Putnam 2014 
Topic: Probability

Problem: Suppose $X$ is a random variable that takes on only nonnegative integer values, with $E\left[ X \right] = 1$, $E\left[ X^2 \right] = 2$, and $E \left[ X^3 \right] = 5$. (Here $E\left[ y \right]$ denotes the expectation of the random variable $Y$.) Determine the smallest possible value of the probability of the event $X=0$.

Solution: Let $a_n = P(X=n)$; we want the minimum value for $a_0$. If we write  $S_k = \sum_{n=1}^\infty n^k a_n$, then the given expectation values  imply that $S_1 = 1$, $S_2 = 2$, $S_3 = 5$. Now define $f(n) = 11n-6n^2+n^3$, and note that $f(0) = 0$, $f(1)=f(2)=f(3)=6$, and  $f(n)>6$ for $n\geq 4$; thus $4 = 11S_1-6S_2+S_3 = \sum_{n=1}^\infty f(n)a_n \geq 6 \sum_{n=1}^{\infty} a_n$.  Since $\sum_{n=0}^\infty a_n = 1$, it  follows that $a_0 \geq \frac{1}{3}$. Equality is achieved  when $a_0=\frac{1}{3}$, $a_1=\frac{1}{2}$, $a_3=\frac{1}{6}$, and $a_n = 0$ for all other $n$,  and so the answer is $\frac{1}{3}$. The answer is $\boxed{\frac{1}{3}}$.