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:

$\sigma^2 + \mu^2 = \int_{0}^{\infty}x^2 f(x) dx$.


\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!

  • $\begingroup$ Well, for any $\mu >0 $ and $a \geq \mu$ there exists a nonnegative random variable $X$, such that $E(X) = \mu$ and $P(X > a) = \frac{\mu}{a}$ (if it's not a continuous RV, you can approximate it as close as you want with continuous ones). Therefore you won't be able to find an upper bound on $P(X > a)$ that improves over Markov inequality uniformly for all $a$ - like in the example you gave, your bound was better for some $a$ and worse for others. What bound you want depends therefore on what do you want to apply it to. $\endgroup$ Dec 28, 2016 at 14:59

1 Answer 1


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 $\alpha=\max(a/\mu, (\sigma / \mu)^2+1)$.

  • 1
    $\begingroup$ Thanks a lot for your detailed answer. I checked your argument and arrived at the same results. In particular, it is now clear why we cannot have a useful bound for (with $\mu = 1$) $a \leq 1 + \sigma^2$: The discrete distribution $\beta = 0$ with probability $q = 1 - \frac{1}{\sigma^2 + 1}$, $\alpha = \sigma^2 + 1$ with probability $p = \frac{1}{\sigma^2 + 1}$ satisfies all criteria and has a partial expectation of $\mu(a) = 1$. $\endgroup$
    – Martin
    Jan 4, 2017 at 11:06

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