How to compute $d X_t$ for given $X_t$?

Question

I am trying to understans, what would $d X_t$ be, according to the Itô's formula, for the following $X_t$, where $W_t$ is a Wiener process:

$$ X_t = f(W_t, t) = 5 + t^2W_t^2 + 5\int\limits_0^tW_u d u + \int\limits_0^tW^2_u d W_u \tag{1} $$

The problem is that I pretty much understand how Itô's formula works, and I know, that in full form: $X_t = X_0 + \int\limits_0^tf'_w d W_u + \int\limits_0^t(f'_u + \frac{1}{2}f''_{ww}) d u$, but I do not understand how to properly compute $f'_w$ here, precisely, I do not understand how to deal with integral terms in (1). First two terms are easy to differentiate with respect to $w$, but how can I compute the derivative of $\int\limits_0^tW_u d u$ or $\int\limits_0^tW^2_u d W_u$ with respect to $w$ (not $t$)?

Any help would be appreciated. Thank you in advance!

Answer 1

By Itô's formula, $d(W_t^2) = 2W_t dW_t + dt$, so by Itô's product rule we have $$d(t^2 W_t^2) = d(t^2)W_t^2 + t^2 d(W_t^2) + d(t^2) d(W_t^2) = (2t W_t^2 +t^2 )dt + 2W_t t^2dW_t$$

It follows that: $$X_t = 5 + \int_0^t (5W_u + 2uW_u^2 +u^2 ) du + \int_0^t (W_u^2 + 2W_u u^2 )dW_u$$

from which you can extract $dX_t$.

Answer 2

You are getting confused. There is no $\partial W_t$. If you have for example a scalar SDE with real values, $f$ is just a function $f:\mathbb R \times [0,\infty) \rightarrow \mathbb R.$ The partial derivatives are like with any other function. For example, if $f(x)=x^2$, then $f'(x)=2x$, $f''(x)=2$ and the partial derivative with respect to $t$ would be zero.

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