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.

1 Answer

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.