# Rate of Convergence: Competing Poisson Processes

I am analyzing one of the properties of the Poisson Process, that of Competing Processes:

Suppose $N_1(t), t \geq 0$ and $N_2(t), t \geq 0$ are independente Poisson processes with respective rates $\lambda_1$ and $\lambda_2$. Let $S_n^i$ be the time of the $n$th event of process $i, i = 1,2$.

We know that $P[S_n^1 < S_m^2] = \sum_{k=n}^{n+m-1} {n+m-1 \choose k} \left(\frac{\lambda_1}{\lambda_1 + \lambda_2}\right)^k \left(\frac{\lambda_2}{\lambda_1 + \lambda_2}\right)^{n+m-1-k}$. I'm interested in the case where $n=m$ and $n \rightarrow \infty$. This converges to 1, and my intuition is that the rate of convergence depends on the ratio $\frac{\lambda_1}{\lambda_2}$, but I'm having a hard time proving it out. Any insights into a closed-form rate of convergence would be great.

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## 1 Answer

First write $$S_n^1 = X_1^1 + X_2^1 + X_3^1 + \cdots ,$$ where the $X_k^1$ are i.i.d. exponential random variables with density function $f_{X_1^1 } (x) = \lambda _1 e^{ - \lambda _1 x}$, $x > 0$, and $$S_n^2 = X_1^2 + X_2^2 + X_3^2 + \cdots ,$$ where the $X_k^2$ are i.i.d. exponential random variables with density function $f_{X_1^2 } (x) = \lambda _2 e^{ - \lambda _2 x}$, $x > 0$, independent also of the $X_k^1$. Now let $Y_k = X_k^1 - X_k^2$. Then, $${\rm P}(S_n^1 < S_n^2 ) = {\rm P}(Y_1 + \cdots + Y_n < 0).$$ Next note that the $Y_k$ are i.i.d. random variables with mean $$\mu = {\rm E}(Y_1) = {\rm E}(X_1^1 ) - {\rm E}(X_1^2 ) = \frac{1}{{\lambda _1 }} - \frac{1}{{\lambda _2 }}$$ and variance $$\sigma^2 = {\rm Var}(Y_1) = {\rm Var}(X_1^1 ) + {\rm Var}(X_1^2 ) = \frac{1}{{\lambda _1^2 }} + \frac{1}{{\lambda _2^2 }}.$$ It holds $$\frac{\mu }{\sigma } = \frac{{\lambda _2 - \lambda _1 }}{{\lambda _1 \lambda _2 }}\frac{{\lambda _1 \lambda _2 }}{{\sqrt {\lambda _1^2 + \lambda _2^2 } }} = \frac{{\lambda _2 - \lambda _1 }}{{\sqrt {\lambda _1^2 + \lambda _2^2 } }}.$$ Now let $\tilde S_n = Y_1 + \cdots + Y_n$ and $Z_n = \frac{{\tilde S_n - n\mu }}{{\sigma \sqrt n }}$. Then, $${\rm P}(Y_1 + \cdots + Y_n < 0) = {\rm P}\bigg(\frac{{\tilde S_n - n\mu }}{{\sigma \sqrt n }} < \frac{{ - n\mu }}{{\sigma \sqrt n }}\bigg) = {\rm P}\bigg(Z_n < \frac{{\lambda _1 - \lambda _2 }}{{\sqrt {\lambda _1^2 + \lambda _1^2 } }}\sqrt n \bigg).$$ By the central limit theorem, $Z_n$ converges in distribution to the ${\rm N}(0,1)$ distribution. (So if $\lambda_1 > \lambda_2$, the last probability, which is equal to ${\rm P}(S_n^1 < S_n^2 )$, tends to $1$ as $n \to \infty$.) This may help you deduce the asymptotic behavior of ${\rm P}(S_n^1 < S_n^2 )$.

EDIT:

So the above gives rise to the approximation $${\rm P}(S_n^1 < S_n^2 ) \approx \Phi \bigg(\frac{{\lambda _1 - \lambda _2 }}{{\sqrt {\lambda _1^2 + \lambda _2^2} }}\sqrt n \bigg),$$ where $\Phi$ is the distribution function of the ${\rm N}(0,1)$ distribution. Of course, the quality of the approximation depends on the parameters involved. We can check it by comparing to results obtained from Monte Carlo simulations. [The probability ${\rm P}(S_n^1 < S_n^2 )$ can be approximated easily and accurately using Monte Carlo simulation, noting that an exponential random variable with density function $f_{X} (x) = \lambda e^{ - \lambda x}$, $x > 0$, can be generated as $-\ln(U)/\lambda$, where $U$ is uniformly distributed on $(0,1)$.] Define the quantity $\zeta = \zeta (n,\lambda_1,\lambda_2)$ by $$\zeta = \frac{{\lambda _1 - \lambda _2 }}{{\sqrt {\lambda _1^2 + \lambda _2^2 } }}\sqrt n ,$$ and let $\hat p = \hat p(n,\lambda_1,\lambda_2)$ denote the Monte Carlo approximation obtained (accurately enough) for the probability ${\rm P}(S_n^1 < S_n^2 )$. The following results were obtained: 1) $n=50$, $\lambda_1=2$, $\lambda_2 = 1$ ($\zeta = \sqrt{10}$): $\hat p = 0.999678$, $\Phi(\zeta) \approx 0.999217$; 2) $n=100$, $\lambda_1=4$, $\lambda_2 = 3$ ($\zeta = 2$): $\hat p = 0.978802$, $\Phi(\zeta) \approx 0.977250$. Finally, letting $\rho = \lambda_1 / \lambda_2$, it holds $$\frac{{\lambda _1 - \lambda _2 }}{{\sqrt {\lambda _1^2 + \lambda _2^2 } }} = \frac{{\rho - 1}}{{\sqrt {\rho^2 + 1} }},$$ and thus the original approximation can be written $${\rm P}(S_n^1 < S_n^2 ) \approx \Phi \bigg(\frac{{\rho - 1 }}{{\sqrt {\rho^2 + 1} }}\sqrt n \bigg).$$

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@Avneesh: So the last approximation agrees with your intuition... – Shai Covo Jun 3 '11 at 4:29
This is excellent stuff, thanks so much Shai! Sorry I can't vote your answer up, seems like I need more reputation points. – Avneesh Jun 3 '11 at 22:00
@Avneesh: Thanks, glad to help! – Shai Covo Jun 5 '11 at 4:10
@Avneesh: You can formally accept Shai's answer, though, by clicking on the little check mark next to it. – Mike Spivey Jun 6 '11 at 22:21
@Mike: Thanks for pointing this out! – Shai Covo Jun 6 '11 at 23:01