Thread: Probability Mass Function that goes to infinity

1. Probability Mass Function that goes to infinity

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 $1 or more. 4- Find the probability the expected payment of the bank. 5- Find the variance of the bank. 6- If it costs a quarter to play, should you play? I'm so lost and not sure what kind of infinite sum this is. Any help will be greatly appreciated!!! 2. Originally Posted by Intsecxtanx In a game the bank will pay nothing with probability 0.8 = f(0), and will pay$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 $a$.

3. Thank you for your response, you beautiful mathematician!

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.

4. Originally Posted by Intsecxtanx
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.