Poisson process and probabilities 
Let $\{X(t); t\ge 0\}$ be a Poisson process with rate $\lambda =2$. Find the probability $\Pr\{X(1)=1, X(2)=3\}$ 

And here is my solution:
As there is a Poisson process then the intervals are independent random variables thus $\Pr\{X(1)=1, X(2)=3\}$=$\Pr\{X(1)=1\}$$\Pr\{ X(2)=3\}$=$\frac{2e^{-2}}{1!} \frac{[2(2)]^3e^{-4}}{3!} =\frac{e^{-6}4^3}{3}$
But the book's answer is $4e^{-4}$. I don't see where is my mistake.
Can someone tell me?
 A: In a Poisson process the increments, not the values themselves, are independent. Thus the increment in the first time interval of length $1$ is $1$ while the increment in the second time interval of length $1$ is $2$.
A: For $s<t$, by the definition of Poisson Process, we have
\begin{eqnarray*}
P(X(t)=n,X(s)=n)&=&P(X(t)-X(s)=n-m,X(s)-X(0)=m-0)\\
& &\qquad\text{(introducing  increments)}\\
&=&P(X(t)-X(s)=n-m)\times P(X(s)-X(0)=m-0),\\
& &\qquad\text{(by ind. inc. prop.)}\\
&=&P(X(t-s)=n-m)\times P(X(s)=m)
\end{eqnarray*}
A: HINT
Write $$ P(X_1 = 1,X_2 = 3) = P(X_1-X_0 = 1, X_2-X_1 = 2).$$
Then use the property that $X_1-X_0$ and $X_2-X_1$ are independent Poisson RVs. (Not $X_2$ and $X_1$... reread the definition.)
A: If you have 3 events in the interval $(0,2)$ and $1$ event in the interval $(0,1)$ then you can conclude that you have $2$ events in the interval (1,2).
Therefore the required probability is
$$\underbrace{2^1\frac{e^{-2}}{1!}}_{\text{interval} (0,1)}\cdot \underbrace{2^2\frac{e^{-2}}{2!}}_{\text{interval} (1,2)}=4\cdot e^{-4}$$ 
