Poisson Distribution with Dependent Events Problem:
The average number of cars arriving at a tollbooth per minute is $\lambda$ and the probability of k cars in the interval $(0,T)$ minutes is:
$$P(k;0,T) = e^{-\lambda T}\frac{[\lambda T]^k}{k!}$$
Consider two disjoint, that is, non-overlapping intervals, say $(0,t_1)$, and $(t_1,T)$. Then for the Poisson law:
$$P[n_1\text{ cars in }(0,t_1)\text{ and }n_2\text{ cars in }(t_1,T)] = P[n_1\text{ cars in }(0,t_1)]P[n_2\text{ cars in }(t_1,T)]$$
Thus events in disjoint intervals are independent.


*

*Show that $P[n_1$ cars in $(0,t_1) | n_1 + n_2$ cars in $(0,T)]$ is not a function of $\lambda$.

*In (a) let $T = 2, t_1 = 1$ and $n_1 = 5, n_2 = 5$. Compute $P[5$ cars in $(0,1) | 10$ cars in $(0,2)]$.
Approach:
Given two arbitrary events, $A$ and $B$, the following statement is true.
$$P[A|B]=\frac{P[A\cap B]}{P[B]}$$
From this statement, we can determine the following about the tollbooth problem.
$$P[n_1 \text{ cars in } (0,t_1) | n_1 + n_2 \text{ cars in } (0,T)] = \frac{P[n_1 \text{ cars in } (0,t_1) \cap n_1 + n_2 \text{ cars in } (0,T)]}{P[n_1 + n_2 \text{ in } (0,T)]}$$
If $n_1 + n_2$ cars pass through the tollbooth in $(0,T)$, and $n_1$ cars pass through in $(0,t_1)$, then $n_2$ cars must pass through in $(t_1,T)$. Therefore, we can form the following equality.
$$P[n_1 \text{ cars in } (0,t_1) \cap n_1 + n_2 \text{ cars in } (0,T)] = P[n_1 \text{ cars in }(0,t_1)\text{ and }n_2\text{ cars in }(t_1,T)]$$
Using the Poisson law from the problem:
$$P[n_1 \text{ cars in } (0,t_1) \cap n_1 + n_2 \text{ cars in } (0,T)] = P[n_1\text{ cars in }(0,t_1)]P[n_2\text{ cars in }(t_1,T)]$$
Substituting this result back into our first formula, we can see the following:
$$P[n_1 \text{ cars in } (0,t_1) | n_1 + n_2 \text{ cars in } (0,T)] = \frac{P[n_1\text{ cars in }(0,t_1)]P[n_2\text{ cars in }(t_1,T)]}{P[n_1 + n_2 \text{ in } (0,T)]}$$
Now, by substituting the formula for the Poisson distribution for each of our probability expressions, we can create a formula for the desired conditional probability.
$$P[n_1 \text{ cars in } (0,t_1) | n_1 + n_2 \text{ cars in } (0,T)] = \frac{e^{-\lambda t_1} \frac{[\lambda t_1]^{n_1}}{n_1!} e^{-\lambda (T-t_1)} \frac{[\lambda (T-t_1)]^{n_2}}{n_2!}}{e^{-\lambda T} \frac{[\lambda T]^{n_1+n_2}}{(n_1+n_2)!}}$$
$$P[n_1 \text{ cars in } (0,t_1) | n_1 + n_2 \text{ cars in } (0,T)] = \frac{e^{-\lambda T}}{e^{-\lambda T}} \cdot \frac{\lambda ^{n_1 + n_2}}{\lambda ^{n_1 + n_2}} \cdot \frac{t_1^{n_1}(T-t_1)^{n_2}}{T^{n_1+n_2}} \cdot \frac{(n_1 + n_2)!}{n_1!n_2!}$$
$$P[n_1 \text{ cars in } (0,t_1) | n_1 + n_2 \text{ cars in } (0,T)] = \frac{t_1^{n_1}(T-t_1)^{n_2}}{T^{n_1+n_2}} \cdot \frac{(n_1 + n_2)!}{n_1!n_2!}$$
This formula shows that the conditional probability is not a function of $\lambda$, thus satisfying part (1).
Using the values from part (2), we can calculate the desired conditional probability.
$$P[5 \text{ cars in } (0,1)|10 \text{ cars in } (0,2)]=\frac{(1)^5\cdot(2-1)^5}{(2)^{(5+5)}}\cdot\frac{(5+5)!}{(5)!(5)!}$$
$$P[5 \text{ cars in } (0,1)|10 \text{ cars in } (0,2)]=\frac{1}{2^{10}}\cdot\frac{10!}{5!5!}=\frac{63}{256}\approx 24.61\%$$
 A: By definition of conditional probability, we have
\begin{align}
\mathbb P(N(t_1)=n_1\mid N(T)=n_1+n_2) &= \frac{\mathbb P(N(t_1)=n_1,N(T)=n_1+n_2)}{\mathbb P(N(T)=n_1+n_2)}\\
&=\frac{\mathbb P(N(T)=n_1+n_2\mid N(t_1)=n_1)\mathbb P(N(t_1)=n_1)}{\mathbb P(N(T)=n_1+n_2)}\\
&=\frac{\mathbb P(N(T)-N(t_1)=n_2)\mathbb P(N(t_1)=n_1)}{\mathbb P(N(T)=n_1+n_2)}\\
&=\frac{e^{-\lambda(T-t_1)}(\lambda(T-t_1))^{n_2}/n_2!e^{-\lambda t_1}(\lambda t_1)^{n_1}/n_1!}{e^{-\lambda T}(\lambda T)^{n_1+n_2}/(n_1+n_2)!}\\
&= \frac{(\lambda(T-t_1))^{n_2}}{(\lambda T)^{n_2}}\cdot \frac{(\lambda t_1)^{n_1}}{(\lambda T)^{n_1}}\cdot\frac{(n_1+n_2)!}{n_1!n_2!}\\
&=\left(1-\frac{t_1}T\right)^{n_2}\left(\frac{t_1}T\right)^{n_1}\left(\frac{(n_1+n_2)!}{n_1!n_2!}\right),
\end{align}
which is not a function of $\lambda$.
Substituting $T=2$, $t_1=1$, and $n_1=n_2=5$, we have
\begin{align}
\mathbb P(N(1)=5\mid N(2)=10) &= \left(1-\frac12\right)^5\left(\frac12\right)^5\left(\frac{(5+5)!}{5!5!}\right)\\
&=\frac1{2^{10}}\left(\frac{10!}{5!5!}\right)\\
&=\frac{63}{256}\approx 0.2460938.
\end{align}
