Probability Mass Function that goes to infinity

Intsecxtanx · #1 $Old$ September 26th, 2009, 06:50 PM

In a game the bank will pay nothing with probability 0.8 = f(0), and will pay $n with p.m.f. (probability mass function) f(n) = af(n-1)/n for n=1,2,3... (infinity)

1- find the constant a.
2- Find the probability, that the bank pays $1.
3- Find the probability the bank pays $1 or more.
4- Find the probability the expected payment of the bank.
5- Find the variance of the bank.
6- If it costs a quarter to play, should you play?

I'm so lost and not sure what kind of infinite sum this is. Any help will be greatly appreciated!!!

mr fantastic · #2 $Old$ September 27th, 2009, 02:49 AM

Quote:

Originally Posted by Intsecxtanx $View Post$

In a game the bank will pay nothing with probability 0.8 = f(0), and will pay $n with p.m.f. (probability mass function) f(n) = af(n-1)/n for n=1,2,3... (infinity)

1- find the constant a.
2- Find the probability, that the bank pays $1.
3- Find the probability the bank pays $1 or more.
4- Find the probability the expected payment of the bank.
5- Find the variance of the bank.
6- If it costs a quarter to play, should you play?

I'm so lost and not sure what kind of infinite sum this is. Any help will be greatly appreciated!!!

I will do 1. The rest are left for you to do (or at least to show some working and then give a clear statement of where you're still stuck).

$f(0) = 0.8$

$f(1) = \frac{a f(0)}{1} = \frac{0.8 a}{1}$

$f(2) = \frac{a f(1)}{2} = \frac{0.8 a^2}{1\times 2} = \frac{0.8 a^2}{2!}$

$f(3) = \frac{a f(2)}{2} = \frac{0.8 a^3}{3 \times 2!} = \frac{0.8 a^3}{3!}$ .

The pattern is clear: $f(n) = \frac{0.8 a^n}{n!}$ .

Therefore you require $1 = f(0) + f(1) + f(2) + .... + f(n) + .... = 0.8 \sum_{i=0}^{\infty}\frac{a^n}{n!} = 0.8 e^a$ .

Solve for $a$ .

Intsecxtanx · #3 $Old$ September 27th, 2009, 10:34 AM

Haha, this was fantastic. It helped so much. I worked out the whole problem now and found E[X] to be the same as the constant a= ln(5/4). However, I can't seem to figure out how to find Var[X] because I'm not sure how to square X? Thank you for helping me out by the way. I really appreciate it.

mr fantastic · #4 $Old$ September 27th, 2009, 02:03 PM

Quote:

Originally Posted by Intsecxtanx $View Post$

Haha, this was fantastic. It helped so much. I worked out the whole problem now and found E[X] to be the same as the constant a= ln(5/4). However, I can't seem to figure out how to find Var[X] because I'm not sure how to square X? Thank you for helping me out by the way. I really appreciate it.

$n = 0 \Rightarrow n^2 = 0$ .

$n = 1 \Rightarrow n^2 = 1$ .

$n = 2 \Rightarrow n^2 = 4$ .

$n = 3 \Rightarrow n^2 = 9$ .

etc.