Approximation or bound of $\operatorname{Pr}(X<\operatorname{E}(X))$

Question

$X$ is a continuous random variable (we can assume some statistic (e.g., mean and variance) are known, but the distribution is unknown). Consider a probability $\operatorname{Pr}(X<\operatorname{E}(X))$.

We know for symmetric distributions, $\operatorname{Pr}(X\leq\operatorname{E}(X))=0.5$. However, for asymmetric distributions, is there any approximation to approximate this probability? is there an upper or a lower bound expression for this probability?

Thanks!

Answer 1

While we know that the difference between the median and the mean can be at most one standard deviation. Just knowing the mean and the variance is insufficient to control $P(X \leq E(X))$.

Let $\delta_x$ be the Delta distribution centred at $x$. Consider the family of probability distributions given by

$$ f_n(x) = \frac{1}{n} \delta_{-\sqrt{n-1}}(x) + \frac{n-1}{n} \delta_{1/\sqrt{n-1}}(x) $$

The expectation values are

$$ \int x f_n(x) dx = \frac{1}{n}(-\sqrt{n-1}) + \frac{n-1}{n} \frac{1}{\sqrt{n-1}} = 0 $$

and the variances are

$$ \int x^2 f_n(x) dx = \frac{1}{n}(n-1) + \frac{n-1}{n} \frac{1}{n-1} = 1$$

But $P(X_n \leq 0)$ can be computed to be

$$ \int_{-\infty}^0 f_n(x) dx = \frac{1}{n} \searrow 0~. $$

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In general, by assuming your distribution taking the form

$$ f(x) = \sum_{m = 1}^M \rho_m \delta_{x_m}(x) $$

you get $2M$ degrees of freedom. The equation $\sum_m \rho_m = 1$ gives one equation, and prescribing the moments $E(X^k)$ from $k = 1$ to $k = K$ provides $K$ more equations. So as long as $2M > K + 1$, one expects to be to find more than one probability distribution fitting the given ansatz that has the prescribed moments. In fact, by choosing $M > K+1$, we can even fix $x_m = m$ fir $m < M$ and $x_M = -n$, and reduce to solving a linear system of underdetermined equations. This should allow the construction of a probability distribution that has mean 0 and a prescribed first $K$ moments, with $P(X\leq E(X))$ arbitrarily small.

Answer 2

The quantity $p_X=\mathrm P(X\lt\mathrm E(X))$ can take every value in $[0,1)$. The case $p_X=0$ is when $X=\mathrm E(X)$ almost surely, for every other $p$ in $(0,1)$ there exists some non degenerate random variables $X$ such that $p_X=p$. For example, consider $X$ which assumes two values, say $\mathrm P(X=\mathrm E(X)-(1-p)x)=p$ and $\mathrm P(X=\mathrm E(X)+px)=1-p$ for some positive $x$, then $p_X=p$.