A version of one-sided Chebyshev's inequality Let $X$ be a real random variable with mean $\mu > 0$ and variance $\mu^2$. Does there exist a non-trivial upper bound on the probability $\Bbb P(X < 0)$ or is there a counterexample that shows that this probability can be as high as possible?
 A: The upper bound on $\Bbb{P}(X<0)$ is $\frac{1}{2}$ for any $\mu > 0$ and we will show how to construct a probability distribution function with $\Bbb{P}(X\leq 0)=\frac12)$ and how to go from there to a function with
$\Bbb{P}(X < 0)=\frac12-\epsilon)$ for arbitrary small positive $\epsilon$.
Let us work to maximize $\Bbb{P}(X\leq 0)$, and call p.d.f.'s with mean $\mu$ and variance $\mu^2$ "good pdfs."  
Lemma 1: In a good pdf with mean $\mu$, $\Bbb E(X^2) = 2\mu^2=2[\Bbb E(X)]^2$.
Proof:  $\sigma^2 = \Bbb E(X^2) -(\Bbb E(X)]^2 = \Bbb E(X^2)-\mu^2$ and for a good pdf $\sigma^2 = \mu^2$.
Definition: For any pdf $f$ define 
$$W[f] \equiv  \Bbb E_f(X^2) - 2 [\Bbb E_f(X)]^2$$
Note that if $f$ is a good pdf, then $W[f]=0$.
Lemma 2:  If $f(x)$ is a any pdf with positive mean and $\Bbb{P}(X < 0)$ (strictly negative $X$) is non-zero, let  $g(x)$ be a pdf which is the same as $f(s)$ except that some probability is shifted from $X=-a$ to $X=0$.  Then $W[g] > W[f]$.
Proof: Let $p$ be the probability shifted form $-a$ to $0$ (this argument can of course be made more rigorous for continuous distributions).  Then 
$$\Delta ( [\Bbb E(X)]^2 ) = 2pa\mu + p^2a^2\\
\Delta ( \Bbb E(X^2) ) = 2pa\mu + pa^2\\
W[g]-W[f] = (p-p^2)a^2 > 0
$$
Lemma 3: If $f(x)$ is a any pdf with positive mean $\mu$ and and $\Bbb{P}(X > \mu)$ is non-zero, let  $g(x)$ be a pdf which is the same as $f(s)$ except that some probability $p$ is shifted from $X=\mu+c$ to $X=0$ for some $c>0$. Then $W[g] < W[f]$.
Proof: 
$$\Delta ( [\Bbb E(X)]^2 ) = -2pc\mu + p^2c^2\\
\Delta ( \Bbb E(X^2) ) = -2pc^2\\
W[g]-W[f] = -2pcu-(p-p^2)c^2 < 0
$$
Corollary 4: If $f(x)$ is a good pdf with positive mean $\mu$ and and $\Bbb{P}(X > 0 = \rho)$ is non-zero, then another good pdf $h(x)$ may be formed with  $\Bbb{P}_h(X > \mu) < \rho$ by moving some of the probability at negative $X$ up to zero, and some of the probability at positive $X > \mu$ down to zero.
Lemma 5: For a given $\mu$, the good pdf with mean $\mu$ and greatest $\rho = \Bbb{P}(X > 0) $ will be formed by concentrating a discrete probability $\rho$ at $X=0$ and the remainder at some $X=\mu+k>\mu$.
This is a consequence of corollary 4; and in fact to give the proper mean, we must have $k=\frac{\mu\rho}{1-\rho}$.  To give the proper variance, we then must have
$$
\Bbb{E}(X-\mu)^2 = \mu^2 \\
\rho \mu^2 + (1-\rho) k^2 = \mu^2 \\
\rho\mu^2 + \frac{\rho^2}{1-\rho}\mu^2 = \mu^2 \\
{1-\rho}\rho + \rho^2= {1-\rho} \\ \rho = \frac12
$$
Theorem 6:  For any specified $\mu > 0$, the pdf $f(X)$ with the largest possible $\Bbb{P_f}(X\leq 0)$ consists of a discrete distribution with 
$$
\Bbb P(X=0) = \Bbb P(X=2\mu) = \frac12
$$
Proof:  By lemma 5, with the expression for $k$ implied by corollary 4, $k = \mu$ and $\rho = \frac12$, giving the above distribution.
Finally, Theorem 6:  $\rho = \frac12$ is a tight upper bound on the probability, for a good pdf, that $X<0$.
Proof:
Start with the distribution in theorem 5.  For any desired small $\delta \rho > 0$, displace $\delta \rho$ of the distribution from $0$ to $2\mu + \alpha$.  The 
two conditions that the mean remain $\mu$ and the variance remain $\mu^2$ provide, for a given fixed $\delta \rho$ two equations which determine the suitable $\epsilon$ and $\alpha$.  You thus construct the desired distribution, with mean and variance $\mu$ and $\Bbb P(X<0) = \frac12 - \delta\rho$.
