Math help!

[Einz]

Ok, I give up on this math problem. Any help would be majorly appreciated.

Sum of:
(p^k)*k*(1-p) where p is a constant from 0 to 1 and k ranges from 0 to infinity

I've already consulted my TI-89, my brain, and my statistics textbook. ATOT is my last resort! Help me!

 

[Ender]

0 to infinity.

 

[Einz]

Originally posted by: Ender
0 to infinity.

Say what now? I'm looking for an analytical solution, like a function in terms of p and k.

 

[RaynorWolfcastle]

expand it out and solve: you'll get a geometric series and another one that's almost geometric but can be solved using the same trick as a geometric series. Looks like your calculating the expectation value of a bernoulli trial.

 

[Brutuskend]

Mandatory response

 

[Einz]

Originally posted by: RaynorWolfcastle
expand it out and solve: you'll get a geometric series and another one that's almost geometric but can be solved using the same trick as a geometric series. Looks like your calculating the expectation value of a bernoulli trial.

I've expanded it out, but the only problem I see is that k varies, so it's not geometric; there is no constant ratio. Any further insights?

 

[Spydermag68]

I could give you the exact equation, but I'll give you this instead.

n = 1, 1 = 1
n = 2, 1 + 2 = 2
n = 3, 1 + 2 + 3 = (1 + 3) + 2 = 4 + 2 = 6
n = 4, 1 + 2 + 3 + 4 = (1+ 4) + (2 + 3) = 5 + 5 = 10
n = m 1 + 2 + 3 + 4 + ... + m = (1 + m) + (2 + m - 1) + (3 + m - 2) ... = 🙂

Hint: You need to factor out something.

 

[TuxDave]

Originally posted by: Spydermag68
I could give you the exact equation, but I'll give you this instead.

n = 1, 1 = 1
n = 2, 1 + 2 = 2
n = 3, 1 + 2 + 3 = (1 + 3) + 2 = 4 + 2 = 6
n = 4, 1 + 2 + 3 + 4 = (1+ 4) + (2 + 3) = 5 + 5 = 10
n = m 1 + 2 + 3 + 4 + ... + m = (1 + m) + (2 + m - 1) + (3 + m - 2) ... = 🙂

Hint: You need to factor out something.

I'm so lost on how that should help solve

p+2p^2+3p^3+4p^4.....

 

[RaynorWolfcastle]

here's the solution:
you have
M = sum(p^k*(1-p)*k, =0..infinity)
M = (1-p)*sum(kp^k)

let's solve the sum alone, let S be the result of the sum, note that for k= 0, the sum is zero, so you can start your sum at 1:
S = sum (kp^k)
S-p*S = sum(k*p^k) - sum(k*p^(k+1))
S(1-p) = lim(k-> infinity) of p^1+2p^2+3p^3+... + kp^k - [p^2 + 2p^3 + 3p^4 + kp^(k+1)]
S(1-p) = lim(k-> infinity) p^1 + p^2+p^3 +... + kp^(k+1)
that last term goes to 0 as k goes to infinity

so you're left with
S(1-p) = sum(p^k, k=1 infinity)
S = p/(1-p)^2

M = (1-p)*p/(1-p)^2 = p/(1-p)

Done like dinner.

 

[TuxDave]

Here's the basis of my solution. So if you ignore the (1-p) thing since it's a constant, you essentially need to solve the summation of.

Y=p+2p^2+3p^3+4p^4...... which equals

Y=p+p^2+p^3+p^4.... +p^2+2p^3+3p^4..... which equals

Y=p+p^2+p^3+p^4.... + p*(p+2p^2+3p^3+4p^4...) which equals

Y=p+p^2+p^3+p^4.... + p*Y

You can figure out the rest....

*Sigh....4 minutes too slow... 🙁

 

[Spydermag68]

n(n+1) / 2

n = 1, 1(1+1) / 2 = 1
n = 5, 5 (5+1) / 2 = 20 / 2 = 15

 

[ActuaryTm]

Should really make use of the IDIC* theorem.

* - Infinite Diversity in Infinite Combinations

 

[TuxDave]

Originally posted by: Spydermag68
n(n+1) / 2

n = 1, 1(1+1) / 2 = 1
n = 5, 5 (5+1) / 2 = 20 / 2 = 15

I think Gauss demonstrated that equation? But the OP's summation is neither a simple geometric nor arithmetic progression.