Let $Z_t=\int{W_s }ds $. Show that $Z_t=\int (t-s) dW_s$

Question

Let $Z_t=\int_{0}^{t} W_s ds$. Use integration by parts to show that $Z_t=\int_{0}^{t} (t-s) dW_s$. I have tried and i can't get the answer.

Answer 1

Note that \begin{align*} d(sW_s) = sdW_s + W_s ds. \end{align*} Then you can integrate on both sides.

Answer 2

For $Z_t = \int_{0}^{t} W_s \,ds$, \begin{align}Z_t &= \bigl[sW_s\bigr]^{t}_{0} - \int_{0}^{t} s\,dW_s\\ &=tW_t - s(W_t) \\ &= (t-s)W_t\\ &=\int_{0}^{t}(t-s) dW_s \end{align} where I used the property that $\int_{a}^{b} c\,dW_t = c(W_b - W_a), \, c$ is an arbitrary constant, and $W_0 = 0$

Comment: please correct me if I am wrong!