I am trying to follow the logic in a proof of a lemma in the paper named in the question above. (It focuses on a version of the Monty Hall problem and can be found here, and a previous question I asked to gain some clarity can be found here.)
I will identify the exact steps I am failing to understand in the "Question" section below.
The Lemma
The lemma (which they call Lemma 4.1) is as follows:
For any $\alpha \in \left[\frac{1}{3},\frac{2}{3}\right]$, let
$(m^\ast(\alpha),b^\ast(\alpha)) = \mathop{\arg\max}\limits_{f(m,b) \le \alpha} \ g(m,b) \tag{11}$
$(\tilde{m}(\beta),\tilde{b}(\beta)) = \mathop{\arg\min}\limits_{g(m,b) \ge \beta} \ f(m,b) \tag{12}$
where
$\beta = g(m^\ast(\alpha),b^\ast(\alpha))$
$g(m,b) = \frac{m}{3} + \frac{2b}{3}$
$f(m,b) = \left(\frac{m+2b}{3}\right) \max \left(\frac{m}{m+2b}, 1 - \frac{m}{m+2b} \right) \\ + \left(\frac{(1-m)+2(1-b)}{3}\right) \max \left(\frac{1-m}{(1-m)+2(1-b)}, 1 - \frac{1-m}{(1-m)+2(1-b)} \right)$
Then, $(m^\ast(\alpha),b^\ast(\alpha)) = (\tilde{m}(\beta),\tilde{b}(\beta))$.
The Proof
Their proof is as follows:
Let $\beta = g(m^\ast(\alpha),b^\ast(\alpha))$, this $\beta$ will appear in the constraint in problem (12), for which we will find $(\tilde{m}(\beta),\tilde{b}(\beta))$.
Observe that by the constraint of problem (11), we have:
$f(m^\ast(\alpha),b^\ast(\alpha)) \le \alpha \tag{13}$If $\beta = 1$, then the only possibility is that $(m^\ast(\alpha),b^\ast(\alpha)) = (1,1)$.
Moreover, $(m, b) = (1,1)$ is the only point which satisfies the constraint in problem (12).
Hence, $(m^\ast(\alpha),b^\ast(\alpha)) = (\tilde{m}(\beta),\tilde{b}(\beta)) = (1,1)$.
Indeed, we get the classical Monty Hall problem and in this case $f(m, b) = \frac{2}{3}$.Let us consider now the case $\beta < 1$.
Take $(m, b)$ such that $g(m, b) > \beta$, then $f(m, b) > \alpha$.
Otherwise, $(m^\ast(\alpha),b^\ast(\alpha))$ did not maximize (11).
If $g(m, b) = \beta < 1$, then $(m, b) \ne (1,1)$.
Since $g$ is continuous and $(m, b)$ is not a local maxima, there exists a sequence $\{(m_n, b_n)\}_{n \ge 1}$ which converges to $(m, b)$ with $g(m_n,b_n) > \beta$.
In particular, $f(m_n,b_n) > \alpha$.So, by continuity of $f$ we have $f(m, b) \ge \alpha$.
So we have seen that in any case that $g(m, b) \ge \beta$ then $f(m, b) \ge \alpha$.
Moreover, we have that $(m^\ast(\alpha),b^\ast(\alpha))$ satisfies:
$g(m^\ast(\alpha),b^\ast(\alpha))=\beta$
and therefore:
$f(m^\ast(\alpha),b^\ast(\alpha)) \ge \alpha$.
Then, by inequality (13), we get $f(m^\ast(\alpha),b^\ast(\alpha))=\alpha$ so is a minimizer for (12) which is unique and so $(m^\ast(\alpha),b^\ast(\alpha))=(\tilde{m}(\beta),\tilde{b}(\beta))$.
Question
The following phrase confuses me:
In particular, $f(m_n,b_n) > \alpha$.
- Are they saying that for any $(m_n,b_n)$ in the sequence described (i.e. $\{(m_n, b_n)\}_{n \ge 1}$ converging to $(m,b)$ with $g(m_n,b_n) > \beta$) this inequality must also be true for $f(m_n,b_n)$?
- If so, why is this true?
This is well beyond my usual level of mathematics and any additional translation into lay-speak would be very much appreciated!
After more careful consideration, I am able to answer my own question:
$\beta$ is defined such that it is the highest value $g(m,b)$ can take while $f(m,b) \le \alpha$ is satisfied.
This means that any $(m',b')$ that also satisfy $f(m',b') \le \alpha$ will return $g(m',b') < \beta$.
Hence, any $(m'',b'')$ which return $g(m'',b'') > \beta$ must not satisfy the inequality. Explicitly, by contrast, it must be the case that $f(m'',b'') > \alpha$.