Suppose $X_t$ is Geometric Brownian motion with $\mu=1$, $\sigma^2=1$. Given $X_1=1$, find the probability of $X_3<3$.
Let's remember that we say that $X_t$ is a Geometric Brownian Motion if for all $t\geq0$
\begin{align} X_t=e^{\left ( \mu - \frac{\sigma^{2}}{2} \right )t+\sigma W_{t}} \end{align}
With $W_{t}$ a standard Brownian motion.
In our case, we have:
\begin{align} X_t=e^{\frac{t}{2}+ W_{t}} \end{align}
We also know that $X_3 - X_1 \sim N(0,2)$, and that $X_3-X_1$ is independent of $X_1$. We can express $X_3$ as
\begin{align} X_3=(X_3 - X_1)+X_1 \end{align}
So, \begin{align} \mathbb{P}(X_3<3)&=\mathbb{P}\left ( (X_3 - X_1)+X_1 < 3\right ) \end{align}
And, \begin{align} \mathbb{P}(X_3<3|X_1=1)&=\mathbb{P}\left ( (X_3 - X_1)+X_1 < 3\right |X_1=1)\\&=\mathbb{P}\left ( (X_3 - X_1)+1 < 3\right |X_1=1)\\&=\mathbb{P}\left ( (X_3 - X_1)< 2\right |X_1=1)\\&=\mathbb{P}\left ( (X_3 - X_1)< 2\right) \end{align}
Since, $X_3-X_1 \sim N(0,2)$. Then, $\mathbb{F}_{X_3-X_1}(2)=0.92$.
But it seems that my answer is wrong. Do you identify what is my mistake? Do you have any other hint to solve it?
This is completely incorrect. It would be true if $X_t$ were were a Brownian motion, but it's not -- it's a Geometric Brownian motion. For Geometric Brownian motion, the corresponding result is that $\dfrac{X_3}{X_1}$ is independent of $X_1$ and $\dfrac{X_3}{X_1}$ has the same distribution as $X_{3-1} = X_2$.
This gives $$\begin{align*}\mathbb{P}(X_3<3|X_1=1) &= \mathbb{P}\left(\frac{X_3}{X_1}<3\middle|X_1=1\right) \\ &= \mathbb{P}\left(\frac{X_3}{X_1}<3\right) \\ &= \mathbb{P}(X_2<3) \\ &= \mathbb{P}(W_2 < \ln(3) - 1)\end{align*}$$ which you then can reference the $N(0,2)$ distribution to approximate.