Deriving a Wiener Integral Identity by Stochastic Integration by Parts

# Integration on Wiener Process


# Integration on Wiener Process












How can I show that below equation holds ?

$\int\limits_{0}^{t} f \left( s \right)W_s ds = W_t \int\limits_{0}^{t}f \left( s \right)ds - \int\limits_{0}^{t}\int\limits_{0}^{s} f\left( u \right)dudW_s $

$W_t$ is regular Wiener process.

## Answer by ir7 (score 6, accepted)

https://quant.stackexchange.com/a/57078

Let

$$g_s = \int_0^s f_u du$$

By Ito-Leibniz product rule:

$$ d(W_sg_s) = W_sdg_s+ g_sdW_s +d[g,W]_s $$

Assuming $f_s$ is deterministic, $d[g,W]_s = 0$ and we get:

$$ d(W_sg_s) = W_sdg_s + g_sdW_s $$ In integral form, this is:

$$ W_tg_t = \int_0^t W_sdg_s + \int_0^t g_sdW_s $$

Getting back to $f$:

$$ W_t \int_0^t f_u du = \int_0^t W_sf_sds + \int_0^t \int_0^s f_u du dW_s $$

## Answer by StackG (score 5)

https://quant.stackexchange.com/a/57079

We can use Stochastic Integration by Parts to show this.

Taking the corollary from the link above \begin{align} X_t Y_t = X_0 Y_0 + \int_0^t X_s dY_s + \int_0 ^t Y_{s-} dX_s \end{align}

We set $X_t$ and $Y_t$ equal to the following: \begin{align} X_t &\to \int_0^t f(u) du\\ Y_t &\to W_t \end{align}

then \begin{align} W_t \int_0^t f(u) du &= W_0 \int_0^0 f(u) du + \int_0^t \Bigl( \int_0^s f(u) du \Bigr) dW_s + \int_0^t W_s d\Bigl( \int_0^s f(u) du \Bigr)\\ \end{align}

The second term is $0$ (since the integration range is $0$ and $W_0 = 0$). The fourth term simplifies via the Fundamental Theorem of Calculus which says $d\Bigl( \int_0^s f(u) du \Bigr) = f(s)ds\\\\$, so: \begin{align} W_t \int_0^t f(u) du &= \int_0^t \int_0^s f(u) du dW_s + \int_0^t W_s f(s)ds\\ \int_0^t W_s f(s)ds &= W_t \int_0^t f(u) du - \int_0^t \Bigl( \int_0^s f(u) du \Bigr) dW_s \end{align}

which is the expression in your question.