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# 13. Locally Unpredictable Stochastic Processes

```{eval-rst}
.. index::
   single: locally unpredictable process
   single: martingale; locally
   pair: smoothness; predictability
```

Lack of mean square differentiability implies that a process is "locally
unpredictable" or is "locally like a martingale." Sims used the concept of
local unpredictability in his work on asset prices, interest rates, and
consumption.

This chapter is where the thread begun in {doc}`02_mean_square_continuity_differentiability`
arrives at its economic payoff. There, mean square differentiability was tied to the existence
of $R''(0)$; in {doc}`09_characterizations_ms_differentiability` that became the condition
$p(0) = 0$ on the Wold kernel; in {doc}`11_linear_sde` it became a count, $n - 1 - m$, read off
the degrees of two polynomials. Here the failure of that condition means something. Over short intervals, a process whose
kernel does not vanish at the origin is unforecastable. We use the following definition:

```{eval-rst}
.. index::
   single: locally unpredictable process; definition
   single: local unpredictability
   single: forecast error variance; over short intervals
```

```{eval-rst}
.. index::
   single: Sims, C. A.
   single: permanent income; and local unpredictability
```

```{prf:definition}
:label: def-locally-unpredictable

A stochastic process $x(t)$ with finite second moments is said to
be *locally unpredictable* if

$$
\lim_{\delta \to 0}\ \frac{E_t (x(t+\delta) - E_t x(t+\delta))^2}{E_t (x(t+\delta) - x(t))^2}\ = 1.
$$
```

This definition makes precise the sense in which $x(t)$ is locally a martingale,
i.e., the sense in which for small $\delta$,

$$
E_t x(t+\delta) \simeq x(t).
$$

We have the following theorem.

```{eval-rst}
.. index::
   single: local unpredictability; kernel criterion
```

```{prf:theorem}
:label: thm-local-unpredictability

Let $x(t)$ be a linearly indeterministic covariance stationary
stochastic process with Wold representation

$$
x(t) = \int^\infty_0 p(\tau)\, w(t-\tau)\, d\tau
$$

where $w(t)$ is a fundamental white noise for the $x(t)$ process, and
$Ew(t) w(t-\tau) = \delta(\tau)$. Assume that $p(\tau)$ is twice continuously
differentiable, with $p'$ and $p''$ square integrable. Then if $x(t)$ is not mean square
differentiable, $x(t)$ is locally unpredictable.
```

```{prf:proof}

From the Wiener-Kolmogorov prediction formula, we have that

$$
\frac{E_t (x(t+\delta) - E_t x(t+\delta))^2}{E_t (x(t+\delta) - x(t))^2}
$$

$$
= \frac{\int^\delta_0 p(s)^2\, ds}{\int^\delta_0 p(s)^2\, ds + \int^\infty_0 (p(s+\delta) - p(s))^2\, ds}
$$

Taking the limit as $\delta \to 0$ gives, after applying l'Hospital's rule,

$$
\lim_{\delta \to 0}\ \frac{p(\delta)^2}{p(\delta)^2 + 2 \int^\infty_0 (p(s+\delta) - p(s))\, p'(s+\delta)\, ds}
$$

If $p(0) \neq 0$, this limit is given by

$$
\frac{p(0)^2}{p(0)^2}\ = 1.
$$

Upon noting that $p(0) = 0$ if and only if $x(t)$ is mean square differentiable
(by {doc}`09_characterizations_ms_differentiability`: with $p'$ square integrable, mean square
differentiability reduces to the single condition $p(0) = \lim_{s\to\infty} s\tilde P(s) = 0$),
we have the desired result that if $x(t)$ is not mean square differentiable, then $x(t)$ is
locally unpredictable.
```

For the rational linear stochastic differential equations of {doc}`11_linear_sde`, this
dividing line is explicit. There $x(t)$ is mean square differentiable $n - 1 - m$ times, where
$n$ and $m$ are the degrees of the denominator operator $\theta(D)$ and the numerator operator
$\psi(D)$; by the initial value theorem, $p(0) = \lim_{s\to\infty} s\tilde P(s) \neq 0$ exactly
when $m = n - 1$. The locally unpredictable members of this family are therefore precisely those
with $n - 1 - m = 0$, a numerator just one degree below the denominator. Every smoother member,
with $m < n - 1$, is locally predictable.

Using the preceding theorem and our formula {eq}`eq-12-gen` for geometric distributed leads from
{doc}`12_prediction`, it is straightforward to
establish that if $x(t)$ is a covariance stationary stochastic process with Wold
representation

$$
x(t) = \int^\infty_0 p(s)\, w(t-s)\, ds = \tilde P (D)\, w(t),
$$ (eq-13-wold)

then for any $\rho < 0$, the geometric sum of future expected $x$'s,

$$
\begin{aligned}
x(t)\ast &\equiv E_t \int^\infty_0 e^{\rho s}\, x(t+s)\, ds = \left[\frac{- \tilde P (D) + \tilde P(-\rho)}{D+\rho}\right]\, w(t) \\
&\equiv \tilde G(D)\, w(t) \equiv \int^\infty_0 g(s)\, w(t-s)\, ds
\end{aligned}
$$

is locally unpredictable, even if $x(t)$ is mean square differentiable. To show
this, we apply the initial value theorem of {doc}`09_characterizations_ms_differentiability`
to the kernel $g$:

$$
\begin{aligned}
g(0) &= \lim_{s\to \infty}\, s \tilde G(s) \\
&= \lim_{s\to \infty}\, \left[ \frac{-s\tilde P(s)}{s+\rho} + \frac{s \tilde P(-\rho)}{s+\rho} \right] \\
&= \tilde P (-\rho) \neq 0.
\end{aligned}
$$

(Here we are using that $\lim_{s\to \infty} s\tilde P (s) = 0$ by the assumption
of mean square differentiability of $x(t)$.) We know that $\tilde P (-\rho) \neq 0$
because $\tilde P(s)$ has no zeroes in the right half plane, by the assumption
that {eq}`eq-13-wold` is a Wold representation.

This last result carries the chapter's economic content. Taking a present value manufactures
local unpredictability. However smooth the
dividend, income, or endowment process $x$ may be, the asset price or permanent income built
from it behaves, over short intervals, like a martingale. Nothing about tastes or market
structure is needed for the conclusion; it follows from the algebra of the annihilation
operator.

```{eval-rst}
.. index::
   single: asset prices; and local unpredictability
   single: random walk; as a local martingale
```

## Exercises

The ratio in {prf:ref}`def-locally-unpredictable` can be computed for any kernel, at any $\delta$, directly from the
expression established in the proof:

```{math}
:label: eq-13-ratio
\varrho(\delta) \;=\;
\frac{\int_0^\delta p(s)^2\, ds}{\int_0^\delta p(s)^2\, ds + \int_0^\infty \big(p(s+\delta) - p(s)\big)^2\, ds}\, .
```

The numerator is the variance of the $\delta$-ahead forecast error; the denominator is the
variance of the actual change $x(t+\delta) - x(t)$. A process is locally unpredictable when the
forecast error accounts for *all* of the movement in the limit.

```{code-cell} ipython3
import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import quad

def varrho(p, delta, upper=60.0):
    """The local-unpredictability ratio at horizon delta, for a one-sided kernel p."""
    num, _ = quad(lambda s: p(s)**2, 0.0, delta, limit=200)
    rev, _ = quad(lambda s: (p(s + delta) - p(s))**2, 0.0, upper, limit=400)
    return num / (num + rev)
```

```{exercise-start}
:label: lu_ex1
```

**The dividing line.** Take $a = 1$ and compare two kernels from the rational family of
{doc}`11_linear_sde`:

$$
p_1(\tau) = e^{-a\tau} \quad (n=1,\ m=0), \qquad
p_2(\tau) = \tau\, e^{-a\tau} \quad (n=2,\ m=0).
$$

The first has $p_1(0) = 1 \neq 0$ and is differentiable $n-1-m = 0$ times; the second has
$p_2(0) = 0$ and is differentiable once.

(a) Compute $\varrho(\delta)$ for each over a range of $\delta$ and verify that
$\varrho \to 1$ for $p_1$ and $\varrho \to 0$ for $p_2$.

(b) Explain the *rates*. Show analytically that for a kernel with $p(0)\neq 0$ the numerator of
{eq}`eq-13-ratio` is $O(\delta)$ while the second term of the denominator is $O(\delta^2)$,
whereas for a kernel with $p(0)=0$ they are $O(\delta^3)$ and $O(\delta^2)$ respectively. Hence
$\varrho(\delta) \to 1$ in the first case and $\varrho(\delta) = O(\delta)\to 0$ in the second.

```{exercise-end}
```

```{solution-start} lu_ex1
:class: dropdown
```

(b) If $p(0) \neq 0$ then $p(s)^2 \approx p(0)^2$ near the origin, so
$\int_0^\delta p^2 \approx p(0)^2\delta$. Meanwhile $p(s+\delta)-p(s)\approx \delta p'(s)$, so
the revision term is $\approx \delta^2\int_0^\infty p'^2$. The ratio is
$p(0)^2\delta / [p(0)^2\delta + O(\delta^2)] \to 1$. If instead $p(0)=0$ then
$p(s)\approx p'(0)s$ near the origin and $\int_0^\delta p^2 \approx p'(0)^2\delta^3/3$, while
the revision term is still $O(\delta^2)$; the ratio is therefore $O(\delta) \to 0$. The
*whole* distinction is the behaviour of $p$ at the origin.

```{code-cell} ipython3
a = 1.0
p1 = lambda s: np.exp(-a*s)          # p(0) = 1  -> locally unpredictable
p2 = lambda s: s*np.exp(-a*s)        # p(0) = 0  -> mean square differentiable

deltas = np.logspace(-3, 0.3, 40)
r1 = np.array([varrho(p1, d) for d in deltas])
r2 = np.array([varrho(p2, d) for d in deltas])

for d, x, y in zip(deltas[::13], r1[::13], r2[::13]):
    print(f"delta={d:7.4f}   varrho(p1)={x:.4f}   varrho(p2)={y:.4f}")
```

```{code-cell} ipython3
wide = np.logspace(-3, 1.2, 60)
R1 = np.array([varrho(p1, d, upper=200.0) for d in wide])
R2 = np.array([varrho(p2, d, upper=200.0) for d in wide])

fig, ax = plt.subplots(figsize=(9, 5))
ax.semilogx(wide, R1, 'o-', lw=2, ms=4, label=r'$p(\tau)=e^{-a\tau}$,  $p(0)\neq 0$')
ax.semilogx(wide, R2, 's-', lw=2, ms=4, label=r'$p(\tau)=\tau e^{-a\tau}$,  $p(0)=0$')
ax.axhline(1, color='0.6', ls=':')
ax.axhline(0, color='0.6', ls=':')
ax.axhline(0.5, color='0.4', ls='--', lw=1,
           label=r'$\varrho \to 1/2$ as $\delta\to\infty$ (any stationary process)')
ax.set_xlabel(r'forecast horizon $\delta$  (log scale)')
ax.set_ylabel(r'local-unpredictability ratio $\varrho(\delta)$')
ax.set_title('Local unpredictability is decided by the kernel at the origin')
ax.legend(fontsize=9); plt.show()
```

The figure shows why the *limit at zero* is the thing to look at. At long horizons the two
curves are indistinguishable: both tend to $\tfrac12$, as they must for any stationary process,
since the forecast error variance tends to $R(0)$ while
$E(x(t+\delta)-x(t))^2 \to 2R(0)$. It is only as $\delta \to 0$ that the processes separate, and
then they separate completely. For $p_1$ nearly all of the movement over $[t, t+\delta]$
is unforecastable; for $p_2$ nearly none of it is. The two kernels differ only by a factor
of $\tau$; that is, only in their behaviour at the origin.

```{solution-end}
```

```{exercise-start}
:label: lu_ex2
```

**Present values are locally unpredictable.** Let $x$ have the *smooth* kernel
$p(\tau) = \tau e^{-a\tau}$ of the previous exercise, so that $x$ is mean square differentiable
and, by {ref}`lu_ex1`, locally *predictable*. Form the present value

$$
x^*(t) = E_t \int_0^\infty e^{\rho s} x(t+s)\, ds, \qquad \rho < 0 .
$$

(a) Show that the kernel of $x^*$ is $g(s) = \int_0^\infty e^{\rho u}\, p(s+u)\, du$, and that
for this $p$,

$$
g(s) = e^{-as}\left[\frac{s}{a-\rho} + \frac{1}{(a-\rho)^2}\right],
\qquad\text{so}\qquad g(0) = \frac{1}{(a-\rho)^2} = \tilde P(-\rho) \neq 0 .
$$

(b) Verify numerically that $\varrho(\delta) \to 1$ for $g$, so that $x^*$ is locally
unpredictable even though $x$ is not. Interpret: if $x$ is a dividend and $x^*$ a stock price,
a smooth dividend process delivers a price that behaves locally like a martingale.

```{exercise-end}
```

```{solution-start} lu_ex2
:class: dropdown
```

(a) Writing $x(t+s) = \int_0^\infty p(\tau)w(t+s-\tau)d\tau$ and keeping only the terms dated
$t$ or earlier, $E_t x(t+s) = \int_0^\infty p(\tau+s) w(t-\tau)d\tau$. Multiplying by
$e^{\rho s}$ and integrating over $s$ gives $x^* = \int_0^\infty g(\tau)w(t-\tau)d\tau$ with
$g$ as stated. For $p(\tau)=\tau e^{-a\tau}$,
$g(s) = e^{-as}\int_0^\infty (s+u)e^{-(a-\rho)u}du$, which integrates to the displayed form.
Setting $s=0$ gives $g(0) = (a-\rho)^{-2}$, and $\tilde P(s) = (s+a)^{-2}$ evaluated at $-\rho$
is the same thing, confirming $g(0) = \tilde P(-\rho)$ of the text.

```{code-cell} ipython3
rho = -0.5
g  = lambda s: np.exp(-a*s)*(s/(a-rho) + 1/(a-rho)**2)

# check (a): the closed form against the defining integral, and against P(-rho)
for s in [0.0, 0.7, 2.0]:
    direct, _ = quad(lambda u: np.exp(rho*u)*p2(s+u), 0, 80, limit=300)
    print(f"s={s:4.1f}:  g(s) closed form = {g(s):.8f},  integral = {direct:.8f}")
print(f"g(0) = {g(0.0):.8f},  P(-rho) = 1/(a-rho)^2 = {1/(a-rho)**2:.8f}")
```

```{code-cell} ipython3
rg = np.array([varrho(g, d) for d in deltas])

fig, ax = plt.subplots(figsize=(9, 5))
ax.semilogx(deltas, r2, 's-', lw=2, color='C1',
            label=r'$x$: kernel $\tau e^{-a\tau}$  (smooth, locally predictable)')
ax.semilogx(deltas, rg, 'D-', lw=2, color='C3',
            label=r'$x^*=E_t\int e^{\rho s}x(t+s)ds$  (locally unpredictable)')
ax.axhline(1, color='0.6', ls=':'); ax.axhline(0, color='0.6', ls=':')
ax.set_xlabel(r'forecast horizon $\delta$  (log scale)')
ax.set_ylabel(r'$\varrho(\delta)$')
ax.set_title('Taking a present value manufactures local unpredictability')
ax.legend(fontsize=9); plt.show()
```

The present-value operator moves the process across the dividing line. The reason is visible in
the closed form for $g$: the operator adds a constant term $(a-\rho)^{-2}$ to a kernel that
previously vanished at the origin. This is the formal content of the observation, due to Sims
and to Hall, that asset prices and consumption should look like martingales at high frequency
even when the fundamentals driving them are smooth. {doc}`18_time_aggregation_var` finds for
that reason that the sampling distortions of Table 1 are *less* severe for present-value
variables than for the smooth processes underlying them.

```{solution-end}
```