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Q&A

How does $\lim\limits_{n \to \infty} \dfrac{p[s + (1 - s)c^n]}{p[ s + (1 - s)c^n] + (1 - p)[s + (1 - s)w^n]} = p?$

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To minimize this post's length, I don't repeat the exercise itself. The title refers to the sentence beside my red line below.

I divide both the numerator and denominator by $\color{red}{c^n}$. Then $P(G|U) = \dfrac{p[ \dfrac{s}{\color{red}{c^n}} + (1 - s)]}{p[ \dfrac{s}{\color{red}{c^n}} + (1 - s)] + \dfrac{s(1 - p)}{\color{red}{c^n}} + (1-p)(1-s)\dfrac{w^n}{\color{red}{c^n}}} $.

The question postulates $0 < w < 0.5 < c < 1$, but this $\iff 0 < \dfrac{w^n}{c^n} < \dfrac{0.5^n}{c^n} < 1 < \dfrac{1}{c^n} $. So if $n \to \infty$, then all terms above containing ${\color{red}{c^n}} \to 0$.

Then $\lim\limits_{n \to \infty} P(G|U) =\dfrac{p(1 - s)}{p(1 - s)} = 1 \neq p$. What did I flub?

Image alt text

Blitzstein, Introduction to Probability (2019 2 ed), Example 2.4.5, p 62.

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So if $n \to \infty$, then all terms above containing $c^n \to 0$.

This is false. It's true that as $n \to \infty$, $c^n \to 0$ (for $0 < c < 1$). But a term like $\frac{s}{c^n}$ will diverge instead of converging to $0$.

You haven't done yourself any favors by dividing everything by $c^n$. Just use the fact that $c^n \to 0$ (and similarly $w^n \to 0$) with the original expression, go slowly, and you should get the right answer.

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Thanks. Can you pls elaborate why terms like $s/c^n$ will diverge instead of converging to $0$? Ca... (2 comments)

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