Find expected value of $$Y_t = W_{1/4}\int_{0}^{1/2}\cos{t}\,dW_t$$
Firstly, I thought that $E[Y_t] = 0$. But then I realized that $ W_{1/4}\int_{0}^{1/4}\cos{t}\,dW_t $ could be dependent.
We can represent $Y_t$ as the sum of two integrals: $$Y_t = W_{1/4} \left ( \int_{0}^{1/4}\cos{t}\,dW_t + \int_{1/4}^{1/2}\cos{t}\,dW_t \right )$$ Then $$ E[Y_t] = E \left [W_{1/4} \int_{0}^{1/4}\cos{t}\,dW_t\right ] + E \left [ W_{1/4}\int_{1/4}^{1/2}\cos{t}\,dW_t \right ] $$ Where second part is $0$
So $$ E[Y_t] = E \left [W_{1/4} \int_{0}^{1/4}\cos{t}\,dW_t\right ] $$ We can represent $W_{1/4}$ as $\sum_{i=1}^{n}(W_{t_{i+1}} - W_{t_i})$
Also: $\int_{0}^{1/4}\cos{t}\,dW_t $ as $\sum_{i=1}^{n}\cos t_i (W_{t_{i+1}} - W_{t_i})$ $$ \begin{align*} E[Y_t] &= E \left [ \sum_{i=1}^{n}(W_{t_{i+1}} - W_{t_i}) \sum_{i=1}^{n}\cos t_i (W_{t_{i+1}} - W_{t_i}) \right ] \\ &= E \left [ \sum_{i=1}^{n}\cos t_i (W_{t_{i+1}} - W_{t_i})^2 \right ] \end{align*} $$ Is it correct?