Seeking Lower Bound for Partition Probability in Random Variable Analysis

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I am reaching out to seek assistance with a probability problem involving random variables.

For each $p$ in $[1,\infty)$, consider positive random variables $X_{1,p}, X_{2,p}, \ldots, X_{n,p}$ such that $\mathbb{P}(X_{i,p} < X_{j,p}) = 1$ for each $i < j$. These variables may not be identically distributed or independent. Similarly, let $Y_1, Y_2, \ldots, Y_n$ be positive random variables satisfying $\mathbb{P}(Y_i < Y_j) = 1$ for each $i < j$, again not necessarily equally distributed or independent. Assume $\mathbb{P}(X_{i,p} \leq Y_i) = 1$ for each $i$.

Furthermore, given $\delta > 0$ and $m > 1$, assume the existence of $p_{n,m,\delta} > 1$ such that for $p \geq p_{n,m,\delta}$: $$ \mathbb{P}\left(Y_{i-1} - X_{i-1,p}\leq \frac{(X_{i,p} - X_{i-1,p})}{m}\right) \geq 1 - \delta \tag{1}$$ for each $i$, where $p_{n,m,\delta}$ is independent of $i$.

The challenge lies in finding an $i$-independent lower bound for the probability: $$ P\left(\left.X_{i-1,p} \leq \varepsilon <X_{i,p}\right| Y_{i-1} \leq \varepsilon < Y_i \right) $$ where $\varepsilon > 0$ is a predetermined value.

Remark: The sequences $X_{1,p}, X_{2,p}, \ldots, X_{n,p}$ and $Y_1, Y_2, \ldots, Y_n$ can be interpreted as partitions of the interval $[0,\infty)$. Condition (1) implies that these partitions can be made nearly identical for sufficiently large $p$. It suggests that for large $p$, the interval $[Y_{i-1}, X_{i,p}]$ is nearly encompassed within $[X_{i-1,p}, X_{i,p})$ and its length is at least $(1 - \frac{1}{m})$ times the length of $[X_{i-1,p}, X_{i,p})$. Intuitively, for large $m$, these intervals almost coincide, and the probability of $\varepsilon$ falling within the same partition is high. However, this intuition requires theoretical substantiation. I conjecture that this lower bound could be $\frac{1}{1-\frac{1}{m}}$, i.e., that one could have $$ P\left(\left.X_{i-1,p} \leq \varepsilon <X_{i,p}\right| Y_{i-1} \leq \varepsilon < Y_i \right) \geq 1-\frac{1}{m}. $$ However, I have not been able to verify it.