Two different answers when integrating with respect to Brownian motion Consider the integral with respect to Brownian motion 
$$\int_{0}^{t}s  \ dB_{s} \ . $$
A textbook I am reading uses integration by parts to rewrite the above integral as 
$$tB_{t}-\int_{0}^{t}B_{s} \ ds.$$
However, I am familiar with the fact that 
$$\int_{0}^{t}g(s) \ dB_{s}\sim \mathcal{N}\left ( 0,
\int_{0}^{t}g^{2}(s) \ ds \right ).$$
If I use this idea, then $g(s)=s$, and 
$$\int_{0}^{t}s  \ dB_{s}\sim \mathcal{N}\left ( 0,
\frac{t^3}{3} \right ).$$
But then, 
$$tB_{t}-\int_{0}^{t}B_{s} \ ds \neq \int_{0}^{t}s  \ dB_{s}\sim \mathcal{N}\left ( 0,
\frac{t^3}{3} \right ).$$
So it seems I am getting two different answers.  Or maybe, the above equation is actually an $\textit{equality}.$  If that is the case, I guess I am not seeing it. Can anyone help me understand this?
Thanks in advance! 
 A: There is no contradiction in the two results which you mention. It does hold true that $$t B_t - \int_0^t B_s \, ds = \int_0^t s dB_s \sim N(0,t^3/3).$$
To prove that $t B_t- \int_0^t B_s \, ds$ is Gaussian (without using the integration by parts formula) we can reason as follows:
Since the Brownian motion $(B_t)_{t \geq 0}$ is a Gaussian process, we know that the random vector  $(B_{s_1},\ldots,s_{m})$ is Gaussian for any choice of $0<s_1 < \ldots < s_m$ and $m \in \mathbb{N}$. This, in turn, implies that sums of the form
$$\sum_{j=1}^m \alpha_j B_{s_j} \tag{1}$$
are Gaussian for real coeficients $\alpha_j \in \mathbb{R}$.If we choose $s_j := t j/m$ for fixed $t>0$, we find in particular that
$$t B_t - \frac{1}{m} \sum_{j=1}^{m-1} B_{t j/m} $$
is Gaussian (simple choose $\alpha_j := -1/m$ for $j=1,\ldots,m-1$ and $\alpha_m := t$ in $(1)$). Noting that
$$\frac{1}{m} \sum_{j=1}^{m-1} B_{tj/m}$$
is a Riemann sums which converges pointwise to the integral $\int_0^t B_s \, ds$ as $m \to \infty$ we conclude that $$t B_t - \int_0^t B_s \, ds$$ is Gaussian as a (pointwise) limit of Gaussian random variables. Applying Fubini's theorem we find that the random variable has expectation zero; showing that the variance equals $t^3/3$ equals a bit more work but is not too difficult using known identities for moments of Brownian motion.
