1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.