partial binomial The following inequality of partial binomial expansions seems to be true:
$\sum_{i = 0}^{r-1} {n \choose i} q^{n - i} (1-q)^i \leq \sum_{i = 0}^{r-1} {{n-r} \choose i} q^{n - r - i} (1-q)^i$
For $q \in [0,1]$ and integers $n, r$. $0 \leq r \leq n$. For example:
For $r=1$: $q^n \leq q^{n-1}$
For $r=2$: $q^n + n q^{n-1}(1-q) \leq q^{n-2} + (n-2) q^{n-3}(1-q)$
Is this true? Can you help me prove it?
 A: I think it's easier to prove something stronger: $$\sum_{i = 0}^{r-1} \binom{m + s}{i} q^{m + s - i} (1-q)^i \leq \sum_{i = 0}^{r-1} \binom{m}{i} q^{m - i} (1-q)^i$$ where $s \ge 0$ and $m \ge 0$.
Intuitively it's easy to see by thinking about pachinko. If we have an accumulator which starts at $0$ and a series of Bernoulli trials where we increment the accumulator (go right) with probability $q$ and leave it alone (go left) with probability $1-q$ then the LHS is the probability that the accumulator is less than $r$ after $m+s$ trials. But since the accumulator only increases, that probability must go down as $m+s$ goes up.
To formalise that,
$$\sum_{i = 0}^{r-1} \binom{m}{i} q^{m - i} (1-q)^i = \sum_{i = 0}^{r-1} \binom{m}{i} q^{m - i} (1-q)^i (q + 1 - q) \\
= \sum_{i = 0}^{r-1} \binom{m}{i} q^{m + 1 - i} (1-q)^i + \sum_{i = 0}^{r-1} \binom{m}{i} q^{m - i} (1-q)^{i+1} \\
= \sum_{i = 0}^{r-1} \binom{m}{i} q^{m + 1 - i} (1-q)^i + \sum_{j = 1}^{r} \binom{m}{j-1} q^{m + 1 - j} (1-q)^{j} \\
= \binom{m}{0} q^{m + 1} + \sum_{i = 1}^{r-1} \binom{m+1}{i} q^{m + 1 - i} (1-q)^{i} + \binom{m}{r-1} q^{m + 1 - r} (1-q)^{r} \\
= \binom{m}{0} q^{m + 1} + \binom{m}{r-1} q^{m + 1 - r} (1-q)^{r} - \binom{m+1}{0} q^{m + 1} + \sum_{i = 0}^{r-1} \binom{m+1}{i} q^{m + 1 - i} (1-q)^{i} \\
= \binom{m}{r-1} q^{m + 1 - r} (1-q)^{r} + \sum_{i = 0}^{r-1} \binom{m+1}{i} q^{m + 1 - i} (1-q)^{i} \\
\geq \sum_{i = 0}^{r-1} \binom{m+1}{i} q^{m + 1 - i} (1-q)^{i}$$
And if $\forall m \in \mathbb{N}: f(m+1) \le f(m)$ then by induction $\forall m,s \in \mathbb{N}: f(m+s) \le f(m)$.
