I never learned stochastic differential equations, and so am trying to do some self study. I've arrive at this question: $tB_t\sim N(0,t^3)$? $B_t$ is standard brownian motion. $B_t\sim N(0,t)$, so just applying standard probability, we get the $t^3$. Is that ok?
But I have derived the formula: $tB_t=\int_0^t sdB_s+\int_0^t B_sds$. I think that both integrals on the right are $N(0,\frac{1}{3}t^3)$ random variables. So I can't just add the variances on the left to get $\frac{4}{3}t^3$.
It seems that the integrals have a non-zero covariance, so this doesn't surprise me, but I can't seem to figure out what I am misunderstanding.
So my question(s): Is $$tB_t=\int_0^t sdB_s+\int_0^t B_sds$$ correct? If so, how am I to see that the left side as a $N(0,t^3)$? Since these are stochastic integrals, am I to think of the $B_s$ in each integral as a distinct random process?
Your stochastic integral formula is correct, as are your values for the variances of the two integrals on the right side. This implies that the covariance of those integrals is $t^3/6$. To see this directly, let $M_t$ denote the first stochastic integral on the right; this is a martingale. Therefore, $$ \Bbb E\left[M_t\cdot\int_0^t B_s\,ds\right]=\int_0^t\Bbb E[M_tB_s]\,ds= \int_0^t \Bbb E[M_sB_s]\,ds. $$ But by the Ito Isometry formula, $$ \Bbb E[M_sB_s]=\int_0^s u \,du= s^2/2, $$ Inserting this into the previous formula we get the covariance: $$ \Bbb E\left[M_t\cdot\int_0^t B_s\,ds\right]=\int_0^t s^2/2\,ds = t^3/6. $$