Variation of Markov's inequality - sharper bound? Let $X$ be a non-negative random variable with continuous density $f(\cdot)$, finite mean $\mu > 0$, and finite variance $\sigma^2$.
I am interested in providing an upper bound for $\int_{a}^{\infty}x f(x)dx$, for $a > \mu$.
Here is a very crude approach:
$\sigma^2 + \mu^2 = \int_{0}^{\infty}x^2 f(x) dx$.
Thus
\begin{align}
\sigma^2 + \mu^2 &= \int_{0}^{\mu}x^2 f(x)dx + \int_{\mu}^{a}x^2 f(x)dx + \int_{a}^{\infty}x^2 f(x) dx\\
&\geq \int_{\mu}^{a}x^2 f(x)dx + \int_{a}^{\infty}x^2 f(x)dx\\
&= \int_{\mu}^{a} x^2 [f(x)]dx + \int_{a}^{\infty}x [x f(x)]dx\\
&\geq \mu^2 \int_{\mu}^{a} f(x)dx + a \int_{a}^{\infty}x f(x)dx\\
&=  \mu^2 [F(a) - F(\mu)] + a \int_{a}^{\infty}x f(x)dx.
\end{align}
Subtracting $\mu^2 [F(a) - F(\mu)]$ and dividing by $a > \mu$ finally yields
\begin{equation}
\int_{a}^{\infty}x f(x)dx \leq \frac{\sigma^2 + \mu^2(1-[F(a)-F(\mu)])}{a}.
\end{equation}
Since it obviously holds that $\int_{a}^{\infty}x f(x)dx \leq \mu$, the above bound is only useful for values of $a$ such that $\frac{\sigma^2 + \mu^2(1-[F(a)-F(\mu)])}{a} < \mu$. In particular, this does not hold for $a$ close above $\mu$.
Intuitively, I would expect that one could obtain a useful bound for any $a > \mu$, although I haven't been able to find one. Has anybody got an idea? Many thanks in advance!
 A: A Markov-like inequality would be a universal upper bound, valid for all distributions $F$, for the partial expectation $$\mu(a)=\int_a^\infty x dF(x)$$ in terms of the given moments $$\mu=\int_0^\infty x dF(x)$$ and $$\mu_2 = \mu^2 + \sigma^2 = \int_0^\infty x^2 dF(x).$$  Such a bound has to be achieved by extremal points in the space of distributions: that is, purely discrete distributions.  It's not hard to see that such a distribution will have support consisting of at most three points: $0$, $\alpha \ge a$, and some other number $\beta$ between $0$ and $\mu$.  Let its probability at $\alpha$ be $p\ge 0$ and at $b$ be $q\ge 0$ with $p+q \lt 1$.  Thus $\mu(a)=\alpha p \le \mu$. 
Simplify the calculation by adopting units of measurement in which $\mu=1$.
Computing the first two moments of this distribution yields
$$\cases{1 = \beta q + \alpha p \\ 1+\sigma^2 = \beta^2q + \alpha^2p.}$$
Maximizing $\mu(a)$ subject to these equality constraints (together with the preceding inequality constraints) yields
$$\eqalign{
\mu(a)&=\frac{\alpha\sigma^2}{\sigma^2+(\alpha-1)^2}; \\
\alpha &= \max(a, \sigma^2+1); \\
\beta &= 1 - \frac{\sigma^2}{\alpha-1}; \\
p &= \frac{\sigma^2}{\sigma^2 + (\alpha-1)^2}; \\
q &= 1-p = \frac{(\alpha-1)^2}{\sigma^2 + (\alpha-1)^2}.
}$$
To convert back to the original units of measurement, multiply $\mu(a)$, $\sigma$, $\alpha$, and $\beta$ by $\mu$. Note that $p+q=1$: the possible atom at $0$ was superfluous and these extremal distributions are all scaled, shifted Bernoulli$(p)$ distributions.
Here are plots of the upper bound versus $a$ for various values of $\sigma$ (all for $\mu=1$).  The tail decreases slowly: it is an inverse quadratic.

When $F$ is continuous with density $f$ and $a \gt \mu$, we have
$$\int_a^\infty x f(x) dx = \mu(a) \le  \frac{\mu\alpha\sigma^2}{\sigma^2 + \mu^2(\alpha-1)^2}$$
where $\alpha=\max(a/\mu, (\sigma / \mu)^2+1)$.
