The Proof is Trivial!

Problem 479 **

A die is thrown until an even number appears. What is the expected value of the sum of all the scores?

Problem 479 **

A die is thrown until an even number appears. What is the expected value of the sum of all the scores?

Your response doesn't appear particularly rigorous to me, it, at best, only seems to consider the cases where the first or second roll is even. Also, your value for E isn't consistent throughout your work. While the first and last equations will give E=7, the middle one doesn't:
$E = 2 + \frac{3}{2} + \frac{3 E}{6}\Rightarrow 6E=12+18+3E\Rightarrow 3E=30\Rightarrow E=10\not=7.$

Last time I checked $\displaystyle 6 \times \dfrac{3}{2} = 9$

Your response doesn't appear particularly rigorous to me, it, at best, only seems to consider the cases where the first or second roll is even. Also, your value for E isn't consistent throughout your work. While the first and last equations will give E=7, the middle one doesn't:
$E = 2 + \frac{3}{2} + \frac{3 E}{6}\Rightarrow 6E=12+18+3E\Rightarrow 3E=30\Rightarrow E=10\not=7.$

His response is perfectly rigorous. It is using the law of total expectation (you may know it as the tower property of conditional expectation, as that is what I was lectured it as):

http://en.wikipedia.org/wiki/Law_of_total_expectation

In this case the partitioning events he is conditioning on are the outcome of the first roll, which is all that's necessary. His argument is actually much cleaner (as he instantly produces a simple linear equation without having to define 3 other things) and much more obviously rigorous than your own.

His response is perfectly rigorous. It is using the law of total expectation (you may know it as the tower property of conditional expectation, as that is what I was lectured it as):

http://en.wikipedia.org/wiki/Law_of_total_expectation

In this case the partitioning events he is conditioning on are the outcome of the first roll, which is all that's necessary. His argument is actually much cleaner (as he instantly produces a simple linear equation without having to define 3 other things) and much more obviously rigorous than your own.

Thanks I thought mine was cleaner, personally, but I suppose it's a matter of taste. For one thing, I didn't have to bother summing any series.

ETA: I learnt it as "iterated expectation", but in my mind it's really just taking expectations on both sides of the Law of Total Probability, rather than being a separate theorem in its own right.

Well, if there is one advantage to mine is that if you haven't done much stats (or didn't pay attention to it) it's a lot easier to follow since they only mildly complicated bit is knowing that $\sum\limits_{i=1}^{\infty}2^{-i}=1$. And in many respects it's not even that complicated in that it's a well known result (or at least from 0 to infinity) and beyond that it's not exactly hard to understand why.

Oh, of course the sum's an easy one to evaluate I probably just didn't write out my answer very fully, so it's perhaps less comprehensible. (The key idea in it is really simple, though: if we get 1,3,5 then we proceed just as before, the expectation of the game being unchanged except for having 1,3,5 appended to it, while if we get 2,4,6 we stop with 2,4,6 appended to the current tally.)

It was rather difficult to follow, all being in one long string didn't help at all, more line breaks and a bit more commentary would definitely have made it easier to follow, and would have highlighted what that you were actually using that law.

Nice, thats the route i took However, there is the small issue that we have implicitly assumed E is a finite number which we do not necessarily know, though it is intuitively obvious (but so is a lot of mathematics haha).

Problem 481*/**

Show that the series

$\displaystyle\sum _{n\geq1} \ln(\dfrac{n+1}{n})$

Is divergent, even though

$\displaystyle \lim_{n \rightarrow \infty} \ln(\dfrac{n+1}{n}) = 0$

If we know that pie is irrational.

how do I prove via contradiction that pie ^2 or pie^3 is also irrational?

If we know that pie is irrational.

how do I prove via contradiction that pie ^2 or pie^3 is also irrational?

Suppose not: both $\pi^2, \pi^3$ is rational. Then what can you say about $\frac{\pi^3}{\pi^2}$?