# 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

## 1 Answer

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

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

• I know this might be off topic but who do you guys read these equations? Is there a guide/book? How can you tell what equation does what? This looks incredibly fascinating. – LogicalBranch Jul 11 '20 at 12:30
• This is just simple algebraic manipulation. Usually the logic is this: you want something to be true, you then work backward to see if the requirements are satisfied. – CABLE Jul 11 '20 at 12:34
• @LogicalBranch You may want to check out this post for potential information sources if you have interest in QF: quant.stackexchange.com/questions/38862/… – amdopt Jul 11 '20 at 13:31
• Thank you so much! That helped a lot! :-) – Wombat Jul 11 '20 at 19:32
• 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