Re: Poisson and exponential distributions
- From: Ray Vickson <RGVickson@xxxxxxx>
- Date: Tue, 12 Aug 2008 23:45:46 -0700 (PDT)
On Aug 12, 8:44 pm, "giganut" <giga...@xxxxxxxxx> wrote:
hello,
i'm having some trouble seeing the relationship between the Poisson and the
exp. distributions. specifically,
if the Poisson with rate L and time t is:
P(k| L,t) = (Lt)^k exp(-Lt) / k!
then i think that the exp. process with rate-parameter L, should basically
be P(0| L, t) = exp(-Lt).
No "think" required; it follows right away from the formula.
why doesn't the normalizing constant (1/L) come
out directly from this to give Lexp(-Lt) ?
Because it is not a probability density, so does not need to integrate
to 1. Here is what is happening: P(k| L,t) = probability that exactly
k "arrivals" occur in the time interval [0,t], so P(0|L,t) =
probability of NO arrivals in [0,t] = probability that the first
arrival occurs after t = P{T > t}, where T = time of first arrival.
Thus, the complementary cumulative distribution of T is exp(-L*t),
hence the probability density of T is f(t) = -(d/dt)P{T > t} = L*exp(-
L*t), as desired.
does this have something to do with the fact that P(0| L, t) as above is a
likelihood and not a probability ?
No, it is a probability, *not* a probability DENSITY function.
if so, then using laws of probability,
can the exponential distribution be derived, i.e.:
P(t|0,L) = p(0|L,t)p(t|L) / p(0|L)
conversely, how would one derive the poisson from the exponential
distribution ?
Let T1, T2, ... be the successive inter-arrival times of the process:
T1 = time of first arrival, T2 = time between first and second
arrival, etc. The Ti are independent, identically distributed random
variables with distribution expl(L). The time of the nth arrival is Sn
= T1 + T2 +...+ Tn, which has the so-called n-Erlang distribution,
with density f_n(t) = L^n t^(n-1)* exp(-L*t)/(n-1)! (can get this by
induction, or by Laplace transform methods). We have
P{>= n arrivals in [0,t]} = P{Sn <= t}, because Sn is <= t if and only
if at least n arrivals occur by t. We have P{Sn <= t} =
integral(f_n(s) ds, s=0 .. t). You can use repeated integration by
parts to show this has the form 1 - sum{k=0 .. (n-1)} (L*t)^k exp(-
L*t)/k!, so P{n arrivals} = P{>= n arrivals} - P{>= n+1 arrivals} =
(L*t)^n exp(-Lt)/n! = Poisson probability.
Alternatively, you can develop a set of differential equations for the
Pn(t) = P{n arrivals in [0,t]}, then solve them sequentially, starting
with P0(t) = P{0 arrivals in [0,t]} = P{T1 > t} = exp(-L*t).
Isn't all this material in your course textbook?
R.G. Vickson
thanks,
-g
.
- References:
- Poisson and exponential distributions
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