Expected Value of Excellent Wine Age I Ran into this question and I can't find the right way to approach it.
We have $n$ different wine bootles numbered $i=1...n$. the first is 1 year old, the second is 2 years old ... the $n$'th bottle is $n$ years old.
Each bottle is still good at probability of $1/i$.
We pick out a random bottle and it is good. what is the expected value of the age of the bottle?
I'm really not sure what the random varable here is and how to aproach the question. I'd be grateful for a lead.
Thanks,
Yaron.
 A: You are asked to compute a conditional expectation (...'given that the bottle is good') for which you need a conditional probability distribution. Let $I$ be the age of the bottle you picked out, and recall that this variable take values over the range $i = 1...n$ By definition
$E[I\,|\, I\mbox{ is good}] = \sum_{i=1}^n i\, P[I=i\,|\, I\mbox{ is good}]$
Now you just need this conditional probability,which can be obtained from Bayes' Rule $P(A|B) = P(A, B)/P(B) $ (which you should read as 'the probability of A given B is the probability that both A and B occur, divided by the probability that B occurs):
$P[I=i\,|\, I\mbox{ is good}] = \frac{P[I = i \,, I \mbox{ is good}]}{P[I\mbox{ is good}]} = \frac{(1/n)(1/i)}{\sum_{j=1}^n (1/n)(1/j)}$
where the numerator is the product of the probability that a particular $i$ is chosen ($1/n$, since there are n bottles) with the probability $1/i$ that it is still good. The denominator is the probability that $I$ is good, but since $I$ can be any bottle, it is the sum over all possible probabilities appearing in the numerator.
Combining these results gives the desired solution:
$E[I\,|\, I\mbox{ is good}] = \frac{\sum_{i=1}^n i (1/n)(1/i)}{\sum_{j=1}^n (1/n)(1/j)} = \frac{1}{(1/n)\sum_{j=1}^n(1/j)} = \frac{n}{\sum_{j=1}^n(1/j)}$
Note that the denominator is just the harmonic series, which scales like $\ln(n)$
A: You are interested in the age of the bottle (the random variable), conditioned on it being good. 
The probability the bottle selected is of age $i$ given it is good is $\dfrac{\frac{1}{i}}{\sum_1^n \frac{1}{j}} = \dfrac{1}{iH(n)}$ where $H(n)$ is a harmonic number.
So the expected age given it is good is $\displaystyle\sum_1^n i\frac{1}{iH(n)} = \dfrac{n}{H(n)}.$ 
