Independent increments of a Poisson process In the following $\{X_t\}$ is a Poisson process.
Assume that I've proved that 
$P(X_s=i,X_t-X_s=k)=P(X_s=i)P(X_t-X_s=k)$ 
so that the two events are independent, does it follow that
$P(X_u-X_s=h,X_t-X_u=k)=P(X_u-X_s=h)P(X_t-X_u=k)$ where $s<u<t$?
 A: No,  independence does not follow merely from this observations. This fact gives you only pairwise independence and this is not sufficient to assure independence of events. (See https://notesofastatisticswatcher.wordpress.com/2012/01/02/pairwise-independence-does-not-imply-mutual-independence/)

We might think that 
\begin{align} \Delta_1 &\perp X_s \\
\Delta_3 &\perp X_s   \Rightarrow \Delta_2 = \Delta_3-\Delta_1 \perp X_s
\end{align}
\begin{align} \Delta_2 &\perp X_s + \Delta_1 \\
\Delta_2 &\perp X_s   \Rightarrow \Delta_2 \perp \Delta_1 
\end{align}
This argument goes wrong for the reason explained in the example, that I shall rephrase here in our context.
Consider a tetrahedron dice where three of the faces are coloured red, green, and blue, respectively. On the fourth face, include all three colours.
Now, roll the dice and define the following indicator variables for
whether a colour appeared on the face the the dice landed on: $A_{1}=\left\{ \mbox{red}\right\} ,A_{2}=\left\{ \mbox{green}\right\} ,A_{3}=\left\{ \mbox{blue }\right\}$.
The way we have defined these, notice that $\mathbf{P}\left(A_{i}\right)=\frac{1}{2}$,
because for any colour, there are two faces that include that colour. 
Now note by $X_i = 1_{A_i}$.
The above implies that $X_1 \perp X_3$ $X_1 \perp X_3$ but note that $X_1 + X_2 = 2 \Rightarrow X_3 = 1$ hence $X_1 + X_2$ is not independent of  $X_3$
