Assuming a random variable X has a strictly decreasing pdf, prove that its mean is strictly greater than its median

Question

Got asked this question in an interview. More generally, I would like to understand for which distributions is the mean greater than the median and vice versa. For example, it is known that for negative skew distributions median is always greater than the mean, but I am unable to provide a more general argument why this should be the case.

Answer 1

Here is a proof in the case that the pdf $f(x)$ is differentiable.

By shifting the distribution of $X$ we may assume the pdf $f(x)$ of $X$ starts at $x=0$. Consider the survival function $$S(x):=P(X> x)=1-F(x),\tag1$$ where $F(x)$ is the CDF of $X$. The median $m$ of $X$ satisfies $$\frac12 = P(X>m)=S(m),\tag2$$ while the mean $\mu$ of $X$ is the area under the survival curve: $$\mu:=E(X)=\int_0^\infty P(X>x)\,dx=\int_0^\infty S(x)\,dx.\tag3$$ From (1) calculate $S'(x)=-f(x)$ and $S''(x) = -f'(x)$ which by assumption is strictly positive for every $x$. One consequence of $S''(x)>0$ is that the graph of $S$ lies above all of its tangent lines. Consequently the area under the survival curve exceeds the area under any tangent line.

This is true in particular for the tangent at $(x,S(x))=(m,\frac12)$. If this tangent line intersects the $y$-axis at $\frac12+h$ (where $0<h<\frac12)$, then the area of the triangle formed by the tangent line and the coordinate axes is $$A(h) = \frac12 \left(h+\frac12\right)^2\frac mh=\frac m2 \left(h+1+\frac1{4h}\right).$$ (This can be verified using similar triangles, or by finding the equation of the tangent line and performing an integration.) Check that $A$ is a strictly decreasing function for $h\in(0,\frac12)$, hence $$\mu=\int_0^\infty S(x)\,dx\ge A(h)> A(1/2) =m.$$

Answer 2

Here is another proof that I have come up with inspired by the answer by @grand_chat, although it uses some assumptions that might be unjustifiable:

Assuming that F is an invertible cdf of X with pdf f(x), we know that $X = F^{-1}(U)$ where $U \sim \text{Uniform}(0,1)$.

Now we compute the second derivative of $F^{-1}(u)$:

$$ \frac{\partial^2 }{\partial u^2} F^{-1} (u) = -\frac{1}{f(F^{-1}(u))^3} * f'(F^{-1}(u)) > 0 $$

Since $f'(u)$ is negative by the assumption of the exercise, and assuming that the pdf is nowhere exactly zero this would imply that $F^{-1}(u)$ is strictly convex. Now we can apply Jensen's inequality:

$$ E[X] = E[F^{-1}(U)] > F^{-1}(E[U]) = F^{-1}(\frac{1}{2}) = \text{median} $$

Thereby concluding the proof.