$X$ is a random variable with binomial distribution parameters $n$ and $p_1$.
$Y$ is a random variable with binomial distribution parameters $n$ and $p_2$.
$p_1 < p_2$
How can I show that $P(X \leqslant k) \geqslant P(Y \leqslant k)$?
Please just give me a hint.
I tried comparing the terms ${n \choose k}p_1^k(1-p_1)^{n-k}$, but this doesn't work (after trying this, it's obvious in hindsight that it doesn't work.
Hint: define the two variables on the same probability space. Specifically, let $X = \sum_{i=1}^n A_i$ and $Y = \sum_{i=1}^n B_i$, where $\{A_i\}$ and $\{B_i\}$ are each collections of independent Bernoulli variables. However, $A_i$ and $B_i$ should be dependent variables; specifically, they should each be defined on the common sample space $[0, 1]$ in the following way: $$(A_i, B_i) \mapsto \begin{cases} (1, 1), & x < p_1\\ (0, 1), & p_1 < x < p_2\\ (0, 0), & x > p_2. \end{cases}$$