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Jonas

Joined: 08 Sep 2005
Posts: 77

Posted: Fri Jun 02, 2006 4:12 pm    Post subject: Mean of Geometric Distribution

I am trying to derive the mean of the geometric distribution but I am
having trouble on one step. I would appreciate it if somebody could
explain what I am not seeing.

Here is what I have:

step1: E(X)=sum from x=1 to infinity of x(1-p)^(x-1) p

step 2: E(X)=(p) sum from x=1 to infinity of x(1-p)^(x-1)

step 3: E(X)=(p) derivative with respect to (1-p) of sum of x=1 to
infinity of (1-p)^x

I don't understand where the x from step 2 went. It seems to me that
without this x in the equation, 1+2+3+4+5+.......infinity is missing.

What am I not understanding. Your suggestions are greatly appreciated.
danheyman@yahoo.com
science forum beginner

Joined: 18 Jul 2005
Posts: 33

Posted: Fri Jun 02, 2006 7:59 pm    Post subject: Re: Mean of Geometric Distribution

You need to know that the derivative of the sum is the sum of the
derivatives (this is always true for finite sums but not always true
for infinite sums - for the geometric sum it is true). So there is a
step 2a which observes that x(1-p)^(x-1) is the derivative of (1-p)^x
wrt x.

Dan Heyman

Jonas wrote:
 Quote: I am trying to derive the mean of the geometric distribution but I am having trouble on one step. I would appreciate it if somebody could explain what I am not seeing. Here is what I have: step1: E(X)=sum from x=1 to infinity of x(1-p)^(x-1) p step 2: E(X)=(p) sum from x=1 to infinity of x(1-p)^(x-1) step 3: E(X)=(p) derivative with respect to (1-p) of sum of x=1 to infinity of (1-p)^x I don't understand where the x from step 2 went. It seems to me that without this x in the equation, 1+2+3+4+5+.......infinity is missing. What am I not understanding. Your suggestions are greatly appreciated.
Jonas

Joined: 08 Sep 2005
Posts: 77

Posted: Fri Jun 02, 2006 8:54 pm    Post subject: Re: Mean of Geometric Distribution

Dan,

I'm sorry but I still don't understand what happened to the x between
step 2 and step 3. It seems to me that step 3 should be:

E(X)=(p)(x) derivative with respect to (1-p) of sum of x=1 to
infinity of (1-p)^x

Where does the x go in the product (p)(x)?

danheyman@yahoo.com wrote:
 Quote: You need to know that the derivative of the sum is the sum of the derivatives (this is always true for finite sums but not always true for infinite sums - for the geometric sum it is true). So there is a step 2a which observes that x(1-p)^(x-1) is the derivative of (1-p)^x wrt x. Dan Heyman Jonas wrote: I am trying to derive the mean of the geometric distribution but I am having trouble on one step. I would appreciate it if somebody could explain what I am not seeing. Here is what I have: step1: E(X)=sum from x=1 to infinity of x(1-p)^(x-1) p step 2: E(X)=(p) sum from x=1 to infinity of x(1-p)^(x-1) step 3: E(X)=(p) derivative with respect to (1-p) of sum of x=1 to infinity of (1-p)^x I don't understand where the x from step 2 went. It seems to me that without this x in the equation, 1+2+3+4+5+.......infinity is missing. What am I not understanding. Your suggestions are greatly appreciated.
Mike1170

Joined: 17 Sep 2005
Posts: 74

Posted: Sat Jun 03, 2006 3:00 am    Post subject: Re: Mean of Geometric Distribution

"Jonas" wrote in message
 Quote: I am trying to derive the mean of the geometric distribution but I am having trouble on one step. I would appreciate it if somebody could explain what I am not seeing. Here is what I have: step 1: E(X)=sum from x=1 to infinity of x(1-p)^(x-1) p step 2: E(X)=(p) sum from x=1 to infinity of x(1-p)^(x-1) step 3: E(X)=(p) derivative with respect to (1-p) of sum of x=1 to infinity of (1-p)^x I don't understand where the x from step 2 went. It seems to me that without this x in the equation, 1+2+3+4+5+.......infinity is missing. What am I not understanding. Your suggestions are greatly appreciated.

Why not just use the basic technique for summing infinite series:
S = a + ar + ar^2 + ar^3 + ...
rS = ar + ar^2 + ar^3 + ...
(1-r) S = a; S = a / (1-r)

In this case (finding the mean of the geometric distribution defined on the
positive integers)
you just have to apply the (same) technique twice since "x" appears as a
factor in each term.
E(x) = 1p + 2p(1-p)^1 + 3p(1-p)^2 + 4p(1-p)^3 + ...
(1-p)E(x) = 1p(1-p)^1 + 2p(1-p)^2 + 3p(1-p)^3 + ...
Subtracting the latter from the former,
pE(x) = 1p + 1p(1-p)^1 + 1p(1-p)^2 + 1p(1-p)^3 + ...
At this point you can divide both sides by p ( p <> 0)
E(x) = 1 + (1-p)^1 + (1-p)^2 + (1-p)^3 + ...
now repeat same technique:
(1-p)E(x) = (1-p)^1 + (1-p)^2 + (1-p)^3 + ...
pE(x) = 1
E(x) = 1/p
Note: some people define the geometric distribution on the non-negative
integers, x = 0, 1, 2, 3, ..
when so defined E(x) = 1/p - 1 = (1-p)/p
Mike1170

Joined: 17 Sep 2005
Posts: 74

Posted: Sat Jun 03, 2006 7:22 pm    Post subject: Re: Mean of Geometric Distribution

"Jonas" wrote
 Quote: I am trying to derive the mean of the geometric distribution but I am having trouble on one step. I would appreciate it if somebody could explain what I am not seeing. Here is what I have: step1: E(X)=sum from x=1 to infinity of x(1-p)^(x-1) p step 2: E(X)=(p) sum from x=1 to infinity of x(1-p)^(x-1) step 3: E(X)=(p) derivative with respect to (1-p) of sum of x=1 to infinity of (1-p)^x I don't understand where the x from step 2 went. It seems to me that without this x in the equation, 1+2+3+4+5+.......infinity is missing. What am I not understanding. Your suggestions are greatly appreciated.

The x went into the exponent of (1-p). Step 3 makes use of the following
result from calculus I:
"the derivative of (x^n) wrt x = n x^(n-1)"

Summation aside, your step 3 amounts to the above in reverse: x (1-p)^(x-1)
= "the derivative wrt (1-p) of (1-p)^x."
The derivative is with respect to "1-p", as you note. (If the derivative
were wrt to p, the sign would change.)

So, yes, the derivative of a sum is the sum of the derivatives but why use
derivative in the first place? You can derive the mean of the geometric
distribution without using derivatives as I outlined in earlier post.
Derivatives do come into play however when using generating functions but
you don't mention them. If the generating function of a distribution is
represented by g(s) (s a dummy variable), then the mean of the distribution
is the first derivative of g(s) at s=1, or g'(1). The generating funtion of
the geometric distribution is g(s) = ps/(1-(1-p)s).
danheyman@yahoo.com
science forum beginner

Joined: 18 Jul 2005
Posts: 33

Posted: Mon Jun 05, 2006 11:48 pm    Post subject: Re: Mean of Geometric Distribution

Let q=1-p. We have E(X)=sum (1 to oo) pxq^(x-1) =sum (1 to 00)
(d/dq)pq^x
=p* (d/dq) sum (1 to oo) q^x and so forth.

Another poster recommended not using derivatives at all, which I agree
is the best way. However, this technique is used widely and is worth
knowing.

Dan Heyman

Jonas wrote:
 Quote: Dan, I'm sorry but I still don't understand what happened to the x between step 2 and step 3. It seems to me that step 3 should be: E(X)=(p)(x) derivative with respect to (1-p) of sum of x=1 to infinity of (1-p)^x Where does the x go in the product (p)(x)? danheyman@yahoo.com wrote: You need to know that the derivative of the sum is the sum of the derivatives (this is always true for finite sums but not always true for infinite sums - for the geometric sum it is true). So there is a step 2a which observes that x(1-p)^(x-1) is the derivative of (1-p)^x wrt x. Dan Heyman Jonas wrote: I am trying to derive the mean of the geometric distribution but I am having trouble on one step. I would appreciate it if somebody could explain what I am not seeing. Here is what I have: step1: E(X)=sum from x=1 to infinity of x(1-p)^(x-1) p step 2: E(X)=(p) sum from x=1 to infinity of x(1-p)^(x-1) step 3: E(X)=(p) derivative with respect to (1-p) of sum of x=1 to infinity of (1-p)^x I don't understand where the x from step 2 went. It seems to me that without this x in the equation, 1+2+3+4+5+.......infinity is missing. What am I not understanding. Your suggestions are greatly appreciated.

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