I'm trying to figure out the connection between the difference between two order statistic variables and exponential distribution. I saw previous topic but it's to complex to understand (higher mathematics than what I currently know). I'm not familiar with Chi-square distribution. What is the connection between two order statistic variables and exponential distribution? when does it holds? In an exam I had the following question:
In a taxi service at $00:00$ there are $8$ cars. Each car getting a phone call with passion distribution of $2$ calls in one hour. Each taxi that gets a call, leaves and does not come back. Find the expected value until the taxi service is empty of cars.
In the solution they had the following equation: $$ \sum_{i=1}^8 E(X_{(i)}-X_{(i-1)})=\sum_{i=1}^8 \frac{1}{i\cdot \lambda} $$
They stated that $X_i$ is the leave time of taxi $i$ and we want $E(X_{(8)})$.
When $X_{(i)}-X_{(i-1)}\sim \exp (i\lambda)$ is true and how do I find $\lambda$?
EDIT: The question is why if $X_1,\ldots X_n \sim Pois(\lambda)$ then $X_{(i)}-X_{(i-1)}\sim \exp (i\lambda)$?
Note that $X_1, \ldots, X_n\overset{\text{i.i.d.}}{\sim} \text{Exponential}(\lambda)$ where $\lambda = 2$. They are not Poisson.
$X_{(1)} = \min\{X_1, \ldots, X_n\}$ is the time of the first departure. It is a common exercise to show that $X_{(1)} \sim \text{Exponential}(n\lambda)$. Thus $E[X_{(1)}] = \frac{1}{n\lambda}$.
Now, use the memorylessness property of the exponential distribution. That is, $P(X_i > t +s \mid X_i > t) = P(X_i > s)$. This implies that conditioned the earliest taxi leaving at time $t$, the additional time until each other taxis' departures is also $\text{Exponential}(\lambda)$. In particular, the time between the first departure and the second departure $X_{(2)}-X_{(1)}$ is the minimum of $n-1$ i.i.d. $\text{Exponential}(\lambda)$ random variables, which is $\text{Exponential}((n-1)\lambda)$ and has mean $\frac{1}{(n-1)\lambda}$.