# A mean that doesn't exist

• Apr 17th 2010, 10:54 PM
rebghb
A mean that doesn't exist
Hello everyone, I've came across this problem and I need to see if my approach is right...

An urn initially contains one Black and one White ball. At each stage, a ball is randomly chosen and then replaced along with another of the same color. Let $\displaystyle X$ denote the selection number of the 1st Black ball chosen.

(a)Find the pdf of $\displaystyle X$.

(b)Show that the $\displaystyle E(X)$ doesn't exist for $\displaystyle X$.

for (a), I thought that if I came accross a black ball I'll replace it so
$\displaystyle P(X=1)=\frac{1}{2},\ P(X=2)=(\frac{1}{2})(\frac{1}{2})$ and for any $\displaystyle k,\ P(X=k)={(\frac{1}{2})}^k$.

As for (b), $\displaystyle E(X)=\sum_k k{(\frac{1}{2})}^k$ which is 2!!
How could this be?!

Any help is very much appreciated!
• Apr 17th 2010, 11:14 PM
CaptainBlack
Quote:

Originally Posted by rebghb
Hello everyone, I've came across this problem and I need to see if my approach is right...

An urn initially contains one Black and one White ball. At each stage, a ball is randomly chosen and then replaced along with another of the same color. Let $\displaystyle X$ denote the selection number of the 1st Black ball chosen.

(a)Find the pdf of $\displaystyle X$.

(b)Show that the $\displaystyle E(X)$ doesn't exist for $\displaystyle X$.

for (a), I thought that if I came accross a black ball I'll replace it so
$\displaystyle P(X=1)=\frac{1}{2},\ P(X=2)=(\frac{1}{2})(\frac{1}{2})$ and for any $\displaystyle k,\ P(X=k)={(\frac{1}{2})}^k$.

As for (b), $\displaystyle E(X)=\sum_k k{(\frac{1}{2})}^k$ which is 2!!
How could this be?!

Any help is very much appreciated!

The probability that the second ball is B given the first is W is $\displaystyle 1/3$ not $\displaystyle 1/4$

In general the probability that the first B ball is seen on the $\displaystyle N$-th selection is:

$\displaystyle P(N)=\frac{1}{N(N-1)}$

CB
• Apr 17th 2010, 11:19 PM
rebghb
Okay but noone implied that there is a conditional probability... The probavility of getting a black ball is the same as that of getting a white ball which is 1/2...
What about the 1st selection, we would have a pole!!
• Apr 18th 2010, 12:22 AM
CaptainBlack
Quote:

Originally Posted by rebghb
Okay but noone implied that there is a conditional probability... The probavility of getting a black ball is the same as that of getting a white ball which is 1/2...

No it is not, you are adding white balls to the urn until you get a black ball, so the probabilities change for every draw. For the N-th draw to be the first B requires that the previous N-1 were all W.

Quote:

What about the 1st selection, we would have a pole!!
it is only valid for N>1, what I wrote was the solution for the recurrence implied in the problem with initial condition p(1)=1/2 (or rather the initial state being (1,1)).

If the first draw is W the urn contains 2W and 1B for the second draw so the probability of B on the second draw given that the first was W is 1/3. So the probability that the first B occurs on the 2-nd draw is (1/2)(1/3)

CB
• Apr 19th 2010, 02:35 AM
rebghb
I would like to start by thanking you (oh I will press the "thanks" button), BUT!...

This is my post:
Quote:

An urn initially contains one Black and one White ball. At each stage, a ball is randomly chosen and then replaced along with another of the same color. Let http://www.mathhelpforum.com/math-he...2dc6b383-1.gif denote the selection number of the 1st Black ball chosen.
This is how you interpreted the problem:
Quote:

you are adding white balls to the urn until you get a black ball, so the probabilities change for every draw. For the N-th draw to be the first B requires that the previous N-1 were all W.
I don't get how you translated what I said (which was completely vague) into something interesting... Thanks though and you can consider it a closed thread...