I was reviewing over my notes and couldn't understand where the underline portion from attached note comes from and why it is a sum from k = 0 to x-1. This is with respect to a Gambler's ruin game with end points 0 and N and g(x) is our expected duration of the game.
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Excuse me, where are the attached notes in this page? – IRO Sep 06 '19 at 09:22
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Write out the summation, substituting all values of $k$ from $k=0$ to $k=x-1$. You'll obtain a telescoping sum, where all terms cancel out except the first and last.
grand_chat
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I understood that part, but I do not understand the initial argument that g(x) equates to the summation of differences. – Mid Feb 18 '15 at 23:24
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The telescoping trick is a device used to take advantage of the previously derived observation that $g(x+1) - g(x) = c - 2x $ for all $x$. – grand_chat Feb 18 '15 at 23:29
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I understand that part, but how would I know that g(x) = the sum of g(k+1)-g(k) from 0 to x-1 to use that trick. – Mid Feb 18 '15 at 23:35
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The telescoping trick is a useful way of deducing $a_n$ when it is easy to calculate $a_k - a_{k-1}$ for each $k$. It's not always possible to do things this way; it's often a matter of luck whether telescoping is possible. Nonetheless, it's a nice device to have in your "bag of mathematical tricks". – grand_chat Feb 18 '15 at 23:40
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What conditions makes the trick applicable to finding the expected duration of a game? I still do not see that we should know to have g(x) equate to a sum, I understand how the middle terms cancel. – Mid Feb 18 '15 at 23:48
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If I understand the notation correctly, $g(x) := E_x(V_0\wedge V_n)$, i.e. $g(x)$ is the expected duration of the game if your starting fortune is $x$. The system of equations (1) and (2) have to be solved to obtain $g(x)$. It happens that equation (2) can be rearranged into a form involving the increment $g(x+1)-g(x)$, which leads to the telescoping sum. – grand_chat Feb 19 '15 at 00:15