# Asymptotic behaviour of max(Z_n)

More on high variance strategies.

I was wondering what the asymptotic behaviour of $$\max(Z_n)$$ was. It seems “obvious” that it should grow without bound, but it turns out that it grows really very slowly. It turns out that for $$n = 10^{10}$$ it’s still less than 7.

I thought I would quickly verify that it does actually grow without bound.

Let $$T_n = \max Z_n$$. Then $$E(T_n) = E(T_n | T_n < 0) P(T_n < 0) + E(T_n | T_n > 0) P(T_n > 0)$$. But $$E(T_n | T_n < 0) \geq -1$$ (because $$E(Z_1 | Z_1 < 0) = -1$$ and $$T_n \geq Z_1$$\) and $$P(T_n < 0) = 2^{-n}$$, so $$E(T_n) \geq (1 – 2^{-n}) E(T_n|T_n > 0) – 2^{-n}$$. So we need only concern ourselves with the positive behaviour.

Let $$S_n$$ be a random variable with the distribution of $$T_n | T_n > 0$$.

Consider $$t > 0$$. We want to show that for sufficiently large $$n$$, $$E(S_n) > t$$.

Now for any $$s$$, $$E(S_n) \geq s P(S_n \geq s)$$, because $$S_n$$ is always positive. So let $$s = 2t$$. Now $$E(S_n) \geq 2t P(S_n \geq 2t)$$.

But $$P(S_n \geq s) \geq P(T_n \geq s) = 1 – F(s)^n$$ where $$F$$ is the cdf of the standard normal distribution. So $$E(S_n) \geq 2t (1 – F(2t)^n)$$.

But if $$n \geq \frac{\log(\frac{1}{2})}{\log(F(2t))}$$ then $$F(2t)^n \leq \frac{1}{2}$$ and so $$E(S_n) \geq 2t (1 – \frac{1}{2}) = t$$.

Thus $$E(S_n)$$ grows without bound and thus so does $$E(T_n)$$.

This could be tightened up a bit to get the asymptotic behaviour of the growth, but I’m not that clear on what the asymptotic behaviour of $$F(t)$$ is so I haven’t worked through the details yet.

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