Characteristic function of the binomial distribution

In summary, the conversation revolves around finding the characteristic function, moments, and cumulants of a binomial variable with parameters n and p. The characteristic function is found to be (1-p+pe^{it})^n, where t is just a parameter. The moments and cumulants are then calculated using the characteristic function, with a different notation being used in the class. The moments can also be calculated using the moment-generating function, but it becomes complicated for higher moments. The conversation also discusses the possibility of finding a general expression for the nth derivative of the characteristic function.
  • #1

fluidistic

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Homework Statement


Hey guys, I'm self studying some probability theory and I'm stuck with the basics.
I must find the characteristic function (also the moments and the cumulants) of the binomial "variable" with parameters n and p.
I checked out wikipedia's article http://en.wikipedia.org/wiki/Characteristic_function_(probability_theory), apparently the solution is [itex](1-p+pe^{it})^n[/itex] though I didn't really understand what t stand for (number of successes?).

Homework Equations


Characteristic function: [itex]\int e^{ikX} P(x)dx[/itex].

The Attempt at a Solution


I'm guessing that I must simply apply the given formula. The k would be wikipedia's t variable. I'm stuck at finding P(x) and X. I've searched and found out the binomial distribution's article in wikipedia and [itex]P(K=k)=\frac{n!p^k (1-p)^{n-k}}{k!(n-k)!}[/itex] which is called the probability mass function. I don't know how how I could "plug" this into the given formula.
Thanks for any tip.
 
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  • #2
fluidistic said:

Homework Equations


Characteristic function: [itex]\int e^{ikX} P(x)dx[/itex].

The definition of the characteristic function of a random variable X is

[tex]E[e^{ikX}][/tex]

How you calculate this expectation depends on what kind of random variable you are dealing with.

The equation you listed is for a continuous random variable, i.e. one that has a probability density function.

For a discrete random variable, with probability mass function P(X = n), the characteristic function would be

[tex]E[e^{ikX}] = \sum_{n} e^{ikn} P(X = n)[/tex]
 
  • #3
fluidistic said:

Homework Statement


Hey guys, I'm self studying some probability theory and I'm stuck with the basics.
I must find the characteristic function (also the moments and the cumulants) of the binomial "variable" with parameters n and p.
I checked out wikipedia's article http://en.wikipedia.org/wiki/Characteristic_function_(probability_theory), apparently the solution is [itex](1-p+pe^{it})^n[/itex] though I didn't really understand what t stand for (number of successes?).

Homework Equations


Characteristic function: [itex]\int e^{ikX} P(x)dx[/itex].


The Attempt at a Solution


I'm guessing that I must simply apply the given formula. The k would be wikipedia's t variable. I'm stuck at finding P(x) and X. I've searched and found out the binomial distribution's article in wikipedia and [itex]P(K=k)=\frac{n!p^k (1-p)^{n-k}}{k!(n-k)!}[/itex] which is called the probability mass function. I don't know how how I could "plug" this into the given formula.
Thanks for any tip.

First tip: do more than consult Wikipedia. Get a good *book*.

Anyway, the characteristic function of the Binomial random variable [itex]X[/itex] is
[tex]\text{ch}_{X}(t) \equiv E\left(e^{iX}\right) = \sum_{k=0}^n {n \choose k} e^{i t k} p^k (1-p)^{n-k} \\
= \sum_{k=0}^n {n \choose k} (pe^{it})^k (1-p)^{n-k} = (1-p + pe^{it})^n,[/tex]
using the binomial expansion
[tex](a+b)^n = \sum_{k=1}^n {n \choose k} a^k b^{n-k}. [/tex]
The "t" has nothing to do with numbers of successes, or anything; it is just a parameter used in the characteristic function.

RGV
 
  • #4
Ok thank you very much guys.
By the way which book(s) would you recommend me? Because I'm totally stuck at finding the moments (I know they are the coefficients of the Taylor's expansion of the characteristic function and the cumulants which are just the logarithm of the moments). Should I just calculate the Taylor's series of [itex](1-p + pe^{it})^n[/itex] with respect to t? Is this the way to go?
 
  • #5
Some news about my tries:
The n'th moment around the point "a" is definied as [itex]\mu _ n (a)=\sum (x-a)^n P(x)[/itex] where P is the probability mass function.
So if I take the binomial distribution where there is N tries with each a probability of "p" to occur, then [itex]\mu _ N (a)= \sum _{k=1}^{N} (x-a)^N {N \choose k} p ^k (1-p)^{N-k}[/itex].
I'm sure I've messed up some variables here. I'm kind of confused. Any help is appreciated.
 
  • #6
fluidistic said:
Some news about my tries:
The n'th moment around the point "a" is definied as [itex]\mu _ n (a)=\sum (x-a)^n P(x)[/itex] where P is the probability mass function.
So if I take the binomial distribution where there is N tries with each a probability of "p" to occur, then [itex]\mu _ N (a)= \sum _{k=1}^{N} (x-a)^N {N \choose k} p ^k (1-p)^{N-k}[/itex].
I'm sure I've messed up some variables here. I'm kind of confused. Any help is appreciated.

It is a bit easier to use the moment-generating function (mgf) instead of the characteristic function. The mgf of any discrete random variable X is
[tex] m_X(t) = \sum_{x} p(x) e^{tx} = E\, e^{tX}.[/tex] To get moments of X about 'a', you need to find [itex] E(X-a)^k[/itex], and the easiest way is to use
[tex] m_{X,a} (t) \equiv E\, e^{(X-a)t} = \sum_{x} p(x) e^{(x-a)t} = e^{-at} m_X(t).[/tex]
We have
[tex] E\,(X-a)^k = \left. \left( \frac{\partial}{\partial t}\right)^k m_{X,a}(t)\right|_{t=0}.[/tex]
For the binomial B(n,p) we have [itex] m_{X,a}(t) = e^{-at} (1-p + pe^t)^n.[/itex] For 'a' different from the mean the computation of the kth moment gets complicated (but do-able) for k > 2.

RGV
 
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  • #7
Ray Vickson said:
It is a bit easier to use the moment-generating function (mgf) instead of the characteristic function. The mgf of any discrete random variable X is
[tex] m_X(t) = \sum_{x} p(x) e^{tx} = E\, e^{tX}.[/tex] To get moments of X about 'a', you need to find [itex] E(X-a)^k[/itex], and the easiest way is to use
[tex] m_{X,a} (t) \equiv E\, e^{(X-a)t} = \sum_{x} p(x) e^{(x-a)t} = e^{-at} m_X(t).[/tex]
We have
[tex] E\,(X-a)^k = \left. \left( \frac{\partial}{\partial t}\right)^k m_{X,a}(t)\right|_{t=0}.[/tex]
For the binomial B(n,p) we have [itex] m_{X,a}(t) = e^{-at} (1-p + pe^t)^n.[/itex] For 'a' different from the mean the computation of the kth moment gets complicated (but do-able) for k > 2.

RGV
I've had my second class in this course, the professor helped us a bit for this exercise.
He uses a different notation than in this thread; to get the first moment I think he chose your way (let me know).
From the characteristic function [itex]\phi _X (k)=(pe^{ik}+1-p)^N[/itex], one gets the moments via the formula [itex]<X^n>=\frac{1}{i^n} \frac{d^n}{dk^n} \phi _X (k) \big | _{k=0}[/itex]. (*)
Where in my case [itex]X=n_1[/itex] where [itex]P_N(n_1)=\frac{N!}{n_1!(N-n_1)!}p^{n_1} (1-p)^{N-1}[/itex]. So in a way I think he expect us to use the formula (*) where our "a" would be 0. Is there a link with the moment generating function you suggested to use?
I'm wondering if I can get a general expression for the nth derivative of the characteristic function. I'll work on this.
 
  • #8
fluidistic said:
I've had my second class in this course, the professor helped us a bit for this exercise.
He uses a different notation than in this thread; to get the first moment I think he chose your way (let me know).
From the characteristic function [itex]\phi _X (k)=(pe^{ik}+1-p)^N[/itex], one gets the moments via the formula [itex]<X^n>=\frac{1}{i^n} \frac{d^n}{dk^n} \phi _X (k) \big | _{k=0}[/itex]. (*)
Where in my case [itex]X=n_1[/itex] where [itex]P_N(n_1)=\frac{N!}{n_1!(N-n_1)!}p^{n_1} (1-p)^{N-1}[/itex]. So in a way I think he expect us to use the formula (*) where our "a" would be 0. Is there a link with the moment generating function you suggested to use?
I'm wondering if I can get a general expression for the nth derivative of the characteristic function. I'll work on this.

That formula is the same one I use, but in a different variable (k instead of t). The reason I use the moment-generating function instead of the characteristic function is just to avoid the annoying factor 1/i^k in front. As to a link: just Google "moment generating function", which turns up numerous articles.

RGV
 
  • #9
Some news about this. I've found the first 2 moments (I used the formula in my previous post) and I'd like to find the first 2 cumulants.
What I've done so far is [itex]N \ln (pe^{ik}+1-p)=\sum _{n=1}^{\infty } C_n \frac{(ik)^n}{n!}[/itex]. I'm stuck here. I know I want [itex]C_1[/itex] and [itex]C_2[/itex] but I don't know how to proceed; at all. Any tip is welcome.
 

What is the binomial distribution?

The binomial distribution is a probability distribution that describes the probability of obtaining a certain number of successes in a fixed number of independent trials, where each trial has a binary outcome (success or failure) and the probability of success is constant.

What is the characteristic function of the binomial distribution?

The characteristic function of the binomial distribution is a mathematical function that fully describes the distribution. It is defined as the expected value of e^(itX), where X is a random variable following the binomial distribution and t is a real number.

How is the characteristic function used in the binomial distribution?

The characteristic function is used to calculate the probability of obtaining a certain number of successes in a given number of trials. It can also be used to find the expected value, variance, and other important parameters of the binomial distribution.

What is the relationship between the characteristic function and the probability generating function?

The probability generating function is a special case of the characteristic function, where t is equal to 1. This means that the probability generating function can be derived from the characteristic function, and vice versa.

How does the binomial distribution differ from other probability distributions?

The binomial distribution is unique in that it only has two possible outcomes (success or failure) and the probability of success remains constant throughout all trials. Other distributions may have more than two outcomes or have varying probabilities for each outcome.

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