# Relaxing some assumptions from the “high variance strategies” post

Advance warning: This is a very boring post.

In my last post I outlined a ludicrously over-simplified model for why you might want to consider high variance strategies.

I’ve been thinking over some of the modelling assumptions and wondering whether it could be made a bit less over-simplified. The only ones that are obviously easy to weaken are the assumptions on the distribution shape.

Here’s an example that shows you need some assumptions on the distribution shape. Consider a standard distribution $$Z$$ with $$P(Z = 1) = P(Z = -1) = \frac{1}{2}$$ and suppose we can choose strategies of the form $$\mu + \sigma Z$$. Note that $$E(Z) = 0$$ and $$\mathrm{Var}(Z) = 1$$ so these really are the mean and standard deviation of our distributions.

But $$E(\max\limits_{1 \leq k \leq n} Z_i) = 1 (1 – 2^{-n}) – 2^{-n} = 1 – 2^{1 – n}$$ (because the maximum takes the value $$-1$$ only if all of the individual values are $$-1$$, which happens with probability $$2^{-n}$$). So $$E(\max\limits_{1 \leq k \leq n} \mu + \sigma Z_i) = \mu + (1 – 2^{1 – n} \sigma$$. $$1 – 2^{1 – n} < 1$$, so you’re always better off raising $$\mu$$ rather than $$\sigma$$.

The interesting feature of this example is that $$P(X_k \leq \mu + \sigma) = 1$$. If this happens then it will always be the case that $$E(\mathrm{max}(X_k) ) \leq \mu + \sigma$$ so there’s no real benefit to raising $$\sigma$$ instead of $$\mu$$ (note: It’s conceivable that there’s some complicated dependency on $$\mu$$ as a parameter, but I’m just going to assume that $$\mu$$ is purely positional and not worry about that).

You only need to go slightly beyond that to show that for some sufficiently large group you’ll always eventually be better off raising $$\sigma$$ rather than $$\mu$$.

Suppose all our strategies are drawn from some distribution $$X = \mu + Z^\sigma$$ with $$E(Z^\sigma) = 0$$. The only dependency on $$\sigma$$ that we care about is that $$P(Z^\sigma \geq (1 + \epsilon)\sigma \geq p$$. for fixed $$\epsilon > 0, 0 < p < 1$$ and all $$\sigma > 0$$ (this is trivially satisfied by the normal distribution for example).

Then we have $$E(\max\limits_{1 \leq k \leq n} X_n) = \mu +E(\max\limits_{1 \leq k \leq n} Z^\sigma_n)$$.

So we now just want to find some lower bounds on $$T_n = E(\max\limits_{1 \leq k \leq n} Z^\sigma_n)$$. We’ll split this up as three variables. Let $$T_n = U_n + V_n + W_n$$ where $$U_n = T_n \mathbb{1}_{T_n \leq 0}$$, $$V_n = T_n \mathbb{1}_{0 < T_n < (1 + \epsilon) \sigma }$$ and $$W_n = T_n \mathbb{1}_{(1 + \epsilon) \sigma \leq T_n }$$.

Because $$U_n \geq 0$$ and $$W_n \leq (1 + \epsilon) \sigma$$ this gives us the lower bound  $$E(T_n) \geq E(U_n) + (1 + \epsilon)\sigma P(W_n \geq (1 + \epsilon) \sigma) \geq E(U_n) + (1 + \epsilon)\sigma (1 – p)^n$$.

We now just need to bound $$U_n$$ below. But $$U_n \geq U_1 \mathbb{1}_{T_k \leq 0, k \geq 2}$$. But these two random variables are independent  so $$E(U_n) \geq E(U_1) P(Z \leq 0)^{n – 1}$$. Therefore $$E(T_n) \geq + (1 + \epsilon)\sigma (1 – p)^n$$

This lower bound lets us show a much less pretty version of our last result:

Given a strategy $$\mu, \sigma$$ being employed by $$n$$ people, and given some increase $$a$$ which could go to either $$\mu$$ or $$\sigma$$ there exists some sufficiently large $$m$$ such that for $$m$$ people, changing the strategy to $$\mu, \sigma + a$$ would beat changing the strategy to $$\mu + a, \sigma$$.

Yeah, that phrasing is kinda gross to me too.

Note though that if we go back to the previous case where $$\sigma$$ is just a scaling parameter and are just dropping the normality strategy, we can use our lower bound on $$E(T_n)$$ to find some $$n$$ for which $$E(T_n|\sigma = 1) > 1$$ and for all $$m \geq n$$ it will be beneficial to increase $$\sigma_n$$.

Note by the way the crucial role of $$\epsilon$$. I think if you consider a distribution that takes with equal probability the values $$\sigma, -\sigma, \sigma + 2^{-\sigma}, -\sigma – 2^{-\sigma}$$ (note that $$\sigma$$ is not the standard deviation here) then it’s not actually helpful to raise $$\sigma$$ instead of $$\mu$$, even though $$P(Z^\sigma > \sigma) = \frac{1}{4}$$. I have not bothered to work out the details.

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