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\begin{align*} p = C\exp(-M) &= \exp(\log \log b + \log(1/2 - \alpha - \gamma)) \exp(-\log b - 1/2 \log \log b) \\ &= b^{-1} \exp(1/2 \log \log b + \log(1/2 - \alpha - \gamma)) \\ &= b^{-1} \omega(1) \end{align*}

\begin{align*} p = \varepsilon \exp(-C) &= b^{-(1/2+\alpha)} b^{-1/2+\alpha+\gamma} \\ &= b^{-1} b^{\gamma} = b^{-1} \omega(1) \end{align*}

\begin{align*} p = C\exp(-M) &= (1/2 - \alpha - \gamma) (\log b) b^{-1} (\log b)^{-1/2} \\ &= b^{-1}(1/2 - \alpha - \gamma) \sqrt{\log b} \\ &= b^{-1} \omega(1) \end{align*}

\begin{align*} \varepsilon \exp(-C) &= b^{-(1/2+\alpha)} b^{-1/2+\alpha+\gamma} = b^{-1} b^{\gamma} = b^{-1} \omega(1) \end{align*}

$$ b \,\mathbf P(Y_i <\varepsilon) = b \int_0^\varepsilon \int_0^\varepsilon \exp(-y) \, dy\,dx = b O(\epsilon^2) = O(b^{-2\alpha}) = o(1). $$

$$ b \,\mathbf P(X_i < M) = b \int_M^\infty \int_x^\infty \exp(-y) \, dy\,dx = b \exp(-M) = 1/\sqrt{\log b} = o(1). $$

$$ p = \mathbf P(X_i < C, Y_i > M) = \int_0^C\int_M^\infty \exp(-y) \, dy\,dx = C\exp(-M) $$

$$ b \,\mathbf P(Y_i <\varepsilon) = b \int_0^\varepsilon \int_x^\varepsilon \exp(-y)\ dy\ dx = b O(\epsilon^2) = O(b^{-2\alpha}) = o(1). $$

$$ b \,\mathbf P(X_i < M) = b \int_M^\infty \int_x^\infty \exp(-y) \ dy\ dx = b \exp(-M) = 1/\sqrt{\log b} = o(1). $$

$$ p = \mathbf P(X_i < C, Y_i > M) = \int_0^C\int_M^\infty \exp(-y) \ dy\ dx = C\exp(-M) $$

Roughly, for (A) we need $\varepsilon = b^{1/2} o(1)$, and $M > \log b + \Omega(1)$. Guaranteeing (B) is a bit tricky. One might hope some interval just covers $(\varepsilon, M)$ with high probability, but AFAIK that's too ambitious. What we can do instead is find a third parameter $C$ for which, with high probability

\begin{align*} p = C\exp(-M) &= \exp(\log \log b + \log(1/2 - \alpha - \gamma)) \exp(-\log b - 1/2 \log \log b) \\ &= b^{-1} \exp(1/2 \log \log b + \log(1/2 - \alpha - \gamma)) \\ &= b^{-1} \Omega(1) \end{align*}

So the probability that none of the $(X_i, Y_i)$ cover $(C, M)$ is

$$ (1-p)^b = (1-\Omega(1)/b)^b < \exp(-\Omega(1)) = o(1), $$

\begin{align*} p = \varepsilon \exp(-C) &= b^{-(1/2+\alpha)} b^{-1/2+\alpha+\gamma} \\ &= b^{-1} b^{\gamma} = b^{-1} \Omega(1) \end{align*}

Roughly, for (A) we need $\varepsilon = b^{1/2} o(1)$, and $M > \log b + \omega(1)$. Guaranteeing (B) is a bit tricky. One might hope some interval just covers $(\varepsilon, M)$ with high probability, but AFAIK that's too ambitious. What we can do instead is find a third parameter $C$ for which, with high probability

\begin{align*} p = C\exp(-M) &= \exp(\log \log b + \log(1/2 - \alpha - \gamma)) \exp(-\log b - 1/2 \log \log b) \\ &= b^{-1} \exp(1/2 \log \log b + \log(1/2 - \alpha - \gamma)) \\ &= b^{-1} \omega(1) \end{align*}

(Here $\omega(1)$ denotes a term that goes to infinity as $b \to \infty$.) So the probability that none of the $(X_i, Y_i)$ cover $(C, M)$ is

$$ (1-p)^b = (1-\omega(1)/b)^b < \exp(-\omega(1)) = o(1), $$

\begin{align*} p = \varepsilon \exp(-C) &= b^{-(1/2+\alpha)} b^{-1/2+\alpha+\gamma} \\ &= b^{-1} b^{\gamma} = b^{-1} \omega(1) \end{align*}