# Expected Shortfall monotonicity

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.

• what's your assumption on $X$? – CABLE Jul 11 '20 at 11:59

$$E(X|X\geq b)=\frac{\int_b^{\infty}X dP}{P(X\geq b)}=\frac{\int_b^{a}X dP+\int_a^{\infty}X dP}{P(X\geq b)} \leq \frac{a\int_b^{a} dP+\int_a^{\infty}X dP}{P(X\geq b)}=\frac{a\int_b^{a} dP+\int_a^{\infty}X dP}{\int_b^{a} dP + P(X\geq a)}$$
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 $$\frac{\frac{\int_a^{\infty}X dP}{P(X\geq a)}\int_b^{a} dP+\int_a^{\infty}X dP}{\int_b^{a} dP + P(X\geq a)} = \frac{\int_a^{\infty}X dP}{P(X\geq a)} = E(X|X\geq a)$$.
• Sorry, can I ask one more thing? Why is $\int_b^a X dP\leq a \int_b^a dP$? – Wombat Jul 11 '20 at 22:21