I have a problem with a solution. I find it hard to understand some of the steps in the solving process. Can someone help me understand whats going on here?
"Calculate the probability P(X(1)=1, X(4)=4|X(2)=2, X(3)=3, X(5)=5) for a Poisson process X(t) with arrival rate λ > 0."
Solution:
\begin{align}
(i) \ \ P(X(1)=1, X(4)=4|X(2)=2, X(3)=3, X(5)=5) \\
\
\end{align}
\begin{align} (ii) = \ \frac{P(X(1)=1, X(4)=4, X(2)=2, X(3)=3, X(5)=5)}{ P(X(2)=2, X(3)=3, X(5)=5)} \\ \ \end{align}
\begin{align} (iii) = \ \frac{P(X(1)=1, X(2)−X(1) = 1, X(3)−X(2) = 1, X(4)−X(3) = 1, X(5)−X(4) = 1)}{ P(X(2)=2, X(3)−X(2) = 1, X(5)−X(3) = 2)} \\ \ \end{align}
\begin{align} (iiii) = \ \frac{[P(X(1)=1)]^5}{P(X(2)=2)P(X(2)=1)P(X(2)=2)} \\ \ \end{align}
\begin{align} = \ \frac{[[λ^1/((1!) · e^λ)]^4}{[(2λ)2/((2!) · e^2λ)]^2} \\ \ \end{align}
\begin{align} = \ \frac{1}{4} \\ \ \end{align}
I don´t understand what happens between (ii)->(iii)
P(X(1)=1, X(4)=4, X(2)=2, X(3)=3, X(5)=5) -> P(X(1)=1,X(2)−X(1)=1,X(3)−X(2)=1,X(4)−X(3)=1,X(5)−X(4)=1)
Is this the independent increments property? I don´t understand why we applying this property here.
Hope someone can explain this solution to me.
No, it's not using the independent increments property, but it's kind of a preparation to use it… let's consider a simpler example.
You want to calculate: $$P(X(1) = 1, X(2) = 3)$$ unfortunately $X(1)$ and $X(2)$ aren't independent so we cannot split this into two probabilities… but we can rewrite this a bit to get increments for which we know they are independent so we can split it!
So let's do an equivalent transformation to get an increment independent of $X(1)$ to use independence. Obviously if holds: $$\begin{align*} && X(2) &= 3 \\ \iff && X(2) - X(1) &= 3 - X(1)\end{align*}$$
we just removed $X(1)$ on both sides… so we get: $$\begin{align*} &\;P(X(1) = 1, X(2) = 3) \\= &\;P(X(1) = 1, X(2) - X(1) = 3 - X(1))\end{align*}$$
But now we have the additional information given, that $X(1) = 1$ so we can replace $X(1)$ on the the far right by $1$ and get: $$\begin{align*} = &\;P(X(1) = 1, X(2) - X(1) = 3 - 1) \\= &\;P(X(1) = 1, X(2) - X(1) = 2)\end{align*}$$
And now we have the probability of independent increments and can split this and use that $X(2) - X(1) \sim X(1)$ to get $$\begin{align*} = &\;P(X(1) = 1)\cdot P(X(2) - X(1) = 2) \\= &\;P(X(1) = 1)\cdot P(X(1) = 2)\end{align*}$$
And nothing else was done in the step from (ii) to (iii) just 4 times more to get 5 independent increments.
So actually it's nothing else then transform a system of equations, e.g. for the nominators:
$$X(1)=1, X(4)=4, X(2)=2, X(3)=3, X(5)=5$$ is equivalent to $$X(1)=1, X(2)−X(1) = 1, X(3)−X(2) = 1, X(4)−X(3) = 1, X(5)−X(4) = 1$$ and so it follows both probabilities are the same, hence $$\begin{align*} &\; P(X(1)=1, X(4)=4, X(2)=2, X(3)=3, X(5)=5 ) \\ =&\; P(X(1)=1, X(2)−X(1) = 1, X(3)−X(2) = 1, X(4)−X(3) = 1, X(5)−X(4) = 1)\end{align*}$$