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----
-title: "Itō calculus"
-firstLetter: "I"
-publishDate: 2021-11-06
-categories:
-- Mathematics
-- Stochastic analysis
-
-date: 2021-11-06T14:34:00+01:00
-draft: false
-markup: pandoc
----
-
-# Itō calculus
-
-Given two [stochastic processes](/know/concept/stochastic-process/)
-$F_t$ and $G_t$, consider the following random variable $X_t$,
-where $B_t$ is the [Wiener process](/know/concept/wiener-process/),
-i.e. Brownian motion:
-
-$$\begin{aligned}
-    X_t
-    = X_0 + \int_0^t F_s \dd{s} + \int_0^t G_s \dd{B_s}
-\end{aligned}$$
-
-Where the latter is an [Itō integral](/know/concept/ito-integral/),
-assuming $G_t$ is Itō-integrable.
-We call $X_t$ an **Itō process** if $F_t$ is locally integrable,
-and the initial condition $X_0$ is known,
-i.e. $X_0$ is $\mathcal{F}_0$-measurable,
-where $\mathcal{F}_t$ is the filtration
-to which $F_t$, $G_t$ and $B_t$ are adapted.
-The above definition of $X_t$ is often abbreviated as follows,
-where $X_0$ is implicit:
-
-$$\begin{aligned}
-    \dd{X_t}
-    = F_t \dd{t} + G_t \dd{B_t}
-\end{aligned}$$
-
-Typically, $F_t$ is referred to as the **drift** of $X_t$,
-and $G_t$ as its **intensity**.
-Because the Itō integral of $G_t$ is a
-[martingale](/know/concept/martingale/),
-it does not contribute to the mean of $X_t$:
-
-$$\begin{aligned}
-    \mathbf{E}[X_t]
-    = \int_0^t \mathbf{E}[F_s] \dd{s}
-\end{aligned}$$
-
-Now, consider the following **Itō stochastic differential equation** (SDE),
-where $\xi_t = \dv*{B_t}{t}$ is white noise,
-informally treated as the $t$-derivative of $B_t$:
-
-$$\begin{aligned}
-    \dv{X_t}{t}
-    = f(X_t, t) + g(X_t, t) \: \xi_t
-\end{aligned}$$
-
-An Itō process $X_t$ is said to satisfy this equation
-if $f(X_t, t) = F_t$ and $g(X_t, t) = G_t$,
-in which case $X_t$ is also called an **Itō diffusion**.
-All Itō diffusions are [Markov processes](/know/concept/markov-process/),
-since only the current value of $X_t$ determines the future,
-and $B_t$ is also a Markov process.
-
-
-## Itō's lemma
-
-Classically, given $y \equiv h(x(t), t)$,
-the chain rule of differentiation states that:
-
-$$\begin{aligned}
-    \dd{y}
-    = \pdv{h}{t} \dd{t} + \pdv{h}{x} \dd{x}
-\end{aligned}$$
-
-However, for a stochastic process $Y_t \equiv h(X_t, t)$,
-where $X_t$ is an Itō process,
-the chain rule is modified to the following,
-known as **Itō's lemma**:
-
-$$\begin{aligned}
-    \boxed{
-        \dd{Y_t}
-        = \bigg( \pdv{h}{t} + \pdv{h}{x} F_t + \frac{1}{2} \pdv[2]{h}{x} G_t^2 \bigg) \dd{t} + \pdv{h}{x} G_t \dd{B_t}
-    }
-\end{aligned}$$
-
-<div class="accordion">
-<input type="checkbox" id="proof-lemma"/>
-<label for="proof-lemma">Proof</label>
-<div class="hidden">
-<label for="proof-lemma">Proof.</label>
-We start by applying the classical chain rule,
-but we go to second order in $x$.
-This is also valid classically,
-but there we would neglect all higher-order infinitesimals:
-
-$$\begin{aligned}
-    \dd{Y_t}
-    = \pdv{h}{t} \dd{t} + \pdv{h}{x} \dd{X_t} + \frac{1}{2} \pdv[2]{h}{x} \dd{X_t}^2
-\end{aligned}$$
-
-But here we cannot neglect $\dd{X_t}^2$.
-We insert the definition of an Itō process:
-
-$$\begin{aligned}
-    \dd{Y_t}
-    &= \pdv{h}{t} \dd{t} + \pdv{h}{x} \Big( F_t \dd{t} + G_t \dd{B_t} \Big) + \frac{1}{2} \pdv[2]{h}{x} \Big( F_t \dd{t} + G_t \dd{B_t} \Big)^2
-    \\
-    &= \pdv{h}{t} \dd{t} + \pdv{h}{x} \Big( F_t \dd{t} + G_t \dd{B_t} \Big)
-    + \frac{1}{2} \pdv[2]{h}{x} \Big( F_t^2 \dd{t}^2 + 2 F_t G_t \dd{t} \dd{B_t} + G_t^2 \dd{B_t}^2 \Big)
-\end{aligned}$$
-
-In the limit of small $\dd{t}$, we can neglect $\dd{t}^2$,
-and as it turns out, $\dd{t} \dd{B_t}$ too:
-
-$$\begin{aligned}
-    \dd{t} \dd{B_t}
-    &= (B_{t + \dd{t}} - B_t) \dd{t}
-    \sim \dd{t} \mathcal{N}(0, \dd{t})
-    \sim \mathcal{N}(0, \dd{t}^3)
-    \longrightarrow 0
-\end{aligned}$$
-
-However, due to the scaling property of $B_t$,
-we cannot ignore $\dd{B_t}^2$, which has order $\dd{t}$:
-
-$$\begin{aligned}
-    \dd{B_t}^2
-    &= (B_{t + \dd{t}} - B_t)^2
-    \sim \big( \mathcal{N}(0, \dd{t}) \big)^2
-    \sim \chi^2_1(\dd{t})
-    \longrightarrow \dd{t}
-\end{aligned}$$
-
-Where $\chi_1^2(\dd{t})$ is the generalized chi-squared distribution
-with one term of variance $\dd{t}$.
-</div>
-</div>
-
-The most important application of Itō's lemma
-is to perform coordinate transformations,
-to make the solution of a given Itō SDE easier.
-
-
-## Coordinate transformations
-
-The simplest coordinate transformation is a scaling of the time axis.
-Defining $s \equiv \alpha t$, the goal is to keep the Itō process.
-We know how to scale $B_t$, be setting $W_s \equiv \sqrt{\alpha} B_{s / \alpha}$.
-Let $Y_s \equiv X_t$ be the new variable on the rescaled axis, then:
-
-$$\begin{aligned}
-    \dd{Y_s}
-    = \dd{X_t}
-    &= f(X_t) \dd{t} + g(X_t) \dd{B_t}
-    \\
-    &= \frac{1}{\alpha} f(Y_s) \dd{s} + \frac{1}{\sqrt{\alpha}} g(Y_s) \dd{W_s}
-\end{aligned}$$
-
-$W_s$ is a valid Wiener process,
-and the other changes are small,
-so this is still an Itō process.
-
-To solve SDEs analytically, it is usually best
-to have additive noise, i.e. $g = 1$.
-This can be achieved using the **Lamperti transform**:
-define $Y_t \equiv h(X_t)$, where $h$ is given by:
-
-$$\begin{aligned}
-    \boxed{
-        h(x)
-        = \int_{x_0}^x \frac{1}{g(y)} \dd{y}
-    }
-\end{aligned}$$
-
-Then, using Itō's lemma, it is straightforward
-to show that the intensity becomes $1$.
-Note that the lower integration limit $x_0$ does not enter:
-
-$$\begin{aligned}
-    \dd{Y_t}
-    &= \bigg( f(X_t) \: h'(X_t) + \frac{1}{2} g^2(X_t) \: h''(X_t) \bigg) \dd{t} + g(X_t) \: h'(X_t) \dd{B_t}
-    \\
-    &= \bigg( \frac{f(X_t)}{g(X_t)} - \frac{1}{2} g^2(X_t) \frac{g'(X_t)}{g^2(X_t)} \bigg) \dd{t} + \frac{g(X_t)}{g(X_t)} \dd{B_t}
-    \\
-    &= \bigg( \frac{f(X_t)}{g(X_t)} - \frac{1}{2} g'(X_t) \bigg) \dd{t} + \dd{B_t}
-\end{aligned}$$
-
-Similarly, we can eliminate the drift $f = 0$,
-thereby making the Itō process a martingale.
-This is done by defining $Y_t \equiv h(X_t)$, with $h(x)$ given by:
-
-$$\begin{aligned}
-    \boxed{
-        h(x)
-        = \int_{x_0}^x \exp\!\bigg( \!-\!\! \int_{x_1}^x \frac{2 f(y)}{g^2(y)} \dd{y} \bigg)
-    }
-\end{aligned}$$
-
-The goal is to make the parenthesized first term (see above)
-of Itō's lemma disappear, which this $h(x)$ does indeed do.
-Note that $x_0$ and $x_1$ do not enter:
-
-$$\begin{aligned}
-    0
-    &= f(x) \: h'(x) + \frac{1}{2} g^2(x) \: h''(x)
-    \\
-    &= \Big( f(x) - \frac{1}{2} g^2(x) \frac{2 f(x)}{g^2(x)} \Big) \exp\!\bigg( \!-\!\! \int_{x_1}^x \frac{2 f(y)}{g^2(y)} \dd{y} \bigg)
-\end{aligned}$$
-
-
-## Existence and uniqueness
-
-It is worth knowing under what condition a solution to a given SDE exists,
-in the sense that it is finite on the entire time axis.
-Suppose the drift $f$ and intensity $g$ satisfy these inequalities,
-for some known constant $K$ and for all $x$:
-
-$$\begin{aligned}
-    x f(x) \le K (1 + x^2)
-    \qquad \quad
-    g^2(x) \le K (1 + x^2)
-\end{aligned}$$
-
-When this is satisfied, we can find the following upper bound
-on an Itō process $X_t$,
-which clearly implies that $X_t$ is finite for all $t$:
-
-$$\begin{aligned}
-    \boxed{
-        \mathbf{E}[X_t^2]
-        \le \big(X_0^2 + 3 K t\big) \exp\!\big(3 K t\big)
-    }
-\end{aligned}$$
-
-<div class="accordion">
-<input type="checkbox" id="proof-existence"/>
-<label for="proof-existence">Proof</label>
-<div class="hidden">
-<label for="proof-existence">Proof.</label>
-If we define $Y_t \equiv X_t^2$,
-then Itō's lemma tells us that the following holds:
-
-$$\begin{aligned}
-    \dd{Y_t}
-    = \big( 2 X_t \: f(X_t) + g^2(X_t) \big) \dd{t} + 2 X_t \: g(X_t) \dd{B_t}
-\end{aligned}$$
-
-Integrating and taking the expectation value
-removes the Wiener term, leaving:
-
-$$\begin{aligned}
-    \mathbf{E}[Y_t]
-    = Y_0 + \mathbf{E}\! \int_0^t 2 X_s f(X_s) + g^2(X_s) \dd{s}
-\end{aligned}$$
-
-Given that $K (1 \!+\! x^2)$ is an upper bound of $x f(x)$ and $g^2(x)$,
-we get an inequality:
-
-$$\begin{aligned}
-    \mathbf{E}[Y_t]
-    &\le Y_0 + \mathbf{E}\! \int_0^t 2 K (1 \!+\! X_s^2) + K (1 \!+\! X_s^2) \dd{s}
-    \\
-    &\le Y_0 + \int_0^t 3 K (1 + \mathbf{E}[Y_s]) \dd{s}
-    \\
-    &\le Y_0 + 3 K t + \int_0^t 3 K \big( \mathbf{E}[Y_s] \big) \dd{s}
-\end{aligned}$$
-
-We then apply the
-[Grönwall-Bellman inequality](/know/concept/gronwall-bellman-inequality/),
-noting that $(Y_0 \!+\! 3 K t)$ does not decrease with time, leading us to:
-
-$$\begin{aligned}
-    \mathbf{E}[Y_t]
-    &\le (Y_0 + 3 K t) \exp\!\bigg( \int_0^t 3 K \dd{s} \bigg)
-    \\
-    &\le (Y_0 + 3 K t) \exp\!\big(3 K t\big)
-\end{aligned}$$
-</div>
-</div>
-
-If a solution exists, it is also worth knowing whether it is unique.
-Suppose that $f$ and $g$ satisfy the following inequalities,
-for some constant $K$ and for all $x$ and $y$:
-
-$$\begin{aligned}
-    \big| f(x) - f(y) \big| \le K \big| x - y \big|
-    \qquad \quad
-    \big| g(x) - g(y) \big| \le K \big| x - y \big|
-\end{aligned}$$
-
-Let $X_t$ and $Y_t$ both be solutions to a given SDE,
-but the initial conditions need not be the same,
-such that the difference is initially $X_0 \!-\! Y_0$.
-Then the difference $X_t \!-\! Y_t$ is bounded by:
-
-$$\begin{aligned}
-    \boxed{
-        \mathbf{E}\big[ (X_t - Y_t)^2 \big]
-        \le (X_0 - Y_0)^2 \exp\!\Big( \big(2 K \!+\!  K^2 \big) t \Big)
-    }
-\end{aligned}$$
-
-<div class="accordion">
-<input type="checkbox" id="proof-uniqueness"/>
-<label for="proof-uniqueness">Proof</label>
-<div class="hidden">
-<label for="proof-uniqueness">Proof.</label>
-We define $D_t \equiv X_t \!-\! Y_t$ and $Z_t \equiv D_t^2 \ge 0$,
-together with $F_t \equiv f(X_t) \!-\! f(Y_t)$ and $G_t \equiv g(X_t) \!-\! g(Y_t)$,
-such that Itō's lemma states:
-
-$$\begin{aligned}
-    \dd{Z_t}
-    = \big( 2 D_t F_t + G_t^2 \big) \dd{t} + 2 D_t G_t \dd{B_t}
-\end{aligned}$$
-
-Integrating and taking the expectation value
-removes the Wiener term, leaving:
-
-$$\begin{aligned}
-    \mathbf{E}[Z_t]
-    = Z_0 + \mathbf{E}\! \int_0^t 2 D_s F_s + G_s^2 \dd{s}
-\end{aligned}$$
-
-The *Cauchy-Schwarz inequality* states that $|D_s F_s| \le |D_s| |F_s|$,
-and then the given fact that $F_s$ and $G_s$ satisfy
-$|F_s| \le K |D_s|$ and $|G_s| \le K |D_s|$ gives:
-
-$$\begin{aligned}
-    \mathbf{E}[Z_t]
-    &\le Z_0 + \mathbf{E}\! \int_0^t 2 K D_s^2 + K^2 D_s^2 \dd{s}
-    \\
-    &\le Z_0 + \int_0^t (2 K \!+\! K^2) \: \mathbf{E}[Z_s] \dd{s}
-\end{aligned}$$
-
-Where we have implicitly used that $D_s F_s = |D_s F_s|$
-because $Z_t$ is positive for all $G_s^2$,
-and that $|D_s|^2 = D_s^2$ because $D_s$ is real.
-We then apply the
-[Grönwall-Bellman inequality](/know/concept/gronwall-bellman-inequality/),
-recognizing that $Z_0$ does not decrease with time (since it is constant):
-
-$$\begin{aligned}
-    \mathbf{E}[Z_t]
-    &\le Z_0 \exp\!\bigg( \int_0^t 2 K \!+\! K^2 \dd{s} \bigg)
-    \\
-    &\le Z_0 \exp\!\Big( \big( 2 K \!+\! K^2 \big) t \Big)
-\end{aligned}$$
-</div>
-</div>
-
-Using these properties, it can then be shown
-that if all of the above conditions are satisfied,
-then the SDE has a unique solution,
-which is $\mathcal{F}_t$-adapted, continuous, and exists for all times.
-
-
-
-## References
-1.  U.H. Thygesen,
-    *Lecture notes on diffusions and stochastic differential equations*,
-    2021, Polyteknisk Kompendie.