Quadratic Variation and Itô’s Rule for Correlated Processes

# 2 Ito processes - $d(X_{t} + X^{'}_{t})^2 = (Y_t + Y^{'}_{t})^2 dt$ why it is true?


# 2 Ito processes - $d(X_{t} + X^{'}_{t})^2 = (Y_t + Y^{'}_{t})^2 dt$ why it is true?












Having two Ito processes

$dX_{t} =z_{1} dt + Y_{t} dB_t $

$dX^{'}_{t} =z^{'}_{1} dt + Y^{'}_{t} dB_t $

I am analyzing a proof of the product rule

$d(X_t X_t^{'})=X_t dX_t^{'}+ X_t^{'} dX_t + Y_t Y_t^{'}dt$

In the equation

$df(X_t)= \frac{\partial f}{\partial X_t} \partial X_t + \frac{1}{2} \frac{\partial^2 f}{\partial X_t^2} (dX_t)^2$

$(dX_t)^2$ was replaced with $Y_t^2 dt$. I don't quite follow this transition.

This property was used later in the proof as

$d(X_{t} + X^{'}_{t})^2 = (Y_t + Y^{'}_{t})^2 dt$

Could anybody clarify me the substitution of $(dX_t)^2$ with $Y_t^2 dt$?

## Answer by Quantuple (score 8, accepted)

https://quant.stackexchange.com/a/25646

$X_t$ being a stochastic process, one cannot use ordinary calculus to express the differential of a (sufficiently well-behaved) function $f$ of $t$ and $X_t$.

Instead one should turn to Itô's lemma, one of the key results of stochastic calculus, which stipulates (assuming $X_t$ is here a continuous, square integrable stochastic process) $$ df(t,X_t) = \frac{\partial f}{\partial t}(t,X_t) dt + \frac{\partial f}{\partial x}(t,X_t) dX_t + \frac{\partial^2 f}{\partial x^2}(t,X_t) d \langle X,X\rangle_t $$ where the quantity $$ \langle X,X \rangle_t $$ represents the quadratic variation of the process $X_t$ over $[0,t]$ defined as $$ \langle X,X\rangle_t := \lim_{\Vert P \Vert \rightarrow 0} \sum_{i=1}^N (X_{t_{i}}-X_{t_{i-1}})^2 $$ with $P$ representing a generic partition $\{t_0 = 0 < \dots < t_N = t\}$ of the interval $[0,t]$ and $\lim_{\Vert P \Vert \rightarrow 0}$ suggests the limit (when it exists) is taken in probability as $\max(\{t_{i}-t_{i-1} \vert i=1,\dots,N\}) \rightarrow 0$.

For Itô processes, that is, stochastic processes of the form $$X_t = X_{0} + \int_0^t \mu(t,X_t) dt + \int_0^t \sigma(t,X_t) dW_t$$ or equivalently in differential form (this is an abusive notation, essentially used for convenience) $$dX_t = \mu(t,X_t) dt + \sigma(t,X_t) dW_t$$ where $X_t$, $\mu(...)$ and $\sigma(...)$ are adapted (generally to the natural filtration of $W_t$) and the integrands verify the usual integrability conditions, it can be demonstrated (cf. any good stochastic calculus book) that: $$ \langle X,X \rangle_t = \int_0^t \sigma^2(t,X_t) dt $$ or in differential form (this is an abusive notation, essentially used for convenience) $$ d\langle X,X \rangle_t = \sigma^2(t,X_t) dt $$

In your case, because $X_t$ is the unique solution of the SDE $$ dX_t = z_1 dt + Y_t dW_t $$ applying the above result with $\mu(t,X_t) = z_1$ and $\sigma(t,X_t) = Y_t$ $$ d\langle X,X \rangle_t = Y_t^2 dt $$ assuming the usual conditions are met.

As far as the second issue is concerned, as @Gordon mentioned, using the commutativity + bilinearity of quadratic variations: \begin{align} d\langle X + X^{'} , X + X^{'} \rangle_t &= d \langle X, X \rangle_t + 2 d \langle X, X^{'} \rangle_t + d \langle X^{'} , X^{'} \rangle_t \\ &= Y_t^2 dt + 2 Y_t Y_t^{'} dt + (Y_t^{'} )^2 dt \\ &= (Y_t + Y_t^{'})^2 dt \end{align}