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Thread Subject:
Limiting case of an exponential distribution

Subject: Limiting case of an exponential distribution

From: Adshak

Date: 10 Jan, 2009 14:10:36

Message: 1 of 6

Dear All,

I have an exponential distribution curve which is given by:

t = (0.0001:0.0001:15);
z = (15625000/567) .*t.^9.*exp(-10.*t);

figure;
plot(t,z);


Now I am told that the limiting case of this exponential distribution
is a gaussian curve.

I am not very good at statistics.

Can somebody kindly explain to me the significane of the limiting case
of this exponential curve and why it is a gaussian curve....and how do
I find anf fit this gaussian curve to this exponential curve (a few
matlab commands might very helpful).

I have tried googling 'limiting case of exponential distribution' but
didnt find any good explanation.

Thanks in advance.

Subject: Limiting case of an exponential distribution

From: John D'Errico

Date: 10 Jan, 2009 14:23:01

Message: 2 of 6

Adshak <adshaikh.hipnet@googlemail.com> wrote in message <5e3bfffd-9cf6-417e-8c92-fe40feedcd71@o40g2000prn.googlegroups.com>...
> Dear All,
>
> I have an exponential distribution curve which is given by:
>
> t = (0.0001:0.0001:15);
> z = (15625000/567) .*t.^9.*exp(-10.*t);
>
> figure;
> plot(t,z);
>
>
> Now I am told that the limiting case of this exponential distribution
> is a gaussian curve.
>
> I am not very good at statistics.
>
> Can somebody kindly explain to me the significane of the limiting case
> of this exponential curve and why it is a gaussian curve....and how do
> I find anf fit this gaussian curve to this exponential curve (a few
> matlab commands might very helpful).
>
> I have tried googling 'limiting case of exponential distribution' but
> didnt find any good explanation.
>
> Thanks in advance.

Perhaps the problem is this is not an exponential
distribution at all. Assuming that you are trying to
plot the pdf, this is a:

http://en.wikipedia.org/wiki/Gamma_distribution

If you wish to approximate this with a normal
distribution, then you could most simply just
use a normal with the same mean and variance
as this gamma.

John

Subject: Limiting case of an exponential distribution

From: ImageAnalyst

Date: 10 Jan, 2009 15:02:49

Message: 3 of 6

Limiting how? Which coefficient are you going to send to the limit of
0 or infinity, and then once you do that you'll observe a Gaussian?
Mathematically I just don't see it.

Subject: Limiting case of an exponential distribution

From: John D'Errico

Date: 10 Jan, 2009 15:43:02

Message: 4 of 6

ImageAnalyst <imageanalyst@mailinator.com> wrote in message <d2fc0576-1aec-418f-9cdd-154728bf35e2@s9g2000prm.googlegroups.com>...
> Limiting how? Which coefficient are you going to send to the limit of
> 0 or infinity, and then once you do that you'll observe a Gaussian?
> Mathematically I just don't see it.

As the shape parameter of the gamma (k - 1 = 9, from
the exponent of t) goes to infinity, the gamma can be
said to approach a normal distribution. See that the
skewness is 2/sqrt(k), and the excess kurtotis is 6/k. As
these two moments go to zero, the distribution starts to
look more and more normal.

But here that parameter is fixed. It is not allowed to
approach +inf. So it makes no sense to talk about the
limiting behavior. As I suggested, you can make the
simple approximation of using a normal with the
indicated mean and variance, but it will not be a
terribly good approximation. It will also yield a
significant probability that such a normal deviate
will be less than zero.

John

Subject: Limiting case of an exponential distribution

From: Roger Stafford

Date: 10 Jan, 2009 17:27:01

Message: 5 of 6

"John D'Errico" <woodchips@rochester.rr.com> wrote in message <gkafm6$9mt$1@fred.mathworks.com>...
> ImageAnalyst <imageanalyst@mailinator.com> wrote in message <d2fc0576-1aec-418f-9cdd-154728bf35e2@s9g2000prm.googlegroups.com>...
> > Limiting how? Which coefficient are you going to send to the limit of
> > 0 or infinity, and then once you do that you'll observe a Gaussian?
> > Mathematically I just don't see it.
>
> As the shape parameter of the gamma (k - 1 = 9, from
> the exponent of t) goes to infinity, the gamma can be
> said to approach a normal distribution. See that the
> skewness is 2/sqrt(k), and the excess kurtotis is 6/k. As
> these two moments go to zero, the distribution starts to
> look more and more normal.
>
> But here that parameter is fixed. It is not allowed to
> approach +inf. So it makes no sense to talk about the
> limiting behavior. As I suggested, you can make the
> simple approximation of using a normal with the
> indicated mean and variance, but it will not be a
> terribly good approximation. It will also yield a
> significant probability that such a normal deviate
> will be less than zero.
>
> John

  I believe, as the gamma distribution is defined, there is no way for it to approximate a Gaussian distribution as a limit. You have to generalize it by shifting the lower limit of its x-range toward minus infinity as you continually adjust its k and t parameters appropriately to accomplish that.

  Define a "generalized" or "shifted" gamma distribution density as:

 f(x;k,t,c) = (x-c)^(k-1)/(t^k*gamma(k))*exp(-(x-c)/t), x > c
            = 0, x <= c

Now suppose you want to approach a standard Gaussian distribution of mean zero, variance one. We follow John's suggestion of holding the above distribution to these same values. Then we have

 c + k*t = 0 <-- equal means
 k*t^2 = 1 <-- equal variances

so this gives

 t = -1/c
 k = c^2

Now let c approach minus infinity as we hold these last two equalities true and the distribution should approach the desired standard distribution. I am admittedly using my intuition on this last claim, since I don't have a rigorous proof of it.

Roger Stafford

Subject: Limiting case of an exponential distribution

From: Adshak

Date: 3 Feb, 2009 11:40:46

Message: 6 of 6

On Jan 10, 5:27=A0pm, "Roger Stafford"
<ellieandrogerxy...@mindspring.com.invalid> wrote:
> "John D'Errico" <woodch...@rochester.rr.com> wrote in message <gkafm6$9m.=
..@fred.mathworks.com>...
> > ImageAnalyst <imageanal...@mailinator.com> wrote in message <d2fc0576-1=
aec-418f-9cdd-154728bf3...@s9g2000prm.googlegroups.com>...
> > > Limiting how? =A0Which coefficient are you going to send to the limit=
 of
> > > 0 or infinity, and then once you do that you'll observe a Gaussian?
> > > Mathematically I just don't see it.
>
> > As the shape parameter of the gamma (k - 1 =3D 9, from
> > the exponent of t) goes to infinity, the gamma can be
> > said to approach a normal distribution. See that the
> > skewness is 2/sqrt(k), and the excess kurtotis is 6/k. As
> > these two moments go to zero, the distribution starts to
> > look more and more normal.
>
> > But here that parameter is fixed. It is not allowed to
> > approach +inf. So it makes no sense to talk about the
> > limiting behavior. As I suggested, you can make the
> > simple approximation of using a normal with the
> > indicated mean and variance, but it will not be a
> > terribly good approximation. It will also yield a
> > significant probability that such a normal deviate
> > will be less than zero.
>
> > John
>
> =A0 I believe, as the gamma distribution is defined, there is no way for =
it to approximate a Gaussian distribution as a limit. =A0You have to genera=
lize it by shifting the lower limit of its x-range toward minus infinity as=
 you continually adjust its k and t parameters appropriately to accomplish =
that.
>
> =A0 Define a "generalized" or "shifted" gamma distribution density as:
>
> =A0f(x;k,t,c) =3D (x-c)^(k-1)/(t^k*gamma(k))*exp(-(x-c)/t), x > c
> =A0 =A0 =A0 =A0 =A0 =A0 =3D 0, x <=3D c
>
> Now suppose you want to approach a standard Gaussian distribution of mean=
 zero, variance one. =A0We follow John's suggestion of holding the above di=
stribution to these same values. =A0Then we have
>
> =A0c + k*t =3D 0 =A0<-- equal means
> =A0k*t^2 =3D 1 =A0 =A0<-- equal variances
>
> so this gives
>
> =A0t =3D -1/c
> =A0k =3D c^2
>
> Now let c approach minus infinity as we hold these last two equalities tr=
ue and the distribution should approach the desired standard distribution. =
=A0I am admittedly using my intuition on this last claim, since I don't hav=
e a rigorous proof of it.
>
> Roger Stafford- Hide quoted text -
>
> - Show quoted text -

Dear All,

So sorry for the late reply and acknowledgement.

But thank you all so much...especially John who had understood my
badly presented question.....

Now in response:

Yes what I had shown above is indeed a Gamma distribution as it is a
sum of 10 independent exponentially distributed random variables
(sorry some how I dint realise this), and the exact equation which I
had mentioned above was achieved by a 9 time convolution of the
probability densities associated with those 10 exponential random
variables.

Now the limiting case of the Gamma distribution is a Gaussian Curve
when you send N (where N-1 =3D 9 in the equation and the exponent power
is exponential Rate R) to infinity.

For this particular equation given above, the perfect gaussian bell
curve is apparent for N>250 or so...as for below N<250 there is a
right sided skewness......

But thanks again all.....for your help

Adshak

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