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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.

kurtosis
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Wombat
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1 Answers1

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

CABLE
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  • 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. – Malekai 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: https://quant.stackexchange.com/questions/38862/what-are-the-quantitative-finance-books-that-we-should-all-have-in-our-shelves – 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
  • Because $x \leq a$ when $x \in [b,a]$. – CABLE Jul 12 '20 at 05:12
  • Thank you. :-) Sorry, that was a stupid question. I'm still not so used to integrating a random variable with respect to the probability measure, but all good. Thanks for clarifying. – Wombat Jul 12 '20 at 09:11