Question

So I have a probability function that satisfies the following:

P(0)=P(1) and P(k+1)=(1/k)P(k) for k = 1, 2, 3...

Okay... so I know that P(0)=P(1)=P(2) as well as
P(3)=(1/2)P(2), P(4)=(1/3)P(3), .....

So my question is... how do I find P(k)?

Answer 1

The whole point of this is that I am trying to solve for P(0)...

I know that \[\sum_{i=0}^{\infty}P(i)=1\]
So therefore \[P(0)+\sum_{k=1}^{\infty}P(k) = 1\]

So I really really really need to find that P(k)... pretty please with sugar on top someone...... anyone........ CRICKETS!!!!

Answer 2

Maybe Zale is on a snack break lol.

Answer 3

@dan815

Answer 4

Or if maybe anyone has a strategy on how to work backwards on this type of problem that might at least help a little... answer choices are

A) ln e
B) e - 1
C) (e+1)^-1
D) e^-1
E) (e-1)^-1

Answer 5

\[p(1)=p(0)\]

\[p(2)=\frac{p(1)}{1}=p(0)\]

\[p(3)=\frac{p(2)}{2}=\frac{p(0)}{2!}\]

\[p(4)=\frac{p(3)}{3}=\frac{p(0)}{3!}\]

...
\[p(n)=\frac{p(0)}{(n-1)!}\]

Answer 6

So then \[P(0)+\sum_{k=1}^{\infty}\frac{ P(0) }{ (k-1)! }=1\]

\[P(0)+P(0)\sum_{k=1}^{\infty}\frac{ 1 }{ (k-1)! }=1\]

\[P(0)+P(0)e=1\]

\[P(0)(1+e)=1\]

\[P(0)=\frac{ 1 }{ (1+e) }\]

Answer 7

I understand how you got p(k) by looking at yours but even me staring at the thing for hours I would of never been able to put that together... :/ thank you for the help though. I appreciate it.