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I will write down a question that I'm confused about:

The time elapsed between arrivals of customers is modeled that $V_k$ represents the time elapsed between the arrivals of the $(k-1)$th and $k$th customer. ($V_1 = $ the time the first customer arrives). Suppose that:
i) $V_1,V_2,\ldots$ are mutually independent
ii) the pdf of each $V_k$ is $f(t) = 0.1e^{-0.1t}$ for $t>0$ and $t$ is measured in minutes

My interpretation:
The second assumption means that
$$V_k \sim \mathrm{Exp}(0.1)$$

This means that the associated Poisson process will have rate $1/0.1$ (to my knowledge, the parameters of exponential and poisson are reciprocals of each other?)

However, the solution says:
Customers arrive according to a Poisson process with rate 0.1 per minute. Denote this process by $\{N(t): t\geq 0\}$ where $N(t)$ is poisson with rate $0.1$. I thought it should be $1/0.1$?

similarityinvariance1
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  • Rate 10 per minute implies roughly 10.N arrivals after N minutes hence interarrival intervals of mean length 1/10 since one wants to decompose [0,N] into 10.N smaller intervals. This is not what you want -- hence the rate is definitely 1/10 per minute, not 10 per minute. – Did Aug 21 '17 at 11:25
  • By rate $0.1$ per minute there will be averagely $1$ customer per $10$ minutes. This agrees with $\mathbb EV_1=1/0.1=10$. – drhab Aug 21 '17 at 11:25
  • So does this mean that $V_k \sim \mathrm{Exp}(10)$ instead? (10 minutes per customer) – Twenty-six colours Aug 21 '17 at 11:30
  • I would think that OP simply is getting mixed up with the "convention" with the parameters. – Twenty-six colours Aug 21 '17 at 11:32
  • The parameters are equal, the expectations are reciprocal. –  Aug 21 '17 at 11:32
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    @Twenty-sixcolours No it does not. $10$ minutes per customer means that $\frac1{\lambda}=\mathbb EV_k=10$ hence $\lambda=0.1$. – drhab Aug 21 '17 at 11:49

1 Answers1

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In the context of a customer arrival process, consistent with Poisson assumptions, you can use a Poisson random variable $X$, to model the number of customers arriving per unit of time you define (minutes, hours, years, etc.) so that the probability mass function (pmf) of $X$ is defined by $$P(X=x)= e^{-\lambda} \frac{\lambda^x}{x!}, x=\{0,1,2,\ldots\}, \lambda >0 \ \ \text{(Poisson)}$$ In this case $\lambda=E(X)$, called the event rate, or just the rate parameter that is the average number of customers per unit of time.

You can also see this same process by the perspective of the elapsed time between each customer arriving (in the same process). Lets represent this random time between arrivals by the random variable $Y$. In this case it is possible to show that $Y$ is an Exponential random variable defined by the probability density function (pdf): $$f_Y(y)=\lambda e^{-\lambda y},\ \ y>0\ \ \text{(Exponential, usual parameterization)}$$ The parameter $\lambda$ is the same rate parameter as in the related Poisson process, but in this case $E(Y)=1/\lambda$, means the average time elapsed between each customer. Therefore if customers are arriving at the rate $\lambda=2$ customers per hour, the average time between each customer is $1/\lambda=0,5$ hour per customer. This is the most common way to represent the Exponential pdf.

It is also possible (but less common) to use an alternative representation for an Exponential random variable, considering:

$$f_Y(y)=\frac{1}{\beta} e^{-\frac{1}{\beta} y},\ \ y>0\ \ \text{(Exponential, alternative parameterization)}$$

In this case $E(Y)=\beta$, and for the same process described in last paragraphs it will hold that $\beta=\frac{1}{\lambda}$, that is the parameter $\beta$ is the reciprocal of the rate parameter $\lambda$. The parameter $\beta$ is sometimes called survival parameter in some applications.

bluemaster
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