Putnam 2014, Problem A-4

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

Solution:
The answer is \(\frac{1}{3}\).  
First solution: Let \(a_n = P(X = n)\). Define \(S_k = \sum_{n=1}^\infty n^k a_n\), so \(S_1 = 1\), \(S_2 = 2\), \(S_3 = 5\). Let \(f(n) = 11n - 6n^2 + n^3\), so \(f(0) = f(1) = f(2) = f(3) = 6\) and \(f(n) > 6\) for \(n \geq 4\). Then \(4 = 11S_1 - 6S_2 + S_3 = \sum_{n=1}^\infty f(n) a_n \geq 6 \sum_{n=1}^\infty a_n = 6(1 - a_0)\), so \(a_0 \geq \frac{1}{3}\). Equality holds when \(a_0 = \frac{1}{3}\), \(a_1 = \frac{1}{2}\), \(a_3 = \frac{1}{6}\), and \(a_n = 0\) otherwise.  
Second solution: (by Tony Qiao) Define the probability generating function \(G(z) = \sum_{n=0}^\infty P(X = n) z^n\). Then \(G(1) = G'(1) = G''(1) = G'''(1) = 1\). By Taylor’s theorem, \(G(x) = 1 + (x-1) + \frac{(x-1)^2}{2!} + \frac{(x-1)^3}{3!} + \frac{G'''(c)}{4!}(x-1)^4\) for some \(c \in [x,1]\). In particular, \(G(0) = \frac{1}{3} + \frac{1}{24} G'''(c)\). Since \(G\) has nonnegative coefficients and \(c \geq 0\), \(G'''(c) \geq 0\), so \(G(0) \geq \frac{1}{3}\).