Let $N \in \mathbb N$ and suppose that $\phi$ is a submeasure on $[1,N] = \{1,2,\dots,N\}$, by which I mean that $\phi$ is a function $\mathcal P ([1,N]) \rightarrow \mathbb R$ such that

i. $A \subseteq B$ implies $\phi(A) \leq \phi(B)$,

ii. $\phi(A \cup B) \leq \phi(A) + \phi(B)$ for any $A, B \subseteq [1,N]$, and

iii. $\phi(\emptyset) = 0$.

Suppose we choose a subset $X$ of $[1,N]$ at random. We do not know anything about the probability distribution used to select $X$, except that for every $i \in [1,N]$ the probability $P(i \in X)$ that $i$ is in the random set $X$ is at least $\varepsilon > 0$.

I would like to have a lower bound for the expected value of $\phi(X)$ in terms of $\varepsilon$ and $\phi([1,N])$. I conjecture that

$$E(\phi(X)) \,\geq\, \frac{1}{2} \cdot \varepsilon \cdot \phi([1,N])$$

but cannot seem to prove it. I would be happy if there is any constant $c > 0$ (independent of $N$, but possibly dependent on $\varepsilon$) such that

$$E(\phi(X)) \,\geq\, c \cdot \varepsilon \cdot \phi([1,N]).$$

(So my conjecture is that taking $c = \frac{1}{2}$ will do, but if you have a proof for some smaller value of $c$ then I would still love to see it.) My question is simply whether either of these bounds is true.

**Comments:**

If $\phi$ is a measure instead of a submeasure, then it follows from the linearity of expectation that $E(\phi(X)) \geq \varepsilon \cdot \phi([1,N])$ (an even better lower bound than the conjectured one above). Indeed, if $\phi$ is a measure then $$E(\phi(X)) = E\left( \sum_{i \in X}\phi(\{i\}) \right) = E\left( \sum_{i \in [1,N]}\phi(\{i\} \cap X) \right) = \sum_{i \in [1,N]}E(\phi(\{i\} \cap X)) = \sum_{i \in [1,N]}P(i \in X)\phi(\{i\}) \geq \varepsilon \sum_{i \in [1,N]}\phi(\{i\}) = \varepsilon \cdot \phi([1,N])$$

Despite the previous comment, the factor of $\frac{1}{2}$ in my conjecture is necessary. For example, $\phi$ could be a submeasure on $[1,N]$ that assigns $\phi(\emptyset) = 0$, $\phi([1,n]) = 2$, and $\phi(A) = 1$ for all other $A \subseteq [1,N]$. If $X$ is selected uniformly at random from among the $N$ sets of the form $[1,N] \setminus \{i\}$, then we have $E(\phi(X)) = 1$ but $\varepsilon \cdot \phi([1,N]) = 2 - \frac{2}{N} \approx 2$.