Convexity Geometric Brownian Motion Let $B(t)$ be a standard Brownian motion with filtration $\{ \mathcal{F}_t: t \geq 0\}$ and $\mu, \sigma >0$ are real parameters. $S(t)$ is modeled by
\begin{align}
S(t) = e^{\mu t + \sigma B(t)- \frac{1}{2}\sigma^2 t}.
\end{align}
In order to find an expression for $\mathbb{P}\big(\ S(2t)>2S(t)\ \big)$, I equal both expressions:
\begin{align}
S(2t) &= 2 S(t) \\
e^{2 \mu t + \sigma B(2t) - \sigma^2 t} &= 2 e^{\mu t + \sigma B(t)- \frac{1}{2}\sigma^2 t} \\
e^{\mu t + \sigma ( B(2t) - B(t))- \frac{1}{2}\sigma^2 t} &= 2 \\
\mu t + \sigma ( B(2t) - B(t))- \frac{1}{2}\sigma^2 t &= \ln(2).
\end{align}
I understand that the increment $B(2t) -B(t)$ is independent and that we are studying the realized volatility. However, I am wondering how we can find an expression for $\mathbb{P}\big(\ S(2t)>2S(t)\ \big)$?
(is convexity the correct name for the property $S(2t)>2S(t)$?)

By taking the logarithm on the left and right side we have 
$P(S(2t)>2S(t))=P(B_{2t}-B_t>\frac{log(2)-\mu t+0.5\sigma^2t}{\sigma})$
We know that $B_{2t}-B_t$ is normally distibuted with mean 0 and variance t, therefore we can replace $B_{2t}-B_t$ by $\sqrt{t}Y$ where Y is centered and normally distributed. 
Finally ,
$P(S(2t)>2S(t))=P(Y>\frac{log(2)-\mu t+0.5\sigma^2t}{\sigma\sqrt{t}})$
$=1-F(\frac{log(2)-\mu t+0.5\sigma^2t}{\sigma\sqrt{t}})$
where F is a cdf of centered normally distributed variable.
