When Expected Shortfall Equals Conditional Tail Loss

Summary

The document examines an alternative expression for Expected Shortfall (ES): the average loss conditional on the loss being at or above its Value at Risk (VaR) quantile. It defines VaR through the inverse cumulative distribution function, then attempts to show that the probability-weighted tail loss divided by the tail probability equals this conditional expectation. The attempted derivation gets stuck at the event-probability step.

The accepted response uses the definition of conditional expectation: the expected loss restricted to the tail event, divided by the probability of that event, is the conditional mean given the event. This establishes the proposed identity when the tail event has probability equal to one minus the confidence level. The excerpt does not discuss distributions with probability mass at the quantile, where the tail-event probability may differ from that value and the simple equality may need adjustment.

Key ideas

• VaR is defined as the quantile of the loss distribution.
• Conditional expected tail loss divides the event-weighted loss by the probability of the tail event.
• The stated ES identity follows when the tail event has probability one minus the confidence level.
• Probability mass at the quantile can complicate the simple conditional-expectation formulation.

Tags

# Expected Shortfall alternative formulation


# Expected Shortfall alternative formulation












Define:

$$q_\alpha(F_L)=F^{\leftarrow}(\alpha)=\inf\lbrace{x\in \mathbb{R}\mid F_L(x)\geq \alpha\rbrace}=VaR_\alpha(L)$$

I want to prove that:

$$ES_\alpha = \frac{1}{1-\alpha}\mathbb{E}[\mathbb{1}_{\lbrace{ L\geq q_\alpha(L)\rbrace}}\cdot L] \overset{!!!}{=}\mathbb{E}[L\mid L\geq q_\alpha(L)] $$

I get stuck as:

$$\mathbb{E}[\mathbb{1}_{\lbrace{ L\geq q_\alpha(L)\rbrace}}\cdot L]= \mathbb{E}[\mathbb{E}[\mathbb{1}_{\lbrace{ L\geq q_\alpha(L)\rbrace}}\cdot L\mid L\geq q_\alpha(L)]] = \mathbb{E}[\mathbb{1}_{\lbrace{ L\geq q_\alpha(L)\rbrace}}\cdot\mathbb{E}[L\mid L\geq q_\alpha(L)]\ ]$$

Now I would like to use that $\Pr(L\geq q_\alpha(L) \ )=1-\alpha$, but I don't know how to proceed.

## Answer by Gordon (score 2, accepted)

https://quant.stackexchange.com/a/19131

Note that \begin{align*} \mathbb{E}\big(L \mid L\geq q_\alpha(L)\big) &= \frac{\mathbb{E}\big(\pmb{1}_{\{L\geq q_\alpha(L)\}} L\big)}{\mathbb{P}\big(L\geq q_\alpha(L) \big)}. \end{align*} The formula follows immediately.

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