Rewriting bionomial sum using partial derivative

[Nikitin]

Hi. Assume there's a probability ##q## for a guy to take a step to the right, and ##p=1-q## to take one to the left. Then the probability to take ##n## steps to the right out of ##N## trials is ##P(n) = {{N}\choose{n} }q^n p^{N-n}##.

Now, what is ##<n>##? My textbook in statistical physics says ##<n> = \sum_{n=0}^N n{{N}\choose{n}} q^n p^{N-n} = q \frac{\partial}{\partial q} \sum_{n=0}^N {{N}\choose{n} }q^n p^{N-n} = q \frac{\partial}{\partial q} (q+p)^N = qN##

OK, fair enough.. But what about the fact that ##(q+p)^N=1## ? Shouldn't this mean you're taking the partial derivative of a constant, and thus it equals 0? What am I missing here?

 

[HallsofIvy]

I can see no place where the "partial derivative" comes into this. Since p+ q= 1, your equation [itex](q+ p)^N= 1[/itex] is simply [itex]1^N= 1[/itex]. What variable are you taking the partial derivative with respect to?

 

[Nikitin]

I'm taking it with respect to ##q##.. Or, rather, my textbook is. I'm kind of confused as well.

The author wants to find ##<n>##, the expected amount of steps to the right, by doing a "trick":

First he rewrites ##<n>## from ##<n> = \sum_{n=0}^N n{{N}\choose{n}} q^n p^{N-n}## into ##<n> = q \frac{\partial}{\partial q} \sum_{n=0}^N {{N}\choose{n} }q^n p^{N-n}##

Then, since

##\sum_{n=0}^N {{N}\choose{n}} q^n p^{N-n} = (q+p)^N##

This must be true:

##<n> = \sum_{n=0}^N n{{N}\choose{n}} q^n p^{N-n} = q \frac{\partial}{\partial q} \sum_{n=0}^N {{N}\choose{n} }q^n p^{N-n} = q \frac{\partial}{\partial q} (q+p)^N =qN (q+p)^{N-1} = qN##

What I don't understand is how it's legal to take the partial derivative of something (in this case, ##(q+p)^N##) that's equal to 1, and not get 0.

EDIT: I also fixed a type in OP: the ##<n>## was misplaced outside the summation sign.

 

[disregardthat]

Here you are treating q as a real variable, and p, n and N as constants. So basically you are defining a function f(q), and taking its derivative.

 

[Stephen Tashi]

The textbook didn't have to enforce the condition p+q = 1 in order to do the trick.

For a function F(x,y) of two variables , there is a difference between [itex]{ \frac{\partial F(x,y)}{\partial x}} _{|_{ x=q,\ y=1-q}}[/itex] and [itex]{ \frac{\partial F(x,1-x)}{\partial{x}}}_{|_{x=q}} [/itex]

 

[Ray Vickson]

Hi. Assume there's a probability ##q## for a guy to take a step to the right, and ##p=1-q## to take one to the left. Then the probability to take ##n## steps to the right out of ##N## trials is ##P(n) = {{N}\choose{n} }q^n p^{N-n}##.

Now, what is ##<n>##? My textbook in statistical physics says ##<n> = \sum_{n=0}^N n{{N}\choose{n}} q^n p^{N-n} = q \frac{\partial}{\partial q} \sum_{n=0}^N {{N}\choose{n} }q^n p^{N-n} = q \frac{\partial}{\partial q} (q+p)^N = qN##

OK, fair enough.. But what about the fact that ##(q+p)^N=1## ? Shouldn't this mean you're taking the partial derivative of a constant, and thus it equals 0? What am I missing here?

First, regard p and q as two separate variables when differentiating ##P(n)## wrt ##q## and summing over n. Then, after finishing you can finally put ##q = 1-p##. This is not really any different from evaluating ##\sum_{n=1}^N n 2^n## by first differentiating ##\sum_{n=1}^N n x^n## wrt ##x##, then putting in ##x=2## later.

 

[Nikitin]

But how is this legal? You are differentiating something that is 1, yet you do not get 0? This seems very un-mathematical.

 

[disregardthat]

But how is this legal? You are differentiating something that is 1, yet you do not get 0? This seems very un-mathematical.

The expression represents something, an expectancy value for a certain binomial situation where the probability is p.

But, the expression can also be treated differently. Assuming q is a real variable, it is no longer the expectancy value of anything, just a function. But, after plugging in q = 1-p, we find that we have the same thing.

Finding that the function is the derivative of a certain other function is just calculus. And this other function is easier to deal with, and we know its value at q = 1-p. By plugging, we see that we end up with the value we wanted to find. This is an example of using calculus just to calculate a certain value, and the calculation does not need any statistical interpretation. We just wanted to find the value.

 

[Ray Vickson]

But how is this legal? You are differentiating something that is 1, yet you do not get 0? This seems very un-mathematical.

Which message are you responding to? Please use the Quote button---which is put there for a good reason.

Have you looked at my response (post #6)? That says it all, but since you seem to be missing something, here goes again. We certainly have
[tex] (x+y)^n = \sum_{k=0}^n {n \choose k} x^k y^{n-k} [/tex]
for ANY ##x## and ##y##. Do you accept that
[tex] x \frac{\partial}{\partial x} (x+y)^n = n x (x+y)^{n-1}?[/tex]
Do you accept that
[tex] x \frac{\partial}{\partial x} \sum_{k=0}^n {n \choose k} x^k y^{n-k}
= \sum_{k=0}^n k {n \choose k} x^k y^{n-k}? [/tex]
If you accept these equations then you are forced---no choice--- to accept the equation
[tex] x n (x+y)^{n-1} = \sum_{k=0}^n k {n \choose k} x^k y^{n-k} [/tex]
for ANY ##x## and ##y##. What is now preventing you from putting ##x=q, y = 1-q##?

 

[Nikitin]

Ray: The calculus in itself was never the issue.

Guys, what confused me was that it's OK to define away the fact that p+q = 1 in the first moment, and then apply it in the next for convenience. But fair enough, I see things from a different perspective now. At the end of the day there's no mathematical reason you can't do it.