Define $$ Y \equiv \int_0^1 \frac{B_s}{\sqrt{s}} ds$$ I want to calculate the mean and variance of $Y$ but I'm not sure if I've done it correctly:
The mean is fairly easy: $$E(Y) = \int \int_0^1 \frac{B_s(\omega)}{\sqrt{s}} ds dP(\omega) = \int_0^1 \int\frac{B_s(\omega)}{\sqrt{s}} dP(\omega) ds = 0$$ where we may exchange integrals by Fubini's theorem noting that $E|B_s| = \sqrt{2s/\pi}$
The variance is where I'm confused. We know that by Ito's lemma $$\sqrt{t}B_t = \frac{1}{2}\int_0^t \frac{B_s}{\sqrt{s}} + \int_0^t\sqrt{s}dB_s$$
There is no quadratic variation term since $\sqrt{t}$ is FV. Set $t = 1$ to obtain $$Y/2 = B_1 - \int_0^1 \sqrt{s} dB_s = \int_0^1 (1- \sqrt{s}) dB_s $$ so by Ito isometry we get $$E(Y^2) = 4 \int_0^t (1-\sqrt{s})^2 ds $$
I'm flat out too lazy to do the integral and finish this, but I just want to double check that I'm doing this question correctly. Thanks if you can confirm!