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.  
 A: 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 
$$
{\rm P}(S_n^1  < S_n^2 ) \approx \Phi \bigg(\frac{{\rho  - 1 }}{{\sqrt {\rho^2  + 1} }}\sqrt n \bigg).
$$
