Assume we randomly take $M$ integers $t_1 \le t_2 \le \dots \le t_M$ from the set of integers $\{ 1, 2, \dots, T \}$ such that $t_M = T$. We further denote $t_0 = 1$ for convention. For each $s \in [1,T]$, denote the smallest $t_i \ge s$ as $\tilde{t}(s)$. Note that $\tilde{t}(s)$ is a random variable for each $s$ because it depends on $t_i$. It is easy to show that $$ \sup_{s \in [1,T]} E[\tilde{t}(s) - s] \ge C_1 \frac{T}{M} $$ where the expectation is taken over the randomness of $t_i$ and $C_1$ is a positive constant independent of$M$ and $T$.

For example, let's take the uniform probability distribution $P$ over $[1,T]$ and consider $\int_{s \in [1,T]} E[\tilde{t}(s) - s] dP(s)$. We have \begin{align*} \int_{s \in [1,T]} E_{t_i}[\tilde{t}(s) - s] dP(s) &= E_{t_i}[ \int_{s \in [1,T]}(\tilde{t}(s) - s) dP(s)] \\ &= E_{t_i}[ \frac{1}{T} \sum_{i=1}^M \frac{(t_i - t_{i-1})^2}{2}]\\ &\ge E_{t_i}[ \frac{1}{T} \sum_{i=1}^M \frac{(T/M)^2}{2}] \quad (*)\\ &= \frac{1}{2} \frac{T}{M} \end{align*} The inequality $(*)$ comes from the fact that $\sum_{i=1}^M (t_i - t_{i-1})^2 $ is minimized when $t_M - t_{M-1}=t_{M-1} - t_{M-2}=t_{M-2} - t_{M-3}= \dots =t_2 - t_1 = \frac{T}{M}$.

Since $\sup_s E[\tilde{t}(s) - s] \ge \int E[\tilde{t}(s) - s] dP(s)$, the integral inequality proves the desired proposition.

I would like to prove the productive version of the previous proposition. The setting of $t_i$ and $\tilde{t}(s)$ stays the same. I want to prove that $$ \sup_{s \in [1,T]} E[\frac{\tilde{t}(s)}{s}] \ge C_2 T^{\frac{1}{M}}$$ where the expectation is taken over the randomness of $t_i$ and $C_2$ is a positive constant independent of$M$ and $T$.

These two propositons have similar structures. The difference is that the first proposition is additive while the second(unproved) proposition is productive. However, the integral argument for the first proposition fails on the second proposition, and I cannot figure out how to prove the second proposion.

Could you please help me? Suggestions or possible references are welcome!