Re: Limiting case of an exponential distribution

"John D'Errico" <woodchips@xxxxxxxxxxxxxxxx> wrote in message <gkafm6$9mt$1@xxxxxxxxxxxxxxxxxx>...
ImageAnalyst <imageanalyst@xxxxxxxxxxxxxx> wrote in message <d2fc0576-1aec-418f-9cdd-154728bf35e2@xxxxxxxxxxxxxxxxxxxxxxxxxxx>...
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

.

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