# How many samples are needed to distinguish the means of two distributions in multi-armed bandits?

In a paper on Multi Armed Bandits, I came across the following statement:

This generalizes the well-known fact that one needs of order $$\frac{1}{\Delta^2}$$ samples to differentiate the means of two distributions with gap $$\Delta$$.

To me, this is not "well known". That is why I would appreciate a hint where this "fact" comes from. In Multi Armed Bandits, a common assumption is that the random variables are bounded between 0 and 1. However, the exact distribution is not known.

This "well-known" fact comes from the fact that the lower bound on the sample complexity (i.e. the number of samples needed to differentiate the means of two arms) is proportional to $$1/\Delta^2$$.
You can actually say more: any algorithm, in order to identify with probability at least $$1-\delta$$ the best arm (i.e. the one having the largest mean) between two distribution, needs a number of samples $$\tau$$ that satisfy
$$\mathbb{E}[\tau] \gtrsim 1/\Delta^2\log(1/\delta),$$