(t)$), and — as shown next — a non-degenerate continuous-time limit. The actual passage to that limit, the resulting SDE $dS_t = \mu S_t\,dt + \sigma S_t\,dW_t$, and the Itô drift correction $-\sigma^2/2$ in the closed-form solution are taken up in § Bridge to Stochastic Differential Equations and made rigorous in the Itô-calculus chapters.

Recall (see § Itô Calculus Applications): the correct GBM solution is $S_t = S_0 \exp[(\mu - \sigma^2/2)t + \sigma W_t]$; the $-\sigma^2/2$ correction arises from $(dW_t)^2 = dt$ and is absent in ordinary calculus.

Summary¶

Deterministic ODEs fail not because their parameters are poorly chosen but because their mathematical structure is incompatible with five fundamental empirical properties of financial returns:

Zero variance — cannot match observed return dispersion.
Differentiable paths — incompatible with nowhere-differentiable price paths.
No volatility clustering — cannot generate time-varying risk.
No heavy tails — cannot account for the frequency of extreme events.
No leverage effect — no mechanism for asymmetric volatility response.

The minimal resolution is multiplicative noise, which forces us out of ordinary calculus and into the framework of stochastic differential equations. Randomness is no longer optional — it is the only way out. The continuous-time construction itself is carried out next.

Next: § Bridge to Stochastic Differential Equations. $\square$

Exercises¶

Exercise 1. Consider the deterministic ODE $dS/dt = \mu S$ with $S(0) = 100$ and $\mu = 0.10$ (per year). Compute the price $S(t)$ at $t = 1$ year and the log return over $[0,1]$. Now compute $\operatorname{Var}(r)$ for this model. Compare with the empirical annualised volatility of a typical equity ($\sigma \approx 0.25$) and explain why no choice of $\mu$ can resolve this discrepancy.

Solution to Exercise 1

The solution to $dS/dt = \mu S$ with $S(0) = 100$ and $\mu = 0.10$ is:

\[ S(t) = S(0)e^{\mu t} = 100 e^{0.10 \times 1} = 100 e^{0.10} \approx 110.52 \]

The log return over $[0,1]$ is:

\[ r = \log\frac{S(1)}{S(0)} = \log e^{0.10} = 0.10 \]

Since $S(t)$ is a deterministic function, the return $r = \mu t$ is a constant (not a random variable). Therefore:

\[ \operatorname{Var}(r) = 0 \]

A typical equity has annualised volatility $\sigma \approx 0.25$, meaning $\operatorname{Var}(r_{\text{ann}}) = \sigma^2 \approx 0.0625$. No choice of $\mu$ can resolve this discrepancy because $\mu$ only controls the level of the deterministic return, not its variability. The variance is identically zero for any value of $\mu$ — whether $\mu = 0.01$ or $\mu = 100$, the model produces a single deterministic path with no randomness. The failure is structural: a deterministic ODE produces exactly one trajectory, so there is no ensemble of outcomes over which variance could be defined. Matching observed return dispersion requires a fundamentally different mathematical framework that incorporates randomness.

Exercise 2. The quadratic variation of a differentiable function $g(t)$ over $[0,T]$ is defined as

\[ [g]_T = \lim_{|\mathcal{P}| \to 0} \sum_{i=0}^{n-1} \bigl(g(t_{i+1}) - g(t_i)\bigr)^2 \]

Show that $[g]_T = 0$ for any continuously differentiable function. Then explain why $[W]_T = T$ for standard Brownian motion $W_t$, and why this non-zero quadratic variation rules out differentiable paths as a model for asset prices.

Solution to Exercise 2

Quadratic variation of a differentiable function is zero. Let $g$ be continuously differentiable on $[0,T]$. By the Mean Value Theorem, for each subinterval:

\[ g(t_{i+1}) - g(t_i) = g'(\xi_i)(t_{i+1} - t_i) \]

for some $\xi_i \in (t_i, t_{i+1})$. Therefore:

\[ \sum_{i=0}^{n-1}(g(t_{i+1}) - g(t_i))^2 = \sum_{i=0}^{n-1}[g'(\xi_i)]^2(t_{i+1} - t_i)^2 \]

Since $g'$ is continuous on $[0,T]$, it is bounded: $|g'(\xi_i)| \leq M$ for some constant $M$. Let $|\mathcal{P}| = \max_i(t_{i+1} - t_i)$ denote the mesh of the partition. Then:

\[ \sum_{i=0}^{n-1}[g'(\xi_i)]^2(t_{i+1} - t_i)^2 \leq M^2 |\mathcal{P}| \sum_{i=0}^{n-1}(t_{i+1} - t_i) = M^2 |\mathcal{P}| \cdot T \]

As $|\mathcal{P}| \to 0$, this upper bound tends to zero, so $[g]_T = 0$.

Quadratic variation of Brownian motion. For standard Brownian motion $W_t$, the quadratic variation over $[0,T]$ is $[W]_T = T$. This can be seen by considering a partition $0 = t_0 < t_1 < \cdots < t_n = T$ with equal spacing $\Delta t = T/n$. The increments $\Delta W_i = W_{t_{i+1}} - W_{t_i}$ are independent with $\Delta W_i \sim \mathcal{N}(0, \Delta t)$, so $(\Delta W_i)^2$ has mean $\Delta t$ and variance $2(\Delta t)^2$. Therefore:

\[ \mathbb{E}\!\left[\sum_{i=0}^{n-1}(\Delta W_i)^2\right] = n\Delta t = T \]

\[ \operatorname{Var}\!\left[\sum_{i=0}^{n-1}(\Delta W_i)^2\right] = n \cdot 2(\Delta t)^2 = 2T^2/n \to 0 \]

By the $L^2$ convergence (mean $T$, variance $\to 0$), the sum converges to $T$ almost surely.

Why this rules out differentiable paths. Real financial data exhibit non-zero quadratic variation: $\sum_i (S_{t_{i+1}} - S_{t_i})^2$ converges to a positive value $\sigma^2 T > 0$ as the partition is refined. Since any differentiable function has zero quadratic variation, ODE solutions cannot reproduce this empirical feature. Only processes with non-differentiable paths (such as Brownian motion) can generate the non-zero quadratic variation observed in real prices.

Exercise 3. Consider the naive additive-noise model $S(t + \Delta t) = S(t)e^{\mu\Delta t} + \varepsilon_t$ with $\varepsilon_t \sim \mathcal{N}(0, \sigma^2\Delta t)$. If $S(t) = 50$, $\mu = 0.05$, and $\sigma = 0.20$, compute the probability that $S(t + \Delta t) < 0$ for $\Delta t = 1/252$ (one trading day). Then repeat for the multiplicative model $S(t+\Delta t) = S(t)\exp(\mu\Delta t + \sigma\sqrt{\Delta t}\cdot Z)$ with $Z \sim \mathcal{N}(0,1)$ and explain why the probability of a negative price is exactly zero.

Solution to Exercise 3

Additive-noise model: $S(t + \Delta t) = S(t)e^{\mu\Delta t} + \varepsilon_t$ with $\varepsilon_t \sim \mathcal{N}(0, \sigma^2\Delta t)$.

With $S(t) = 50$, $\mu = 0.05$, $\Delta t = 1/252$:

\[ S(t)e^{\mu\Delta t} = 50 \cdot e^{0.05/252} \approx 50 \cdot 1.000198 \approx 50.0099 \]

The standard deviation of $\varepsilon_t$ is $\sigma\sqrt{\Delta t} = 0.20\sqrt{1/252} = 0.20/15.875 \approx 0.01260$.

For $S(t+\Delta t) < 0$, we need $\varepsilon_t < -50.0099$. The number of standard deviations below the mean:

\[ z = \frac{-50.0099}{0.01260} \approx -3969 \]

This is approximately 3969 standard deviations below zero. While $P(\varepsilon_t < -50.0099)$ is astronomically small (effectively zero for practical purposes), it is strictly positive. The Gaussian distribution has unbounded support, so there is always a non-zero probability of any real value. In principle, $P(S(t+\Delta t) < 0) > 0$, making the model economically inadmissible.

Multiplicative model: $S(t+\Delta t) = S(t)\exp(\mu\Delta t + \sigma\sqrt{\Delta t}\cdot Z)$.

Since $S(t) = 50 > 0$ and the exponential function satisfies $e^x > 0$ for all $x \in \mathbb{R}$, regardless of the value of $Z$:

\[ S(t+\Delta t) = 50 \cdot \exp(\mu\Delta t + \sigma\sqrt{\Delta t}\cdot Z) > 0 \]

The probability of a negative price is exactly zero. This is the fundamental advantage of multiplicative noise: the exponential structure guarantees price positivity for all realisations of the driving random variable, making it consistent with the economic requirement that asset prices cannot be negative.

Exercise 4. In the multiplicative noise model, consider the normalised log-increment

\[ \frac{\log S(t+\Delta t) - \log S(t)}{\sqrt{\Delta t}} = \mu\sqrt{\Delta t} + \sigma Z, \qquad Z \sim \mathcal{N}(0,1) \]

Compute the mean and variance of this quantity. Take the limit as $\Delta t \to 0$ and show that it converges in distribution to $\mathcal{N}(0, \sigma^2)$. Explain why this convergence justifies the SDE notation $d(\log S_t) = \mu\,dt + \sigma\,dW_t$.

Solution to Exercise 4

The normalised log-increment is:

\[ Y_{\Delta t} = \frac{\log S(t+\Delta t) - \log S(t)}{\sqrt{\Delta t}} = \mu\sqrt{\Delta t} + \sigma Z, \quad Z \sim \mathcal{N}(0,1) \]

Mean:

\[ \mathbb{E}[Y_{\Delta t}] = \mu\sqrt{\Delta t} + \sigma \cdot \mathbb{E}[Z] = \mu\sqrt{\Delta t} + 0 = \mu\sqrt{\Delta t} \]

Variance:

\[ \operatorname{Var}(Y_{\Delta t}) = \sigma^2 \operatorname{Var}(Z) = \sigma^2 \cdot 1 = \sigma^2 \]

(The term $\mu\sqrt{\Delta t}$ is a constant and does not contribute to variance.)

Limit as $\Delta t \to 0$: As $\Delta t \to 0$, the mean $\mu\sqrt{\Delta t} \to 0$ and the variance remains $\sigma^2$. Therefore:

\[ Y_{\Delta t} = \mu\sqrt{\Delta t} + \sigma Z \xrightarrow{d} \sigma Z \sim \mathcal{N}(0, \sigma^2) \]

The convergence is in distribution (in fact, also in $L^2$ and almost surely, since only the deterministic shift vanishes).

This convergence justifies the SDE notation $d(\log S_t) = \mu\,dt + \sigma\,dW_t$ as follows. Over an infinitesimal interval $dt$, the log-price increment is $d(\log S_t) = \mu\,dt + \sigma\,dW_t$, where $dW_t \sim \mathcal{N}(0, dt)$. Dividing by $\sqrt{dt}$:

\[ \frac{d(\log S_t)}{\sqrt{dt}} = \mu\sqrt{dt} + \sigma\frac{dW_t}{\sqrt{dt}} \]

Since $dW_t/\sqrt{dt} \sim \mathcal{N}(0,1)$, this matches $Y_{\Delta t}$ in the limit. The SDE notation encodes the distributional structure of the limiting rescaled increment.

Exercise 5. The correct GBM solution is $S_t = S_0\exp[(\mu - \sigma^2/2)t + \sigma W_t]$, whereas naive integration gives $S_t = S_0\exp[\mu t + \sigma W_t]$. For $S_0 = 100$, $\mu = 0.08$, $\sigma = 0.30$, and $t = 5$ years, compute $\mathbb{E}[S_t]$ under both formulas. Which formula gives $\mathbb{E}[S_t] = S_0 e^{\mu t}$? Verify that the correct formula satisfies $\mathbb{E}[S_t] = S_0 e^{\mu t}$ by using the moment generating function of the normal distribution.

Solution to Exercise 5

Naive formula: $S_t = S_0\exp[\mu t + \sigma W_t]$

Since $W_t \sim \mathcal{N}(0, t)$, we use the moment generating function $\mathbb{E}[e^{aW_t}] = e^{a^2 t/2}$:

\[ \mathbb{E}[S_t^{\text{naive}}] = S_0 e^{\mu t}\mathbb{E}[e^{\sigma W_t}] = S_0 e^{\mu t} \cdot e^{\sigma^2 t/2} = S_0 e^{(\mu + \sigma^2/2)t} \]

With $S_0 = 100$, $\mu = 0.08$, $\sigma = 0.30$, $t = 5$:

\[ \mathbb{E}[S_5^{\text{naive}}] = 100\exp\!\left[(0.08 + 0.045) \times 5\right] = 100\exp(0.625) \approx 186.82 \]

Correct formula: $S_t = S_0\exp[(\mu - \sigma^2/2)t + \sigma W_t]$

\[ \mathbb{E}[S_t^{\text{correct}}] = S_0 e^{(\mu - \sigma^2/2)t}\mathbb{E}[e^{\sigma W_t}] = S_0 e^{(\mu - \sigma^2/2)t} \cdot e^{\sigma^2 t/2} = S_0 e^{\mu t} \]

With the given parameters:

\[ \mathbb{E}[S_5^{\text{correct}}] = 100\exp(0.08 \times 5) = 100\exp(0.40) \approx 149.18 \]

The correct formula gives $\mathbb{E}[S_t] = S_0 e^{\mu t}$. The naive formula gives $\mathbb{E}[S_t] = S_0 e^{(\mu + \sigma^2/2)t}$, which overestimates the expected price.

Verification using the MGF: For the correct formula, $\log S_t = \log S_0 + (\mu - \sigma^2/2)t + \sigma W_t$, so $S_t = S_0 \exp[(\mu - \sigma^2/2)t + \sigma W_t]$. Taking expectations:

\[ \mathbb{E}[S_t] = S_0 e^{(\mu - \sigma^2/2)t} \cdot \mathbb{E}[e^{\sigma W_t}] \]

Since $\sigma W_t \sim \mathcal{N}(0, \sigma^2 t)$, the MGF gives $\mathbb{E}[e^{\sigma W_t}] = e^{\sigma^2 t / 2}$. Therefore:

\[ \mathbb{E}[S_t] = S_0 e^{(\mu - \sigma^2/2)t} \cdot e^{\sigma^2 t/2} = S_0 e^{\mu t} \]

The $-\sigma^2/2$ in the exponent of the GBM solution exactly cancels the $+\sigma^2/2$ from the MGF, yielding the clean result $\mathbb{E}[S_t] = S_0 e^{\mu t}$.

Exercise 6. For each of the five structural failures of deterministic models (zero variance, smooth paths, no volatility clustering, no heavy tails, no leverage effect), state which feature of the GBM SDE $dS_t = \mu S_t\,dt + \sigma S_t\,dW_t$ addresses it and which failures require extensions beyond basic GBM. Organise your answer as a table with columns: Failure, Addressed by GBM?, Required Extension.

Solution to Exercise 6

Failure	Addressed by GBM?	Required Extension
Zero variance	Yes. The $\sigma S_t\,dW_t$ term gives $\operatorname{Var}(r_t) = \sigma^2\Delta t > 0$	None
Smooth paths	Yes. Brownian motion $W_t$ has continuous but nowhere-differentiable paths, so $S_t$ is also nowhere differentiable	None
No volatility clustering	No. GBM has constant $\sigma$, so $\operatorname{Corr}(r_t^2, r_{t+k}^2) = 0$	Stochastic volatility SDE (e.g., Heston: $dV_t = \kappa(\theta - V_t)\,dt + \xi\sqrt{V_t}\,dW_t^V$)
No heavy tails	No. GBM produces log-normal returns with excess kurtosis 0 for log returns	Jump-diffusion models (Merton) or stochastic volatility (which generates excess kurtosis through mixing)
No leverage effect	No. GBM has a single Brownian motion; there is no mechanism linking returns to future volatility	Correlated Brownian motions with $\rho < 0$ in a stochastic volatility model (Heston with $\rho < 0$)

In summary, GBM resolves the two most fundamental failures of deterministic models (zero variance and smooth paths) by introducing Brownian motion. However, it cannot capture the three stylized facts that involve the structure of volatility (clustering, heavy tails, leverage), which require the volatility itself to be a stochastic process correlated with the price.

Failure	Addressed by GBM?	Required Extension
Zero variance	Yes. The \(\sigma S_t\,dW_t\) term gives \(\operatorname{Var}(r_t) = \sigma^2\Delta t > 0\)	None
Smooth paths	Yes. Brownian motion \(W_t\) has continuous but nowhere-differentiable paths, so \(S_t\) is also nowhere differentiable	None
No volatility clustering	No. GBM has constant \(\sigma\), so \(\operatorname{Corr}(r_t^2, r_{t+k}^2) = 0\)	Stochastic volatility SDE (e.g., Heston: \(dV_t = \kappa(\theta - V_t)\,dt + \xi\sqrt{V_t}\,dW_t^V\))
No heavy tails	No. GBM produces log-normal returns with excess kurtosis 0 for log returns	Jump-diffusion models (Merton) or stochastic volatility (which generates excess kurtosis through mixing)
No leverage effect	No. GBM has a single Brownian motion; there is no mechanism linking returns to future volatility	Correlated Brownian motions with \(\rho < 0\) in a stochastic volatility model (Heston with \(\rho < 0\))