Re: Probability with bus..

On Mar 31, 4:41 am, "mina_world" <mina_wo...@xxxxxxxxxxx> wrote:
Hello teacher~

A passenger arrives at a bus-stop at some arbitrary point in time.
Buses arrive according to a uniform distribution on [0,1]. (Namely, per 1
min.)

What is the mean waiting time until the next bus ?

-----------------------------------------------
Sorry. I need your advice.

Arrival of buses is a "renewal process" with a uniform inter-arrival
time X. If you arrive after many buses have already gone by (that is,
at a large time), the waiting-time distribution is, essentially, the
limiting waiting-time distribution, whose probability density is g(x)
= G(x)/EX, where G(x) is the complementary cumulative distribution of
X and EX is the mean. We have H(x) = 1-x and EX = 1/2, so the density
is g(x) = 2*(1-x) for 0 <= x <= 1, and g(x) = 0 for x < 0 or x > 1.
The expected waiting time is int(x*g(x),x=0..1) = 2*int(x*(1-
x),x=0..1) = 2*(1/6) = 1/3. Of course, this is larger than the "naive"
value that you might get if you simply reason that you come, on
average, halfway between two buses, and buses arrive spaced 1/2 minute
on average, so you ought 1/4 minute on average. That would be wrong,
but the reasons are subtle.

You might also be interested in the expected time Y between the two
buses that "straddle" your arrival at the bus stop; that is, the time
between the last bus before you come and the next bus after you come.
The (limiting, or equilibrium) density of Y is g(y) = y*f(y)/EX, where
f = inter-arrival time density. The mean is EY = E(X^2)/E(X), which
gives EY = 2/3 for X~U(0,1). Again, this is larger than the average
time EX = 1/2 minute between two buses, even though it IS the time
between two specific buses---namely, the one that came before you and
the one that comes after you.

If your arrival time is not large, you need to solve the renewal
process integral equation to get the waiting time density and expected
waiting time.

R.G. Vickson

.

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