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Comments on In $w_{k + 1} - w_k = (\frac{1 - p}{p})^{exponent}(w_1 - w_0)$, why isn't exponent $k + 1$?

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In $w_{k + 1} - w_k = (\frac{1 - p}{p})^{exponent}(w_1 - w_0)$, why isn't exponent $k + 1$?

+1
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Please see the $r^k$ underlined in red, which is $(\frac{1 - p}{p})^k$ as defined by the green underlines.

  1. How do you deduce that the exponent must be $k$? Why isn't the exponent $k + 1$?

  2. Is this question related to the Fence Post Error? Have I committed it?

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Tsitsiklis, Introduction to Probability (2008 2e), p 63.

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If you set $k=0$ the equation becomes $$w_{0+1} - w_0 = r^{\textrm{exponent}}(w_1-w_0)$$ For this equation to hold (in general) the exponent must be zero and not one. Thus an exponent of $k+1$ must be wrong.

I think the problem is not a fence-post problem but the proper base case for the induction. Try to write down the step of going from $$w_{k+1} - w_k = r (w_k-w_{k-1})$$ to $$w_{k+1} - w_k = r^k (w_1-w_0)$$ as a proper proof by induction.

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"For this equation to hold (in general) the exponent must be k = 0 and not k + 1 = 1". What do you me... (3 comments)
"For this equation to hold (in general) the exponent must be k = 0 and not k + 1 = 1". What do you me...
DNB‭ wrote over 3 years ago

"For this equation to hold (in general) the exponent must be k = 0 and not k + 1 = 1". What do you mean These are equivalent. $k = 0 \iff k + 1 = 1$.

JRN‭ wrote over 3 years ago

leovt‭ means the exponent must be zero, so it should be k (which is equal to zero) and it should not be k+1 (which is equal to one).

leovt‭ wrote over 3 years ago

JRN‭l yes exactly, thanks. I hope the edited answer is clearer now.