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Let $_$ and $Y_t$ be two independent Poisson Process with rate parameters $\lambda_1$ and $\lambda_2$ respectively, measuring the number of customers arriving in stores 1 and 2. What is the probability that a customer arrives in store 1 before any customers arrive in store 2?

I have found some solutions for this question that I don't understand, and would like some explanation for. Firstly let me explain my thinking. Let $T_1, T_2$ be the first times a customer arrives at stores 1 and 2 respectively. Fix any $t$, then we can consider $P(T_2 > t) \cdot P(T_1 = t).$ Then we can integrate this from $t = 0$ to $\infty$ and that should give the answer. I think this is similar to the method outline in this question where the integral

$$\int_0^\infty P(T_2>t)dP(T_1=t)=\int_0^\infty e^{-\lambda_2 t}\lambda_1 e^{-\lambda_1 t} dt$$

is calculated. I think this is just me being bad with integration/continuous distributions, but I don't see how my intuition leads to the integral on the LHS. Specifically, what is the $dP(T_1=t)$ term doing? Secondly, how does this translate to the integral on the right? The derivative of $P(T_1 > t) = \lambda_1 e^{-\lambda_1 t}$ so how does this arise from $P(T_1 = t)$ which should be 0?

Airdish
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  • You have two rates and memorylessness so the answer should be $\dfrac{\lambda_1}{\lambda_1+\lambda_2}$ – Henry May 28 '20 at 22:04
  • @Henry Thanks, but I am asking for an explanation of the method outlined rather than an answer to the question – Airdish May 28 '20 at 22:05
  • Would you find it easier to consider the density for $T_1$ the first arrival of the first process being $f_1(t)$ and then $\int\limits_0^\infty P(T_2>t) , f_1(t), dt$ ? – Henry May 28 '20 at 22:11
  • @Henry I know that that is essentially what the integral is, but I want to know why the density appears in the integral. I guess I am not so comfortable with the intuition of densities vs. probabilities in this situation since they are being used at the same time. – Airdish May 28 '20 at 22:13
  • If $T_1$ had a discrete distribution then you might want to calculate $\sum\limits_t P(T_2 > t) \cdot P(T_1 = t) = \sum\limits_t P(T_2 > t) \cdot \left(P(T_1 \le t) - P(T_1 \lt t)\right)$. But here $T_1$ has a continuous distribution and $P(T_1 = t) =0$ for all $t$ so you need to turn the sum into an integral and $P(T_1 \le t) - P(T_1 \lt t)$ into $\frac{d}{dt}P(T_1 \le t), dt$ i.e. $dP(T_1 \le t)$ – Henry May 28 '20 at 22:21

1 Answers1

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The integral on the LHS is essentially considering all possible times $T_1$ can happen and integrates over $t$. It can be re-written as

$$ \int_0^\infty P(T_2>t)f_{T_1}(t)dt $$

where $f_{T_1}(t)$ is the density of $T_1$. Some manipulation yields

$$ = \int_0^\infty S_{T_2}(t) \cdot f_{T_1}(t)dt = \int_0^\infty e^{-t\lambda_2} \cdot \lambda_1 e^{-t\lambda_1}dt $$

where $S_{T_2}(t) = 1 - F_{T_2}(t)$ is the survival function of $T_2$.

Bryden C
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  • Again, I know that there are other methods to do this. My question is asking for a formalization/explanation of the little details of the method I outlined, not another method – Airdish May 28 '20 at 22:14
  • My bad, I misunderstood. I've updated my answer. – Bryden C May 28 '20 at 22:21