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    (Original post by HeavisideDelts)
    Problem 479 **

    A die is thrown until an even number appears. What is the expected value of the sum of all the scores?
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    Let mean score be E; then E = \frac{1}{6} (2+4+6) + \frac{1}{6} (1+E)+\frac{1}{6}(3+E)+\frac{1}{  6}(5+E), since with probability 1/6 each we add 2 or 4 or 6 to the score and then terminate, while with probability 1/6 each we add 1 or 3 or 5 and then repeat. That is, E = 2 + \frac{3}{2} + \frac{3 E}{6}, so \frac{1}{2} E = \frac{7}{2} and E is 7.
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    (Original post by HeavisideDelts)
    Problem 479 **

    A die is thrown until an even number appears. What is the expected value of the sum of all the scores?
    I reached the same conclusion as Smaug, but used a slightly different method (or at least, imo, a cleaner looking version of his)
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    So, we shall consider three things: First, given an even number appears the expected value, which we shall shall E_{even}; second, given an odd number appears the expected value, which we shall denote E_{odd}; and finally the expected number of odd rolls before we get an even, which we shall denote E_{roll}.

    The expected sum, denoted by E is given by : E=E_{even}+E_{roll}E_{odd}.

    E_{even} is given by: \frac{2}{3}+\frac{4}{3}+\frac{6}  {3}=4
    E_{odd} is given by: \frac{1}{3}+\frac{3}{3}+\frac{5}  {3}=3
    Finally, E_{roll} is given by: \sum\limits_{i=1}^{\infty} 2^{-i}=1

    So, finally, substituting these values into the formula for E gives: E=4+1\cdot3=7


    I might add, I'm not quite sure why it's a ** question instead of a *.
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    (Original post by Smaug123)
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    Let mean score be E; then E = \frac{1}{6} (2+4+6) + \frac{1}{6} (1+E)+\frac{1}{6}(3+E)+\frac{1}{  6}(5+E), since with probability 1/6 each we add 2 or 4 or 6 to the score and then terminate, while with probability 1/6 each we add 1 or 3 or 5 and then repeat. That is, E = 2 + \frac{3}{2} + \frac{3 E}{6}, so \frac{1}{2} E = \frac{7}{2} and E is 7.
    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.
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    (Original post by Jammy Duel)
    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
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    (Original post by ThatPerson)
    Last time I checked  \displaystyle 6 \times \dfrac{3}{2} = 9
    My mistake, read it the other way up
    Doesn't change that it wouldn't appear to cover all cases.
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    (Original post by Jammy Duel)
    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.
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    (Original post by DJMayes)
    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.
    Fair dos, can't really say I paid much attention in stats (when I actually turned up) though so when covered it all went in one ear and out the other
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    (Original post by DJMayes)
    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.
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    (Original post by Smaug123)
    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.
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    (Original post by Jammy Duel)
    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.)
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    (Original post by Smaug123)
    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.
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    (Original post by Jammy Duel)
    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.
    Fair enough - to be honest, I considered it to be so trivial that it didn't merit explaining. In a similar way, I'd use sequential compactness interchangeably with compactness in a metric space, without bothering to state the (suitably generalised) Bolzano-Weierstrass theorem and its converse. I felt it was one of those instances where spelling out the rigour would aesthetically mar the solution. (I may, of course, have been wrong.)

    ETA: That's not to say I didn't like the puzzle! It was fun to solve, and I thought the argument was quite pleasing.
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    (Original post by Smaug123)
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    Let mean score be E; then E = \frac{1}{6} (2+4+6) + \frac{1}{6} (1+E)+\frac{1}{6}(3+E)+\frac{1}{  6}(5+E), since with probability 1/6 each we add 2 or 4 or 6 to the score and then terminate, while with probability 1/6 each we add 1 or 3 or 5 and then repeat. That is, E = 2 + \frac{3}{2} + \frac{3 E}{6}, so \frac{1}{2} E = \frac{7}{2} and E is 7.
    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).
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    (Original post by HeavisideDelts)
    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).
    It is finite: the game ends with probability 1/2 at each step, so it ends almost surely :P but yes, good spot.
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    Problem 480 ** (not much probability theory in A-Level)

    Let  X_1, X_2,...,X_n be i.i.d r.v's with uniform distribution on  [1,2] .

    Determine a unique value,  \Delta , such that \lim_{n\to \infty} Pr(a <(X_1X_2...X_n)^{1/n} < b) = 1 iff  a< \Delta <b
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    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
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    (Original post by Arieisit)
    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
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    The partial sums are \displaystyle \sum_{n \geq 1}^{N} \log(n+1) - \log(n). But that telescopes to \log(N+1) - \log(1) = \log(N+1); this diverges. Hence the infinite sum is divergent.

    However, \displaystyle \lim_{n \rightarrow \infty} \ln(\dfrac{n+1}{n}) = \lim_{n \to \infty} \log(1+\frac{1}{n}); by continuity of \log, this is \log(1+\lim_{n \to \infty} \frac{1}{n}) = \log(1) = 0.
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    (Original post by PAPADAPADOPOLOUS)
    If we know that pie is irrational.

    how do I prove via contradiction that pie ^2 or pie^3 is also irrational?
    You cannot deduce that just from pi being irrational (e.g. square root of 2 is irrational but rational when squared). You could use it being transendental though.
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    (Original post by PAPADAPADOPOLOUS)
    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}?
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    (Original post by Smaug123)
    Suppose not: both \pi^2, \pi^3 is rational. Then what can you say about \frac{\pi^3}{\pi^2}?
    Surely if \pi^2 is rational then \pi^3 is irrational?
 
 
 
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