TITLE: Missing mass conjecture
QUESTION [6 upvotes]: Let $n,t$ be positive integers and $p_1,p_2,\ldots,p_n$ positive numbers summing to 1. Conjecture:
$$
\sum_{i=1}^n p_i (1-p_i)^t \le \frac{n(1-1/n)^n}{t}
$$
always holds.
The motivation comes from my missing mass question; the quantity $\sum_{i=1}^n p_i (1-p_i)^t$ is precisely the expected unseen mass after $t$ draws from the distribution $(p_i)$ on $n$ objects.

REPLY [7 votes]: Since the single-variable optimization that David mentions still requires some work, I will present another solution.  Let $f(p) = p(1-p)^t$.  Define the function $g$ that is equal to $f$ on $[0, 1/t]$, and on $[1/t, 1]$ is a linear interpolation between the points $(1/t, f(1/t))$ and $(1, 0)$.  Then we can check that $g$ is concave on all of $[0, 1]$ and $g \ge f$ on all of $[0, 1]$.  (I learned this concept of "concave majorants" or "convex minorants" from Steele's book called The Cauchy-Schwarz Master Class.)  Applying Jensen's inequality, we have $\sum f(p_i) \le \sum g(p_i) \le n g(1/n)$.
We now split into two cases, $t \le n -1$ or $t \ge n$.  First, suppose that $t \le n - 1$.  Then $g(1/n) = f(1/n) = (1/n) (1 - 1/n)^t$.  So we need to show that $(1 - 1/n)^t$ is at most $n (1 - 1/n)^n / t$.  That's equivalent to showing $t(1 - 1/n)^t \le n (1 - 1/n)^n$.  We can check that the left side is an increasing function of $t$ for $t \le n - 1$, and when $t = n - 1$ we have an equality.  So we have established the inequality in this case.
Next suppose that $t \ge n$.  Then by linear interpolation, we find $g(1/n) = (1 - 1/t)^{t-1} (1 - 1/n) / t$.  So we need to show that $(1 - 1/t)^{t - 1} (n - 1) \le n(1 - 1/n)^n$.  That's equivalent to $(1 - 1/t)^{t-1} \le (1 - 1/n)^{n - 1}$.  By taking reciprocals, that's equivalent to $(1 + 1/(t - 1))^{t - 1} \ge (1 + 1/(n-1))^{n - 1}$.  The left side is an increasing function of $t$, and we have an equality when $t = n$.  So we have established the inequality in the second case too.