What is the Error in My Approach to the Poisson Process Problem?

[rsq_a]

I'm a bit frustrated with this one...

Let $(X_t)_{t\geq 0}$ be a Poisson Process with rate $\lambda$

Each time an 'arrival' happens, a counter detects the arrival with probability $p$ and misses it with probability $1-p$. What is the distribution of time, $T$ until the first particle is detected?

I thought I could do it like this:

$$P(T = t) = \sum_{n=1}^\infty P(T = t | X_t = n) P(X_t = n) = \sum_{n=1}^\infty (1-p)^{n-1}p \frac{e^{-\lambda t}(\lambda t)^n}{n!}$$

Unfortunately, this does give me the right answer (T should be Exponentially distributed with parameter $\lambda p$). I've stared at my process, but I can't seem to find anything wrong with it!

Can someone point out the error in my approach?

 

[mmwave]

Hi there,

I can understand your frustration with this problem. It's always frustrating when we think we have a solution, but it turns out to be incorrect. Let's take a closer look at your approach and see if we can find the error.

First, let's define T as the time until the first particle is detected. This means that T can take on values of 0, 1, 2, 3, ... (since the Poisson process has a countable number of arrivals). Therefore, we can rewrite your equation as follows:

P(T = t) = \sum_{n=1}^\infty P(T = t | X_t = n) P(X_t = n) = \sum_{n=1}^\infty (1-p)^{n-1}p \frac{e^{-\lambda t}(\lambda t)^n}{n!} = \sum_{n=1}^\infty (1-p)^{n-1}p e^{-\lambda t} \frac{(\lambda t)^n}{n!}

Now, let's take a closer look at the term (1-p)^{n-1}p. This term represents the probability that the particle is detected on the first arrival (n=1), but not on any of the subsequent arrivals (n-1). However, this does not take into account the possibility that the particle is not detected at all. In other words, this term only considers the cases where T = 0 or T = 1. But we know that T can take on values of 0, 1, 2, 3, ... Therefore, we need to include the case where the particle is not detected at all, which has a probability of (1-p)^0 = 1. This means that the correct expression should be:

P(T = t) = \sum_{n=0}^\infty (1-p)^{n}p e^{-\lambda t} \frac{(\lambda t)^n}{n!}

Now, let's simplify this expression:

P(T = t) = p e^{-\lambda t} \sum_{n=0}^\infty \frac{(\lambda t)^n}{n!} (1-p)^{n} = p e^{-\lambda t} e^{\lambda t (1-p)} = p e^{-\lambda p t}

And this