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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.

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  • $\begingroup$ what's your assumption on $X$? $\endgroup$
    – CABLE
    Commented Jul 11, 2020 at 11:59

1 Answer 1

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

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

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  • $\begingroup$ 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. $\endgroup$
    – Malekai
    Commented Jul 11, 2020 at 12:30
  • $\begingroup$ 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. $\endgroup$
    – CABLE
    Commented Jul 11, 2020 at 12:34
  • $\begingroup$ @LogicalBranch You may want to check out this post for potential information sources if you have interest in QF: quant.stackexchange.com/questions/38862/… $\endgroup$
    – amdopt
    Commented Jul 11, 2020 at 13:31
  • $\begingroup$ Thank you so much! That helped a lot! :-) $\endgroup$
    – Wombat
    Commented Jul 11, 2020 at 19:32
  • $\begingroup$ Sorry, can I ask one more thing? Why is $\int_b^a X dP\leq a \int_b^a dP$? $\endgroup$
    – Wombat
    Commented Jul 11, 2020 at 22:21

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