Given two independent Poisson processes X and Y I am trying to calculate what the probability is of X ever, at any given point, has been above Y as time approaches infinity. ¨
As an example lets assume lambda 2 for X and 3 for Y, process Y starts at 4 and process X starts at 0. A real world illustration, we have two bridges X and Y, average frequency of cars passing per hour is 2 for bridge X and 3 for bridge Y. There have already passed 4 cars on bridge Y, what is the probability that given infinite time more cars at any given point drove over bridge X than bridge Y.
I, with the help of some others, came up with a function that describes this for any given time, however not for all time.
$P(X(t)>Y(t)+n)=1-\sum_{k=0}^{\infty} \bigg [P(X(t)+n\cap Y(t)=k) \bigg] \\ \\ \\ P(X(t)>Y(t)+n)=1-\sum_{k=0}^{\infty} \bigg [\underbrace{P(X(t)\leq k+n)}_{Poisson\: CDF}*\underbrace{P(Y(t)=k)}_{Poisson \: pmf} \bigg] \\$
My initial idea was to take the product integral of this function, however there is not independence between X(t) and X(t-1) and Y(t) and Y(t-1).
Any help would be much appreciated as I have been stuck on this for quite some time.
We can model this as a (biased) simple random walk; take a step right for each arrival in $\{X_t\}$ and a step left for each arrival in $\{Y_t\}$. Let $Z_0=0$ and consider the Markov chain $\{Z_n:n\in\mathbb N_0\}$ on $\mathbb Z$ with transition probabilities \begin{align} \mathbb P(Z_{n+1}=i+1\mid Z_n=i) &= \frac{\lambda_X}{\lambda_X+\lambda_Y}\\ \mathbb P(Z_{n+1}=i-1\mid Z_n=i) &= \frac{\lambda_Y}{\lambda_X+\lambda_Y}, \end{align} where $\lambda_X$ and $\lambda_Y$ are the intensities of $\{X_t\}$ and $\{Y_t\}$, respectively. Then the first time at which $\{X_t\}$ surpasses $\{Y_t\}$, $$\tau := \inf\{t>0: X_t>Y_t\} $$ and the first time $\{Z_n\}$ is positive, $$\sigma:=\inf\{n>0:Z_n=1\} $$ satisfy $$\{\tau=\infty\} = \{\sigma=\infty\}. $$ It is known from elementary random walk theory that when $\lambda_X<\lambda_Y$, $$\mathbb P(\sigma<\infty) = \left(\frac{\lambda_X}{\lambda_X+\lambda_Y}\right)/\left(\frac{\lambda_Y}{\lambda_X+\lambda_Y}\right) = \frac{\lambda_X}{\lambda_Y}, $$ and otherwise $\mathbb P(\sigma<\infty)=1$.