Context: The following question is coming from this paper of Ellis and Narayanan. For a positive integer $n$, let $[n]=\{1, 2, \dots, n\}$ and let $2^{[n]}$ be its power set. The $p-$biased measure on $2^{[n]}$, $\mu_p$ is defined as $\mu_p(\{A\})=p^{|A|}(1-p)^{n-|A|}$ and $\mu_p(\mathcal A)=\sum_{A\in\mathcal A}\mu_p(\{A\})$. And a family $\mathcal A\subseteq 2^{[n]}$ is said to be increasing if for every $A, B$ with $A\in\mathcal A$ and $A\subseteq B\Rightarrow B\in\mathcal A$.
On page 4 of the paper, the authors write that "if $\mathcal A$ is increasing, then $\mu_p$ is a non-decreasing monotone function of $p$". I am trying to prove this statement and I have computed $\mu_p(\mathcal A)$ explicitly and verified the statement for some families but haven't been able to prove the general case. It looks differentiating is not the way to go as the expression for the family $\mathcal A=\{A: |A|\geqslant n-2\}$ already looks pretty complex for differentiation.
Could you please help me with this?
The easiest way to see this is computation-free and purely probabilistic. One way to sample $A$ from the $p$-biased measure is to sample $U_1,\ldots,U_n$ uniformly on $[0,1]$ and let $A_p=\{i\in [n]: U_i\leq p\}$. Note $A_p$ has distribution $\mu_p$. You can do this for all $p\in [0,1]$ simultaneously using the same $U_1,\ldots,U_n$, so you get a coupling of all $p$-biased measures for $p\in [0,1]$ which surely satisfies $A_p\subseteq A_{p'}$ for $p\leq p'$ and each $A_p$ has the right distribution. By monotonicity in $\mathcal{A}$, this means that if $p\leq p'$, the event $A_p\in \mathcal{A}$ implies the event $A_{p'}\in \mathcal{A}$. This means the probability must be nondecreasing.