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I am standing in front of a restaurant. Every time a customer goes in the restaurant, I want to make a prediction of dt (dt is the time interval between the next customer and the current customer). What distribution should I use to predict dt? Thanks!

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There are a few distributions you can use. You can use an exponential distribution $X\sim \textrm{Exponential}(\lambda)$

$$ f(x;\lambda) =\begin{align}\begin{cases} \lambda e^{-\lambda x} & x \geq 0 \\ \\ 0 & x <0 \end{cases} \end{align} \tag{1}$$

which describes the time between events in a Poisson process. You can also use a Gamma distribution. $X \sim \textrm{Gamma}(\alpha, \beta) $ with mass function

$$ f(x;\alpha, \beta ) = \frac{\beta^{\alpha} x^{\alpha-1} e^{-\beta x} }{\Gamma(\alpha)} \tag{2} $$ where $ x > 0 , \alpha ,\beta >0$

This is an example problem I found.

Example Students arrive at a local bar and restaurant according to an approximate Poisson process at a mean rate of 30 students per hour. What is the probability that the bouncer has to wait more than 3 minutes to card the next student?

Ok, I'm pulling this from here. . It notes if $X$ is the number of students then the poisson mean $\lambda$ we have $30$ per hour or $\frac{1}{2}$ . Denoting $W$ the waiting time between students we would expect the $\theta =\frac{1}{\lambda} = 2$ minutes between every arrival. Since $W$ is exponentially distributed we use the pdf above $ W \sim \textrm{Exponetial}(\frac{1}{2})$

$$ f(w;\frac{1}{2}) =\begin{align}\begin{cases} \frac{1}{2} e^{-\frac{1}{2} w} & w \geq 0 \\ \\ 0 & w <0 \end{cases} \end{align} \tag{3}$$

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  • $\begingroup$ Is $x$ here corresponding to my $dt$ ? And is $\lambda$ the expected event counts within a given time? Could you please elaborate how do I choose $\lambda$ in this example? Thanks! $\endgroup$
    – Edamame
    Sep 13, 2018 at 22:36
  • $\begingroup$ the mean is $\frac{1}{\lambda}$ it is describing the lengths of time between a poisson process. $x$ is describing how the probability of how long you would have to wait. $\endgroup$
    – user3417
    Sep 13, 2018 at 22:41
  • $\begingroup$ So if the expected customer number is 24 per day (1440 mins), is the average time length between customer 60 min? So $mean$ = 60? and $1/\lambda$ = 0.01667? $\endgroup$
    – Edamame
    Sep 13, 2018 at 22:46
  • $\begingroup$ one second let me write something, it is an example problem. $\endgroup$
    – user3417
    Sep 13, 2018 at 22:47
  • $\begingroup$ If the poisson mean $\lambda $ for the day is 24 we divide can divide for the poisson rate for the how $ \frac{\lambda}{24}$ then do $\frac{1}{\lambda_{1}}$ So there should be one person arriving every hour and you're right on the last part. $\endgroup$
    – user3417
    Sep 13, 2018 at 22:57

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