This is a question from Gallager's book "Stochastic Processes - Theory for Applications"
Q) The voters in a given town arrive at the place of voting according to a Poisson process of rate $\lambda=100$ voters per hour. The voters independently vote for candidate A and candidate B each with probability $\frac{1}{2}$. Assume that the voting starts at time $0$ and continues indefinitely.
Conditioned on $1000$ voters during the first $10$ hours, find the probability that candidate A receives $n$ votes in the first 4 hours of voting.
This question has been answered here. I have a follow-up question.
My idea is, let $V_A^i$ represent the number of voters voting for candidate A in first $i$ hours. Similarly let $V_B^i$ be the number of votes candidate B gets in first $i$ hours. $V^i$ is the number of total votes in first $i$ hours.
I started with,
$$\mathbb{P} (V_A^4 = n| V^{10} = 1000) = \frac{\mathbb{P} (\{V_A^4 = n\} \cap \{V^{10} = 1000\})}{\mathbb{P} (V^{10} = 1000)}$$
Now, if we condition on the event $V^4 = m$, we can write
$$\mathbb{P} (V_A^4 = n| V^{10} = 1000) = \sum_{m=n}^{1000} \frac{\mathbb{P} (\{V_A^4 = n\} \cap \{V_B^{4} = m-n\} \cap \{V^{10}-V^4 = 1000-m\})}{\mathbb{P} (V^{10} = 1000)}$$
Now, the voting process for candidate A is independent of that for candidate B. The composite voting process after 4th hour is independent of whatever happens before it. Using stationary increments property of Poisson processes, $V^{10}-V^4$ is statistically the same as $V^6$. Using the poisson PDF, we write
$$\mathbb{P} (V_A^4 = n| V^{10} = 1000) = \frac{\mathbb{P} (\{V_A^4 = n\})}{\mathbb{P} (\{V^{10} = 1000\})} \cdot \sum_{m=n}^{1000} \mathbb{P} (\{V_B^{4} = m-n\}) \mathbb{P} (\{V^{6} = 1000-m\})$$
$$= \frac{\frac{\left(\frac{100}{2} \cdot 4\right)^n e^{-\left(\frac{100}{2} \cdot 4\right)}}{n!}}{\frac{\left(100 \cdot 10\right)^{1000} e^{-\left(100 \cdot 10\right)}}{1000!}} \cdot \sum_{m=n}^{1000} \frac{\left(\frac{100}{2} \cdot 4\right)^{m-n} e^{-\left(\frac{100}{2} \cdot 4\right)}}{(m-n)!} \frac{\left(100 \cdot 6\right)^{1000-m} e^{-\left(100 \cdot 6\right)}}{(1000-m)!}$$
$$=\frac{200^n}{1000^{1000}} \cdot \frac{e^{-200}}{e^{-1000}} \cdot \frac{1000!}{n!} \cdot e^{-200} \cdot e^{-600} \cdot \sum_{m=n}^{1000} \frac{(200)^{(m-n)} \cdot (600)^{(1000-m)}}{(m-n)!(1000-m)!}$$
This simplifies to $$\mathbb{P} (V_A^4 = n| V^{10} = 1000) = \frac{1000!}{n!} \left(\frac{3}{5}\right)^{1000}\cdot \sum_{m=n}^{1000} \left(\frac{1}{3}\right)^m \cdot \frac{1}{(1000-m)!(m-n)!}$$
This looks very different from the answer we have. I am not sure if what I did is correct. I would highly appreciate if you could guide me through correct way of using this approach. Thanks for your time!
I'm sorry if my issue here is trivial, I'm new in this subject and still learning.
This equation has a typo: $$\mathbb{P} (V_A^4 = n| V^{10} = 1000) = \frac{\mathbb{P} (\{V_A^4 = n\})}{\mathbb{P} (\{V_A^{10} = 1000\})} \sum_{m=n}^{1000} \mathbb{P} (\{V_B^{4} = m-n\}) \mathbb{P} (\{V^{6} = 1000-m\})$$
In your second equation, the denominator was $P(\{V^{10}=1000\})$, but in this equation, the denominator is $P(\{V_A^{10}=1000\})$. Your second equation was right, the denominator should be $P(\{V^{10}=1000\})$, so it should be: $$\mathbb{P} (V_A^4 = n| V^{10} = 1000) = \frac{\mathbb{P} (\{V_A^4 = n\})}{\mathbb{P} (\{V^{10} = 1000\})} \sum_{m=n}^{1000} \mathbb{P} (\{V_B^{4} = m-n\}) \mathbb{P} (\{V^{6} = 1000-m\})$$
Now, since voters come at a rate of $\lambda=100$ per hour, $V^6$ has a $\lambda=6\cdot 100=600$ while $V^{10}$ has a $\lambda=10\cdot 100=1000$. With $V_A^4$, since voters come at a rate of $\lambda=100$ per hour and there is probability $1/2$ that a voter is for candidate A, voters for candidate A come at a rate of $100\cdot 1/2=50$ per hour, so $V_A^4$ is Poisson with $\lambda=4\cdot 50=200$. Similarly, $V_B^4$ is Poisson with $\lambda=4\cdot 50=200$. Ergo, we have: $$ \mathbb{P} (V_A^4 = n| V^{10} = 1000) = \frac{200^n e^{-200}/n!}{1000^{1000} e^{-1000}/1000!} \sum_{m=n}^{1000} \frac{200^{m-n} e^{-200}}{(m-n)!} \frac{600^{1000-m} e^{-600}}{(1000-m)!} $$ Now, all the powers of $e$ cancel out, which I believe you already noticed in your algebra. This gives us: $$ \mathbb{P} (V_A^4 = n| V^{10} = 1000) = \frac{1000!}{n!}\frac{200^n}{1000^{1000}} \sum_{m=n}^{1000} 200^{m-n} 600^{1000-m} \frac{1}{(1000-m)! (m-n)!} $$ Now, this part of the sum: $$ \frac{1}{(1000-m)! (m-n)!} $$ looks kind of like a binomial coefficient, except it's missing a $(1000-n)!$ on the top. Let's put $(1000-n)!$ at the top of this fraction, and then divide by $(1000-n)!$ outside the sum to cancel it out so the total expression stays the same: $$ \begin{align*} \mathbb{P} (V_A^4 = n| V^{10} = 1000) &= \frac{1000!}{(1000-n)!n!}\frac{200^n}{1000^{1000}} \sum_{m=n}^{1000} 200^{m-n} 600^{1000-m} \frac{(1000-n)!}{(1000-m)! (m-n)!} \\ &= \frac{1000!}{(1000-n)!n!}\frac{200^n}{1000^{1000}} \sum_{m=n}^{1000} 200^{m-n} 600^{1000-m} \binom{1000-n}{m-n} \end{align*} $$ That sum is starting to look like a binomial theorem sum, which would be of the form $\sum_{m=0}^N a^m b^{N-m} \binom{N}{m}$. In order to get this sum into that form, I'll change the indices of the sum from $m=n$ to $1000$ to $m=0$ to $1000-n$: $$ \begin{align*} \mathbb{P} (V_A^4 = n| V^{10} = 1000) &= \frac{1000!}{(1000-n)!n!}\frac{200^n}{1000^{1000}} \sum_{m=0}^{1000-n} 200^{m} 600^{1000-(m+n)} \binom{1000-n}{m} \\ &= \frac{1000!}{(1000-n)!n!}\frac{200^n}{1000^{1000}} \sum_{m=0}^{1000-n} 200^{m} 600^{1000-n-m} \binom{1000-n}{m} \end{align*} $$ Now, it is a biomial theorem sum: By the binomial theorem, this sum is equivalent to $(200+600)^{1000-n}$: $$ \mathbb{P} (V_A^4 = n| V^{10} = 1000) = \frac{1000!}{(1000-n)!n!}\frac{200^n}{1000^{1000}} (200+600)^{1000-n} $$ Now, let's split $1000^{1000}$ into $1000^n$ and $1000^{1000-n}$: $$ \mathbb{P} (V_A^4 = n| V^{10} = 1000) = \frac{1000!}{(1000-n)!n!}\left(\frac{200}{1000}\right)^n \left(\frac{200+600}{1000}\right)^{1000-n} $$ Simplify the fractions: $$ \mathbb{P} (V_A^4 = n| V^{10} = 1000) = \frac{1000!}{(1000-n)!n!}\left(\frac{1}{5}\right)^n \left(\frac{4}{5}\right)^{1000-n} $$ We can replace the $1000!/[(1000-n)!n!]$ with just $\binom{1000}{n}$, and we can also expand out the powers of the fractions to find that: $$ \mathbb{P} (V_A^4 = n| V^{10} = 1000) = \binom{1000}{n} \frac{4^{1000-n}}{5^n} $$ On that post you linked, this is the same answer as the second answer someone gave, which shows we likely got the right answer. Unfortunately, the first answer on that post is wrong. Using the notation from that post, the first answer forgot to put $P(N_4=n)$ into their infinite sum.