Everything is explained right above (3.12) on top of page
138. The mean of the time between renewals is \mu.
The mean of the exponential variable in M+D is 1/lambda. See the
first equation in (3.12).
Good luck!
Best wishes,
Professor Whitt
---------------------------------------------------------------------------
Ward Whitt
Dept. of Industrial Engineering and
304
Ashraf, Manzur - ashmy009 wrote:
Hi,
I am curious to know:
If PDF of an inter-arrival time to a system is a shifted
exponential distributed ( f(x)=L exp( -L(x-d)) where
L=arrival rate, d=constant ………..(1)
The Mean of that distribution is Y= (1/L)exp(Ld)
[by derivation]
Can we say that ‘arrivals’ occur in that system in Poisson with
mean arrival rate of 1/Y = L exp(-Ld) ?
And can we apply the derived rate (1/Y) straight to the Poisson equation, N(t) ?
Otherwise if I try to convolute large number of equations 1 to get
Poisson equation, its become complicated (or not!)
Best wishes
Manzur
Manzur, if z is exponential with mean 1/L,
then the density of z is
f(z) = L exp(-Lz),
z >=0
The mean of z is 1/L.
Now translate to x = z+d, The
density of x is
f(x) = L exp(-L(x-d), x>= d
The mean of x is the mean of z plus d, which is 1/L + d.
The Poisson rate would be 1/L. I hope this helps.
Jack
-----------------------------------------------------------------------------------------
Manzur, for (1),
E(X) = Int_d^\infty[xLexp(-L(x-d)] dx.
To evaluate the
integral, put w = L(x-d). w goes from zero to
infinity.
Then x = w/L + d
and dx = dw/L.
E(X) = Int_0^\infty[(w/L
+ d) L exp(-w] dw/L = Int_0^\infty[(w/L + d) exp(-w] dw =
(1/L)* Int_0^\infty[w exp(-w] dw + d*Int_0^\infty[exp(-w]
dw
= 1/L + d.
In the evaluation
of those last two integrals, I used the integral representation of the gamma
function.
Int_0^\infty[w^(a-1)*exp(-w] dw =
Gamma(a) for a = 1 and 2. Gamma(1) = 1 and
Gamma(2) = 1.
Regarding (2), now
that I think of it, the inverse; that is, the number of events in a fixed time
interval would not have a Poisson distribution. That's because the hazard
rate is no longer constant.
Jack
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Jack Tomsky <jtomsky@ix.netcom.com> 09/18/2006 06:25 AM
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-------- Original Message --------
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Subject:
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RE: Shifted Exponential distribution |
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Date:
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Mon, 18 Sep 2006 15:51:17 +0930 |
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From:
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Ashraf, Manzur - ashmy009 <Manzur.Ashraf@postgrads.unisa.edu.au>
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To:
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Jack Tomsky <jtomsky@ix.netcom.com> |
Hi
Jack,
Thanks
for the useful reply to me. Your answer seems to me reasonable. Can you give me
a good reference /links for that (if any)?
But
I tried as E[X]=\int_0^\infty [x L exp(-L(x-d))] dx = (1/L) exp(L d) = mean of X (when f(x) is a shifted
exp. variable)
Where’s
the wrong – I’m confused!
2)
Its well-known that If I mix a number of exponential
variables, the arrival process is Poisson. But if the components are
‘hyper-exponential ( a mixture of dependent weighted
exponentials) or shifted exponentials, Isnt it the
resultant arrival process is Poisson?
I
am asking that because I tried to convolute a number of ‘shifted exp. variables’
and its going to be complicated.
Once
again thanks a lot to reply me. Best wishes always
Manzur
ITR,
At 06:24 18/09/2006, you wrote:
Hi,
I am curious to know:
If PDF of an inter-arrival time to a system is a shifted exponential
distributed ( f(x)=L exp( -L(x-d)) where L=arrival rate, d=constant ………..(1)
The Mean of that distribution is Y= (1/L)exp(Ld) [by derivation]
the mean is d+1/L
Can we say that ‘arrivals’ occur in that system in Poisson
with mean arrival rate of 1/Y = L exp(-Ld) ?
And can we apply the derived rate (1/Y) straight to the
Poisson equation, N(t) ?
No (even putting the mean right) it is not a
Poisson process
Otherwise if I try to convolute large number of equations 1 to get
Poisson equation, its become complicated (or not!)
If X=d+Z where Z is an ordinary exponential with
parameter L (as you have in (1)) then it has moment generating function (mgf)
E{exp(-dX)} = Lexp{-ds}/(L+s) ,
so the mgf of a sum of n intervals is just
[ Lexp{-ds}/(L+s) ] to the power n
Moments are easily obtained, though the density of the sum (by inverting the
transform) isn't simple
Hope this helps.
Best wishes
Manzur
Manzur, just a follow-up. The density function is
f(x) =
0
if x < d
= L*exp(-L(x-d) if x >=d
Given
that x <=T, where T >=d, the conditional density is
f(x|x <= T) = L*exp(-L(x-d)/[1-exp(-L(T-d)],
which
is not uniform.
Also,
the sum of N shifted exponentials is Nd plus L times
a chi-square with 2N degrees of freedom.
Jack
----- Forwarded by Jack Tomsky/sj/us/DadeInt
on 09/20/2006 09:06 AM -----
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Jack Tomsky/sj/us/DadeInt 09/20/2006 08:25 AM
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If x >= d, then
H(x) = L, as you derived. But if x < d, then the hazard
rate is 0/1 = 0.
Jack
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"Ashraf,
Manzur - ashmy009" <Manzur.Ashraf@postgrads.unisa.edu.au> |
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