Limits- legal and illegal algebraic manipulation.

Question

I am doing probability and am using the strong law of large numbers to get after a bit of irrelevant extra algebra which I wont mention $\frac{\log(C_n)}{n} \rightarrow K$ where $C_i$ are a sequence of random variables, $K$ is some constant and $n$ can only take natural number values. I want to do the following:

$\Rightarrow$ $\log(C_n) \rightarrow nK$

$\Rightarrow$ $(C_n) \rightarrow e^{nK}$

This is definitely not allowed I know- but I'm trying to figure out how $C_n$ behaves and I don't know what to do??

Answer 1

While you can't say that $\log(C_n)\to nK$ and $C_n \to e^{nK}$, your intuition is right -- you can make it precise.

The idea behind your assertions makes sense. If $\log(C_n)/n$ approaches $K$, then for large $n$, we should have $\log(C_n)$ approximately equal to $nK$. Thus we should have $\log(C_n)$ approximately equal to $e^{nK}$.

More precisely, what does this mean? It means that there is a function $\epsilon(n)$ such that $\epsilon(n)/n \to 0$ and $$ K - \epsilon(n)/n \leq \log(C_n)/n \leq K + \epsilon(n)/n. $$ Now we can do algebra all day long. $$ Kn - \epsilon(n) \leq \log(C_n) \leq nK + \epsilon(n) $$ and exponentiating (which is monotone) $$ e^{Kn}e^{-\epsilon(n)} \leq C_n \leq e^{Kn}e^{\epsilon(n)} $$

This actually gives more information than you asked for. It tells you not just that $C_n$ behaves like $e^{nK}$ for large $n$, it tells you how quickly $C_n$ tends to behave like that.

Answer 2

If $\frac{\log(C_n)}{n} \rightarrow K $, then $|\frac{\log(C_n)}{n} - K| < \epsilon $ for all large enough $n$.

Multiplying by $n$, $|\log(C_n) - nK| < n\epsilon $, or $-n\epsilon <\log(C_n) - nK < n\epsilon $.

Exponentiating, $e^{-n\epsilon} <C_ne^{- nK} < e^{n\epsilon} $, or $e^{nK-n\epsilon} <C_n < e^{nK+n\epsilon} $, or $e^{n(K-\epsilon)} <C_n < e^{n(K+\epsilon)} $, and that is about all you can do.

It would be nice is you could first choose $n$ and then make $\epsilon$ small, but it does not work that way. You first choose $\epsilon$ and then you find the values of $n$ for which the inequality holds.