Probability?

Question

Using geometric distribution p(n)=q^(n-1)p
with p=1/10
Calculate E(n^3)
If you can, please show some of the last steps.
Thank you very much.

I've been working on this problem and come up with this:
e(n^3)=e[n(n-1)(n-2)]+3e(n^2)-2e(n)

e[(n(n-1)(n-2)]=6q/p^3
e(n^2)=2q/p^2+1/p
e(n)=1/p
I came up with 4870, which is not the right answer (5410, not so sure if this is right either).

Answer 1

I would strongly recommend you to use either probability generating functions (pgf) or moment generating functions (mgf). Otherwise it is really nasty (you need to differentiate the infinite series of p(n) three times and do some technical tricks).

The mgf of a discrete random variable X, which can only take on positive integer values, is defined as
m(t) = sum(n=1..infinity) of exp(t*n) * p(n).

For the geometric distribution, it becomes
sum(n=1..infinity) of exp(t*n)*q^(n-1)*p =
= exp(t)*p* sum(n=1..infinity) of (exp(t)*q)^(n-1) =
= exp(t)*p*1/(1-exp(t)*q). (By using the formula for the sum of the geometric series.)

From the mgf, you get E(X^r) by differentiating m(t) r times and substituting t=0 into the r-th derivative.

In our case, you need to differentiate m(t) 3 times to get
something rather nasty and then plug in t=0 to arrive at:
(p^2-6*p+6)/p^3. (I substituted 1-p=q because it looked better that way.)
For p=1/10, it gives 5410.

I did it in Maple. Ask me if you still don't understand it or you'd like me to explain it without using moment generating functions.

ADDED: The problem with your solution is that
E[n*(n-1)*(n-2)] = 6*q^2/p^3 and not 6*q/p^3. Check your solution again. It must be some minor miscalculation.

Answer 2

I thought that it might not converge but I just made a speadsheet to check and evidently it does.

E(n^3) = sum k=1 to inf { k^3q^(k-1)p}

For your p=1/10 here are some values:
(C&Ped from a spreadsheet so spacing is bad)

nn^3n^3*q^(n-1)n^3*q^(n-1)*p
1110.1
287.20.72
32721.872.187
46446.6564.6656
512582.01258.20125
6216127.5458412.754584
7343182.28426318.2284263
8512244.888012824.48880128
9729313.810596131.38105961
101000387.42048938.7420489
111331464.091003846.40910038
121728542.2647154.226471

It peaks out around n=28, for large n:
the left column is n, starting at 32 and increasing by a factor of 2 each row:

32327681250.166127125.0166127
64262144343.414016234.34140162
12820971523.239131710.323913171
256167772163.60213E-053.60213E-06
5121342177285.56842E-165.56842E-17
102410737418241.66336E-381.66336E-39
204885899345921.85527E-841.85527E-85
4096687194767362.8851E-1772.8851E-178
81925.49756E+1100
163844.39805E+1200
327683.51844E+1300
655362.81475E+1400
1310722.2518E+1500
2621441.80144E+1600
5242881.44115E+1700
10485761.15292E+1800
20971529.22337E+1800
41943047.3787E+1900

The zeros indicate underflow. So, one of the necessary conditions for convergence is satisfied.

I don't recall how to do that summation.

Answer 3

I actually saw that you added this to a question you previously asked and I answered. Here, again, is my response. I do believe that 5410 is correct.

The skewness of a geometric is

(2-p)/sqrt(1-p)

So applying that to this,

skewness = 1.9/sqrt(0.9)

skewness is calculated as E(n-mu)^3/Var(n)^(3/2)

so

E(n-mu)^3 = E(n^3 - 3*mu*n^2 + 3*n*mu^2 - mu^3)
= E(n^3) - 3*mu*E(n^2) + 3*mu^2*E(n) - mu^3
= E(n^3) - 3*10*(90+10^2) + 3*10^3 - 10^3
= E(n^3) - 5700 + 3000 - 1000
= E(n^3) - 3700

Var(n)^(3/2) = 90^(3/2)
= sqrt(729000)
= 270*sqrt(10)

So

1.9/sqrt(0.9) = (E(n^3) - 3700)/(270sqrt(10))

E(n^3) - 3700 = 1710

E(n^3) = 5410