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Having an estimate $hat{p}$ of $p$, what can we say about $1-hat{p}$?

Mathematics Asked on December 25, 2021

First, let us define:

  • $epsilon$-additive approximation:
    $$
    hat{p} in big[ p -epsilon, p + epsilon big]
    $$

  • $epsilon$-multiplicative approximation:
    $$
    hat{p} in big[ p cdot (1-epsilon), p cdot (1+epsilon) big]
    $$


Suppose we have a randomized procedure which estimates (in a multiplicative sense) the probability $p$ of obtaining heads:
$$
Pr bigg[ hat{p} in big[ pcdot(1-epsilon), p cdot (1+epsilon) big] bigg] geq 1 – delta
$$

Naturally, the probability of obtaining tails is $1 – p$.

What can we say about the estimate of the probability of tails: $1-hat{p}$?

More generally, if we have a multiplicative estimate of some probability quantity $hat{p}$, what is the behaviour of $1-hat{p}$, that is, when is $1-hat{p}$ still a multiplicative estimate, and when it becomes an additive estimate?

Can we derive a threshold, in terms of $epsilon$ to specify this? For instance:
$$
begin{cases}
1-hat{p}: text{multiplicative estimate, if} hat{p} leq T(epsilon) \
1-hat{p}: text{additive estimate, if} hat{p} > T(epsilon)
end{cases}
$$

One Answer

Claim:

If $hat{p}$ is an $epsilon$-multiplicative estimate of $p$, then if $p leq 1/2$, we have that $1-hat{p}$ is also a multiplicative estimate of $1-p$.

Proof:

If $hat{p}$ is an $epsilon$-multiplicative estimate of $p$, then: $$ hat{p} in big[ pcdot(1-epsilon), pcdot(1+epsilon) big] implies 1-hat{p} in big[ 1 - pcdot(1+epsilon), pcdot(1-epsilon) big] $$

For $1-hat{p}$ to be an $epsilon$-multiplicative estimate of $1-p$, the following must hold: $$ (1-p)cdot(1-epsilon) leq 1-hat{p} leq (1-p)cdot(1+epsilon) $$

Considering $1-hat{p} = 1-pcdot(1+epsilon)$, we get: $$ (1-p)cdot(1-epsilon) leq 1-pcdot(1+epsilon) leq (1-p)cdot(1+epsilon) $$ From the first inequality we get: $$ 1-epsilon-p+epsilon p leq 1-p -epsilon p \ 2epsilon p leq epsilon \ color{blue}{p leq 1/2} $$ From the second inequality we get: $$ 1-p -epsilon p leq 1+epsilon-p-epsilon p \ color{blue}{0 leq epsilon} $$

Analogous for $1-hat{p} = 1-pcdot(1-epsilon)$.

$blacksquare$


More generally it can be shown that $1-hat{p}$ is a $lambda$-multiplicative estimate of $1-p$, given that $hat{p}$ is an $epsilon$-multiplicative estimate of $p$, if: $$ p leq frac{lambda}{lambda + epsilon} $$

Answered by Alexandru Dinu on December 25, 2021

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