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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 $X_{(i)}-X_{(i-1)}\sim \exp (i\lambda)$?

StubbornAtom
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  • When $X_1,\ldots,X_n$ are i.i.d exponential with rate $\lambda$, the spacings $X_{(i)}−X_{(i−1)},,i=1,\ldots,n$ (taking $X_{(0)}=0$) are independent exponential with rate $(n−i+1)\lambda$. So, $$\sum_{i=1}^n E(X_{(i)}−X_{(i−1)})=\sum_{i=1}^n \frac1{(n−i+1)\lambda}=\sum_{i=1}^n \frac1{i\lambda}$$

    As $\sum_{i=1}^n (X_{(i)}−X_{(i−1)})=X_{(n)}$, the above is also the mean of $X_{(n)}$.

    See https://math.stackexchange.com/q/509816/321264, https://math.stackexchange.com/q/80475/321264.

     – StubbornAtom Aug 23 '20 at 18:40

1 Answers1

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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 $\frac{1}{(n-1)\lambda}$.

angryavian
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  • But it says " Each car getting a phone call with passion distribution of 2 calls in one hour" so why $X_i$ is not possion? –  Aug 23 '20 at 08:08
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    @abuka123 Given what the solution is, I am fairly certain they mean "Poisson process with rate $2$ calls per hour." A Poisson process is a stochastic process that is related to Poisson distributions and exponential distributions. – angryavian Aug 23 '20 at 17:55
  • Oh I see, thank you. So if $X_{1},\ldots,X_{2}\sim\exp(\lambda)$ then $X_{(i)}-X_{(i-1)}\sim\exp(i\lambda)$? I'm trying to extract a general formula from that question. –  Aug 23 '20 at 18:30