Message Boards Message Boards

Obtain the moments of a probability mass function?

GROUPS:

I am trying to obtain the moments of this function via Probability Generating Function (PGF) or Moment Generating Function (MGF) but having problem with it. The pdf is

((Binomial[n, x])^v*\[Theta]^x*(1 - \[Theta])^(n - x))/\!\(\*UnderoverscriptBox[\(\[Sum]\), \(i = 1\), \(n\)]\(\*SuperscriptBox[\((Binomial[n, i])\), \(v\)]*
\*SuperscriptBox[\(\[Theta]\), \(i\)]*\*SuperscriptBox[\((1 - \[Theta])\), \(n - i\)]\)\)

I want to find the derivative of the PGF, with respect to t as expressed below and set t=1: that will give first moment. differentiating the second time and set t=1 gives second moment .......

D[\!\(\*UnderoverscriptBox[\(\[Sum]\), \(x = 1\), \(n\)]\*FractionBox[\(\*SuperscriptBox[\(t\), \(x\)]\ \*SuperscriptBox[\((1 - \[Theta])\), \(n - x\)]\ \*SuperscriptBox[\(\[Theta]\), \(x\)]\ \*SuperscriptBox[\(Binomial[n, x]\), \(v\)]\), \(\*UnderoverscriptBox[\(\[Sum]\), \(i = 1\), \(n\)]\*SuperscriptBox[\((1 - \[Theta])\), \(\(-i\) + n\)]\ \*SuperscriptBox[\(\[Theta]\), \(i\)]\ \*SuperscriptBox[\(Binomial[n, i]\), \(v\)]\)]\), t] 

I document is attached for details if this code is not clear enough. Thanks for your kindness.

Olorire

Attachments:
POSTED BY: Seyi Olorire
Answer
1 month ago

If you set specific integer values for v, you'll get results and be able to see the pattern.

POSTED BY: Jim Baldwin
Answer
1 month ago

Alright, we can choose specific values for v, like -10, -5, 1, 5, 10 and see the pattern. I will also try it here.

Thanks for your ideal.

POSTED BY: Seyi Olorire
Answer
1 month ago

Playing around with this a bit and trying different values of k suggests that the general formula for the i-th moment is given by

m[i_, n_,  k_, \[Theta]_] := - FactorialPower[n, i]^k \[Theta]^i  HypergeometricPFQ[ConstantArray[i - n, k], 
    ConstantArray[i + 1, k - 1], (-1)^(k - 1) \[Theta]/(-1 + \[Theta])]/
    ((i!)^(k - 1) (-1 + \[Theta])^i (-1 + HypergeometricPFQ[ConstantArray[-n, k], ConstantArray[1, k - 1], 
    (-1)^(k - 1) \[Theta]/(-1 + \[Theta])]))
POSTED BY: Jim Baldwin
Answer
1 month ago

Brilliant idea but complicated somehow.

Thanks for your effort.

POSTED BY: Seyi Olorire
Answer
1 month ago

The brute force method is just to write a single function covering all moments. For the k-th momement one could use

moment[k_, n_, v_, \[Theta]_] := (D[\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(x = 1\), \(n\)]
\*FractionBox[\(
\*SuperscriptBox[\(t\), \(x\)]\ 
\*SuperscriptBox[\((1 - \[Theta])\), \(n - x\)]\ 
\*SuperscriptBox[\(\[Theta]\), \(x\)]\ 
\*SuperscriptBox[\(Binomial[n, x]\), \(v\)]\), \(
\*UnderoverscriptBox[\(\[Sum]\), \(i = 1\), \(n\)]
\*SuperscriptBox[\((1 - \[Theta])\), \(\(-i\) + n\)]\ 
\*SuperscriptBox[\(\[Theta]\), \(i\)]\ 
\*SuperscriptBox[\(Binomial[n, i]\), \(v\)]\)]\), {t, k}]) /. t -> 1

which in Mathematica looks like

Function for moments

That can be used to check the function I defined previously.

POSTED BY: Jim Baldwin
Answer
1 month ago

Group Abstract Group Abstract