Expected Shortfall monotonicity

Question

I have to show monotonicity for a more general case than the expected shortfall.

I have to show that $E(X|X \geq a) \geq E(X|X \geq b), \forall a,b \in \mathbb{R}$ so that $a\geq b$ and $F_X(a-)<1$.

This is how I started:

$E(X|X\geq b)=\frac{\int_b^{\infty}X dP}{P(X\geq b)}=\frac{\int_b^{a}X dP+\int_a^{\infty}X dP}{P(X\geq b)} \leq \frac{\int_b^{a}X dP+\int_a^{\infty}X dP}{P(X\geq a)}=E(X|X\geq a)+ \frac{\int_b^{a}X dP}{P(X\geq a)}$, which does not help, because $\int_b^a X dP$ is positive.

Do you have any hints for me? I would appreciate it a lot.

Answer 1

$E(X|X\geq b)=\frac{\int_b^{\infty}X dP}{P(X\geq b)}=\frac{\int_b^{a}X dP+\int_a^{\infty}X dP}{P(X\geq b)} \leq \frac{a\int_b^{a} dP+\int_a^{\infty}X dP}{P(X\geq b)}=\frac{a\int_b^{a} dP+\int_a^{\infty}X dP}{\int_b^{a} dP + P(X\geq a)}$

Now since $a \leq \frac{\int_a^{\infty}X dP}{P(X\geq a)}$, the right hand side of the equation above is smaller than or equal to $\frac{\frac{\int_a^{\infty}X dP}{P(X\geq a)}\int_b^{a} dP+\int_a^{\infty}X dP}{\int_b^{a} dP + P(X\geq a)} = \frac{\int_a^{\infty}X dP}{P(X\geq a)} = E(X|X\geq a)$.