Proof That Conditional Tail Means Decrease as the Threshold Rises

Summary

The document proves that for thresholds a ≥ b, the expected value of a random variable conditional on being at least a is no smaller than its expected value conditional on being at least b, assuming the higher-threshold event has positive probability. This is a monotonicity property of conditional tail averages, relevant to understanding expected shortfall and related risk measures.

The proof splits the lower-threshold tail into observations between b and a and those at or above a. Values in the first group are bounded above by a, while the conditional mean of the upper tail is at least a. Comparing these contributions shows that adding the lower-valued group cannot raise the conditional average. The argument is concise and does not require a particular distribution, but it relies on the conditional expectations being defined and finite; it does not discuss how the result extends to alternative tail conventions or more general risk measures.

Key ideas

• A conditional tail mean at a higher threshold is at least as large as one at a lower threshold.
• The lower tail segment contains values no greater than the higher threshold.
• The mean of values above the higher threshold is itself at least that threshold.
• Combining the two groups shows that adding the lower segment cannot increase the average.

Tags

# Expected Shortfall monotonicity


# 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.

## Answer by CABLE (score 2, accepted)

https://quant.stackexchange.com/a/55588

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

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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