Inequality of binomial probabilities I need to show the following:
I have two binomial random variables $X \sim BIN(m,p_1)$ and $Y \sim BIN(m,p_2)$, where $p_2 \geq p_1$. I want to show for any fixed constant $c \in \{0,...,m-1\}$ that
$P(Y \geq c \mid Y < m) \geq P(X \geq c \mid X < m)$
This holds with equality when $c=0$ (easy to see), and also holds easily when the condition of being less than $m$ is removed. It makes intuitive sense: the variable with the larger success probability should have at least as high probability of taking on larger values. Nonetheless, I don't see a good way to prove this explicitly. Perhaps there is a clever way to do so.
Any help would be appreciated! Thank you!
 A: Here is my attempt at a solution, it proves the inequality in the other direction for the probability of the complement event from how I originally stated the problem.

A: The probabilities can be computed as
$$P(X \geq n \vert X < m) = \frac{P(n \leq X \leq m)}{P(X < n)+P(n \leq X \leq m)} $$
or
$$P(X \geq n \vert X < m)^{-1} = 1 + \frac{P(X < n)}{P(n \leq X \leq m)}$$
When we increase the value of $p$ then this term $$\frac{P(X < n)}{P(n \leq X \leq m)}$$ is monotonically decreasing. This is because the terms in the denominator and numerator change at different rates.

*

*We can express the probability of the one as a function of the other
$$\frac{P(Y=k)}{P(X=k)} = \left(\frac{p_2/(1-p_2)}{p_1/(1-p_1)} \right)^k = f(k)$$
where this term $f(k)$ is a monotonic increasing function of $k$ if $p_2 > p_1$.


*We can express related inequalities like
$$\frac{P(Y < n)}{P(X < n)} \leq f(n)$$
$$\frac{P(n \leq Y \leq m)}{P(n \leq X\leq m)} \geq f(n)$$
or
$$\frac{P(n \leq X \leq m)}{P(n \leq Y\leq m)} \leq f(n)^{-1}$$


*If we multiply the two, the following follows
$$\frac{\left(\frac{P(Y < n)}{P(n \leq Y \leq m)}\right)}{\left(\frac{P(X < n)}{P(n \leq X \leq m)}\right)} \leq 1$$
and
$$\frac{\left(1+\frac{P(Y < n)}{P(n \leq Y \leq m)}\right)}{\left(1+\frac{P(X < n)}{P(n \leq X \leq m)}\right)}  = \frac{P(X \geq n \vert X < m)}{P(Y \geq n \vert Y < m)} \leq 1$$
from which follows
$$P(Y \geq n \vert Y < m) \geq  P(X \geq n \vert X < m)$$
A: \begin{align*}
P(Y\geq c| Y<m) - P(X \geq c|X<m) &=
\sum_{t=c}^{m-1}{m \choose t}\big[p_2^t(1-p_2)^{m-t}-p_1^t(1-p_1)^{m-t}\big]
\end{align*}
Consider $f(t) = p_2^t(1-p_2)^{m-t}$ and $g(t)  = p_1^t(1-p_1)^{m-t}$
\begin{align*}
h(t) &= \log(f(t))- \log(g(t))\\
&= t(\log(p_2)-\log(p_1)) +(m-t)(\log(1-p_2)-\log(1-p_1))
\end{align*}
\begin{align*}
h'(t) &= \log(p_2)-\log(p_1) - (\log(1-p_2)-\log(1-p_1))\\
&= \log\big(\frac{p_2(1-p_1)}{p_1(1-p_2)}\big)
\end{align*}
Given $\frac{p_2}{p_1} \geq 1$[$p_1 \neq 0$] $\implies \frac{1-p_1}{1-p_2}\geq 1$
Thus, $\frac{p_2(1-p_1)}{p_1(1-p_2)} \geq 1$
$\implies \log\big(\frac{p_2(1-p_1)}{p_1(1-p_2)}\big)$
$\implies h'(t) \geq 0 \implies \ h(t)$ is non-decreasing
$\implies h(t) \geq 0 \ \forall t \ \implies \frac{f(t)}{g(t)}\geq 1\ \implies P(Y\geq c| Y<m) - P(X \geq c|X<m) \geq 0$ 
A: Hint: Define a multinomial random vector $(Z_1, Z_2, Z_3)$ with parameters $m$ and $(p_1, p_2-p_1, 1-p_2)$. Then $X$ has the same distribution as $Z_1$ and $Y$ has the same distribution as $Z_1+Z_2$. 
