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+---
+title: "Itō integral"
+firstLetter: "I"
+publishDate: 2021-11-06
+categories:
+- Mathematics
+
+date: 2021-10-21T19:41:58+02:00
+draft: false
+markup: pandoc
+---
+
+# Itō integral
+
+The **Itō integral** offers a way to integrate
+a time-indexed [random variable](/know/concept/random-variable/)
+$G_t$ (i.e. a stochastic process) with respect
+to a [Wiener process](/know/concept/wiener-process/) $B_t$,
+which is also a stochastic process.
+The Itō integral $I_t$ of $G_t$ is defined as follows:
+
+$$\begin{aligned}
+    \boxed{
+        I_t
+        \equiv \int_a^b G_t \dd{B_t}
+        \equiv \lim_{h \to 0} \sum_{t = a}^{t = b} G_t \big(B_{t + h} - B_t\big)
+    }
+\end{aligned}$$
+
+Where have partitioned the time interval $[a, b]$ into steps of size $h$.
+The above integral exists if $G_t$ and $B_t$ are adapted
+to a common [filtration](/know/concept/sigma-algebra) $\mathcal{F}_t$,
+and $\mathbf{E}[G_t^2]$ is integrable for $t \in [a, b]$.
+If $I_t$ exists, $G_t$ is said to be **Itō-integrable** with respect to $B_t$.
+
+
+## Motivation
+
+Consider the following simple first-order differential equation for $X_t$,
+for some function $f$:
+
+$$\begin{aligned}
+    \dv{X_t}{t}
+    = f(X_t)
+\end{aligned}$$
+
+This can be solved numerically using the explicit Euler scheme
+by discretizing it with step size $h$,
+which can be applied recursively, leading to:
+
+$$\begin{aligned}
+    X_{t+h}
+    \approx X_{t} + f(X_t) \: h
+    \quad \implies \quad
+    X_t
+    \approx X_0 + \sum_{s = 0}^{s = t} f(X_s) \: h
+\end{aligned}$$
+
+In the limit $h \to 0$, this leads to the following unsurprising integral for $X_t$:
+
+$$\begin{aligned}
+    \int_0^t f(X_s) \dd{s}
+    = \lim_{h \to 0} \sum_{s = 0}^{s = t} f(X_s) \: h
+\end{aligned}$$
+
+In contrast, consider the *stochastic differential equation* below,
+where $\xi_t$ represents white noise,
+which is informally the $t$-derivative
+of the Wiener process $\xi_t = \dv*{B_t}{t}$:
+
+$$\begin{aligned}
+    \dv{X_t}{t}
+    = g(X_t) \: \xi_t
+\end{aligned}$$
+
+Now $X_t$ is not deterministic,
+since $\xi_t$ is derived from a random variable $B_t$.
+If $g = 1$, we expect $X_t = X_0 + B_t$.
+With this in mind, we introduce the **Euler-Maruyama scheme**:
+
+$$\begin{aligned}
+    X_{t+h}
+    &= X_t + g(X_t) \: (\xi_{t+h} - \xi_t) \: h
+    \\
+    &= X_t + g(X_t) \: (B_{t+h} - B_t)
+\end{aligned}$$
+
+We would like to turn this into an integral for $X_t$, as we did above.
+Therefore, we state:
+
+$$\begin{aligned}
+    X_t
+    = X_0 + \int_0^t g(X_s) \dd{B_s}
+\end{aligned}$$
+
+This integral is *defined* as below,
+analogously to the first, but with $h$ replaced by
+the increment $B_{t+h} \!-\! B_t$ of a Wiener process.
+This is an Itō integral:
+
+$$\begin{aligned}
+    \int_0^t g(X_s) \dd{B_s}
+    \equiv \lim_{h \to 0} \sum_{s = 0}^{s = t} g(X_s) \big(B_{s + h} - B_s\big)
+\end{aligned}$$
+
+For more information about applying the Itō integral in this way,
+see the [Itō calculus](/know/concept/ito-calculus/).
+
+
+## Properties
+
+Since $G_t$ and $B_t$ must be known (i.e. $\mathcal{F}_t$-adapted)
+in order to evaluate the Itō integral $I_t$ at any given $t$,
+it logically follows that $I_t$ is also $\mathcal{F}_t$-adapted.
+
+Because the Itō integral is defined as the limit of a sum of linear terms,
+it inherits this linearity.
+Consider two Itō-integrable processes $G_t$ and $H_t$,
+and two constants $v, w \in \mathbb{R}$:
+
+$$\begin{aligned}
+    \int_a^b v G_t + w H_t \dd{B_t}
+    = v\! \int_a^b G_t \dd{B_t} +\: w\! \int_a^b H_t \dd{B_t}
+\end{aligned}$$
+
+By adding multiple summations,
+the Itō integral clearly satisfies, for $a < b < c$:
+
+$$\begin{aligned}
+    \int_a^c G_t \dd{B_t}
+    = \int_a^b G_t \dd{B_t} + \int_b^c G_t \dd{B_t}
+\end{aligned}$$
+
+A more interesting property is the **Itō isometry**,
+which expresses the expectation of the square of an Itō integral of $G_t$
+as a simpler "ordinary" integral of the expectation of $G_t^2$
+(which exists by the definition of Itō-integrability):
+
+$$\begin{aligned}
+    \boxed{
+        \mathbf{E} \bigg( \int_a^b G_t \dd{B_t} \bigg)^2
+        = \int_a^b \mathbf{E} \big[ G_t^2 \big] \dd{t}
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-isometry"/>
+<label for="proof-isometry">Proof</label>
+<div class="hidden">
+<label for="proof-isometry">Proof.</label>
+We write out the left-hand side of the Itō isometry,
+where eventually $h \to 0$:
+
+$$\begin{aligned}
+    \mathbf{E} \bigg[ \sum_{t = a}^{t = b} G_t (B_{t + h} \!-\! B_t) \bigg]^2
+    &= \sum_{t = a}^{t = b} \sum_{s = a}^{s = b} \mathbf{E} \bigg[ G_t G_s (B_{t + h} \!-\! B_t) (B_{s + h} \!-\! B_s) \bigg]
+\end{aligned}$$
+
+In the particular case $t \ge s \!+\! h$,
+a given term of this summation can be rewritten
+as follows using the *law of total expectation*
+(see [conditional expectation](/know/concept/conditional-expectation/)):
+
+$$\begin{aligned}
+    \mathbf{E} \Big[ G_t G_s (B_{t + h} \!-\! B_t) (B_{s + h} \!-\! B_s) \Big]
+    = \mathbf{E} \bigg[ \mathbf{E} \Big[ G_t G_s (B_{t + h} \!-\! B_t) (B_{s + h} \!-\! B_s) \Big| \mathcal{F}_t \Big] \bigg]
+\end{aligned}$$
+
+Recall that $G_t$ and $B_t$ are adapted to $\mathcal{F}_t$:
+at time $t$, we have information $\mathcal{F}_t$,
+which includes knowledge of the realized values $G_t$ and $B_t$.
+Since $t \ge s \!+\! h$ by assumption, we can simply factor out the known quantities:
+
+$$\begin{aligned}
+    \mathbf{E} \Big[ G_t G_s (B_{t + h} \!-\! B_t) (B_{s + h} \!-\! B_s) \Big]
+    = \mathbf{E} \bigg[ G_t G_s (B_{s + h} \!-\! B_s) \: \mathbf{E} \Big[ (B_{t + h} \!-\! B_t) \Big| \mathcal{F}_t \Big] \bigg]
+\end{aligned}$$
+
+However, $\mathcal{F}_t$ says nothing about
+the increment $(B_{t + h} \!-\! B_t) \sim \mathcal{N}(0, h)$,
+meaning that the conditional expectation is zero:
+
+$$\begin{aligned}
+    \mathbf{E} \Big[ G_t G_s (B_{t + h} \!-\! B_t) (B_{s + h} \!-\! B_s) \Big]
+    = 0
+    \qquad \mathrm{for}\; t \ge s + h
+\end{aligned}$$
+
+By swapping $s$ and $t$, the exact same result can be obtained for $s \ge t \!+\! h$:
+
+$$\begin{aligned}
+    \mathbf{E} \Big[ G_t G_s (B_{t + h} \!-\! B_t) (B_{s + h} \!-\! B_s) \Big]
+    = 0
+    \qquad \mathrm{for}\; s \ge t + h
+\end{aligned}$$
+
+This leaves only one case which can be nonzero: $[t, t\!+\!h] = [s, s\!+\!h]$.
+Applying the law of total expectation again yields:
+
+$$\begin{aligned}
+    \mathbf{E} \bigg[ \sum_{t = a}^{t = b} G_t (B_{t + h} \!-\! B_t) \bigg]^2
+    &= \sum_{t = a}^{t = b} \mathbf{E} \Big[ G_t^2 (B_{t + h} \!-\! B_t)^2 \Big]
+    \\
+    &= \sum_{t = a}^{t = b} \mathbf{E} \bigg[ \mathbf{E} \Big[ G_t^2 (B_{t + h} \!-\! B_t)^2 \Big| \mathcal{F}_t \Big] \bigg]
+\end{aligned}$$
+
+We know $G_t$, and the expectation value of $(B_{t+h} \!-\! B_t)^2$,
+since the increment is normally distributed, is simply the variance $h$:
+
+$$\begin{aligned}
+    \mathbf{E} \bigg[ \sum_{t = a}^{t = b} G_t (B_{t + h} \!-\! B_t) \bigg]^2
+    &= \sum_{t = a}^{t = b} \mathbf{E} \big[ G_t^2 \big] h
+    \longrightarrow
+    \int_a^b \mathbf{E} \big[ G_t^2 \big] \dd{t}
+\end{aligned}$$
+</div>
+</div>
+
+Furthermore, Itō integrals are [martingales](/know/concept/martingale/),
+meaning that the average noise contribution is zero,
+which makes intuitive sense,
+since true white noise cannot be biased.
+
+<div class="accordion">
+<input type="checkbox" id="proof-martingale"/>
+<label for="proof-martingale">Proof</label>
+<div class="hidden">
+<label for="proof-martingale">Proof.</label>
+We will prove that an arbitrary Itō integral $I_t$ is a martingale.
+Using additivity, we know that the increment $I_t \!-\! I_s$
+is as follows, given information $\mathcal{F}_s$:
+
+$$\begin{aligned}
+    \mathbf{E} \big[ I_t \!-\! I_s | \mathcal{F}_s \big]
+    = \mathbf{E} \bigg[ \int_s^t G_u \dd{B_u} \bigg| \mathcal{F}_s \bigg]
+    = \lim_{h \to 0} \sum_{u = s}^{u = t} \mathbf{E} \Big[ G_u (B_{u + h} \!-\! B_u) \Big| \mathcal{F}_s \Big]
+\end{aligned}$$
+
+We rewrite this [conditional expectation](/know/concept/conditional-expectation/)
+using the *tower property* for some $\mathcal{F}_u \supset \mathcal{F}_s$,
+such that $G_u$ and $B_u$ are known, but $B_{u+h} \!-\! B_u$ is not:
+
+$$\begin{aligned}
+    \mathbf{E} \big[ I_t \!-\! I_s | \mathcal{F}_s \big]
+    &= \lim_{h \to 0} \sum_{u = s}^{u = t}
+    \mathbf{E} \bigg[ \mathbf{E} \Big[ G_u (B_{u + h} \!-\! B_u) \Big| \mathcal{F}_u \Big] \bigg| \mathcal{F}_s \bigg]
+    = 0
+\end{aligned}$$
+
+We now have everything we need to calculate $\mathbf{E} [ I_t | \mathcal{F_s} ]$,
+giving the martingale property:
+
+$$\begin{aligned}
+    \mathbf{E} \big[ I_t | \mathcal{F}_s \big]
+    = \mathbf{E} \big[ I_s | \mathcal{F}_s \big] + \mathbf{E} \big[ I_t \!-\! I_s | \mathcal{F}_s \big]
+    = I_s + \mathbf{E} \big[ I_t \!-\! I_s | \mathcal{F}_s \big]
+    = I_s
+\end{aligned}$$
+
+For the existence of $I_t$,
+we need $\mathbf{E}[G_t^2]$ to be integrable over the target interval,
+so from the Itō isometry we have $\mathbf{E}[I]^2 < \infty$,
+and therefore $\mathbf{E}[I] < \infty$,
+so $I_t$ has all the properties of a Martingale,
+since it is trivially $\mathcal{F}_t$-adapted.
+</div>
+</div>
+
+
+
+## References
+1.  U.H. Thygesen,
+    *Lecture notes on diffusions and stochastic differential equations*,
+    2021, Polyteknisk Kompendie.