Consider the SDE $$dX_t = \sigma X_t dB_t,$$ where $B_t$ is Brownian motion. Ive calculated the solution to this SDE as $$X_t = \exp(\sigma B_t-\frac{1}{2}\sigma^2t)X_0.$$ Then the expectation is $$\mathbb{E}(X_t) = \exp(\sigma^2 t)X_0.$$ Now I want to calculate the variance. If I use the definition $Var(X_t)= \mathbb{E}[(X_t-\mathbb{E}(X_t))^2]$ then I get the following result: \begin{align} Var(X_t)&= \mathbb{E}[(\exp(\sigma B_t-\frac{1}{2}\sigma^2t)X_0 -\exp(\sigma^2 t)X_0)^2]\\ &=\mathbb{E}[X_0^2(\exp(2\sigma B_t - \sigma^2 t)-2^2 \exp(\frac{1}{2}\sigma^2 t + \sigma B_t)+ \exp(2\sigma^2 t))]\\ &= X_0^2\exp(\sigma^2 t)(\exp(\sigma^2 t) -1)). \end{align}
But if I use the definition $Var(X_t)=\mathbb{E}(X_t^2) - \mathbb{E}(X_t)^2$ then I get a reverse result, that is: \begin{align} Var(X_t)&=\mathbb{E}[(\exp(2\sigma B_t-\sigma^2t)X_0^2] - (\exp(\sigma^2 t)X_0)^2\\ &= X_0^2\exp(2\sigma^2 t - \sigma^2 t) - X_0^2 \exp(2 \sigma^2 t)\\ &= X_0^2(\exp(\sigma^2 t)-\exp(2\sigma^2 t))\\ &= X_0^2 \exp(\sigma^2 t)(1-\exp(\sigma^2 t)). \end{align}
How can that be? Where am I making the mistake? Im thankfull for any help..
If $dX_t=\sigma X_tdB_t$, then from Itoh's Lemma $d\log(X_t)=-\frac12 \sigma^2 dt + \sigma dB_t$ and $\log(X_t)$ is normally distributed with mean, $\mu=-\sigma^2/2$ and variance equal to $\sigma^2$.
The $k$'th moments of $X_t$ are given by $X_0^k e^{k(2\mu+k\sigma^2)/2}$. Hence, $\mathbb{E}(X_t)=X_0$ and $\mathbb{E}(X_t^2)=X_0^2 e^{\sigma^2t}$
We can calculate the variance of $X_t$ as
$$\begin{align} Var(X_t)&=\mathbb{E}((X_t-X_0)^2)\\\\ &=\mathbb{E}(X_t^2)-2X_0\mathbb{E}(X_t)+X_0^2\\\\ &=X_0^2e^{\sigma^2t}-2X_0^2+X_0^2\\\\ &=X_0^2(e^{\sigma^2t}-1) \end{align}$$
Alternatively, we can compute the variance of $X_t$ as
$$\begin{align} Var(X_t)&=\mathbb{E}(X_t^2) -X_0^2\\\\ &=X_0^2 e^{\sigma^2t}-X_0^2\\\\ &=X_0^2(e^{\sigma^2t}-1) \end{align}$$
as expected!