I am really struggling with the Poisson Process. I am hoping someone out there can give me few hints or point me in the right direction with the following exercise.
Let $\{X(t)\}$ be a Poisson process with arrival rate $\lambda>0$. Consider a "bridge" for this Poisson process. It is known that a Poisson process, $\{X(t)\}$, has initial value $X(0) = 0$. Suppose it is also known that $X(t) = n$ for some $t > 0$ and $n$ a positive integer. Consider the probability function for a fixed earlier time point s such that $0<s<t$.
(a) What are the feasible outcomes for $X(s)$?
(b) Compute the conditional probability, $P(X(s) = x|X(t) = n)$.
(c) For a fixed s with $0<s<t$. State the conditional distribution of $X(s)|X(t) = n$ and hence check whether the result in (b) is a valid probability function. (Check over the feasible outcomes specified in (a)).
While I am having trouble with how the Poisson process works in general, here are my thoughts so far
For part (a) we have some interval $\zeta=(0,t]$ and a smaller interval $\xi=(0,s]$ such that $\xi\subseteq\zeta$. I believe we want all the possible outcomes for $\xi.$ This means all outcomes greater than $t$ are not feasible? This leads me to conclude that the feasible outcomes are given by
$$s_0,s_1,s_2,\ldots,s_n.$$
However, this does not feel very concrete to me.
For part (b) we want to solve the conditional probability $P(X(s) = x|X(t) = n).$ Here let us denote $P(X(s) = x$ by our interval $\xi$ and $X(t) = n)$ by our interval $\zeta$. So we want to find $P(\zeta|\xi)$. However, since the time between each event is indecent we need only find $P(\xi)$. Is this correct so far?
Also, when actually doing the calculation do I use
$$P(n,\lambda)=\sum_{\lambda=0}\dfrac{e^{-\lambda t}(\lambda t)^{n}}{n!}.$$
Somehow if feels to me as though I do not have enough information, which usually means I am missing a component of the question.
For part (c) I am confused about stating the conditional distribution. How is this different from part b. As for showing it is a probability function, I believe I just need to show my equation in part (b) is nonnegative and sums to zero.
Thank you all in advance you help and suggestions with this one. i am so far completely lost with this exercise.
Best Answer
(a) The tell you that by time $t$, $X(t) = n$. In other words, you could say there have been $n$ arrivals by time $t$. What are the possible values of $X(s)$?
(b) If there are $n$ by time $t$, and you want $x$ by time $s$, then how many must occur in between $s$ and $t$? Can you say something special about these two events? Are they disjoint? Independent?
(c) Name the distribution found in (b), and check against the possible values of (a).
(a) Since there are $n$ occurrences by time $t$, then the possible values of $X(s)$ are $$\{0,1,2,\dotsc,n\}$$ since $X(s)$ is counting the number of arrivals by time $t, 0< s<t$.
(b) Notice that $$\{X(s) = x, X(t) = n\}\iff \{X(s) = x, X(t)-X(s) =n-x\},$$ $X(t)-X(s)\sim\text{Pois}((t-s)\lambda)$, and that $X(s)$ is independent of $X(t)-X(s)$.
Then we proceed as usual, assuming some $\lambda >0$, \begin{align*} P(X(s)=x|X(t) = n)&=\frac{P(X(s) = x,X(t) = n)}{P(X(t) = n)}\\ &=\frac{P(X(s) = x, X(t)-X(s) =n-x)}{P(X(t) = n)}\\ &=\frac{P(X(s) = x)P( X(t)-X(s) =n-x)}{P(X(t) = n)}\\ &=\frac{e^{-s\lambda}\frac{(s\lambda)^x}{x!}\cdot e^{-(t-s)\lambda}\frac{[(t-s)\lambda]^{n-x}}{(n-x)!}}{e^{-t\lambda}\frac{(t\lambda)^n}{n!}}\\ &=\frac{n!}{x!(n-x)!}\cdot\frac{s^x(t-s)^{n-x}}{t^x\cdot t^{n-x}}\\ &=\binom{n}{x}\left(\frac{s}{t}\right)^x\left(1-\frac{s}{t}\right)^{n-x}. \end{align*}
(c) This is a binomial distribution $$X(s)|X(t) = n\sim\text{Binomial}(n, p = s/t).$$