Expected Shortfall Formula in terms of P

Question

Let $X$ be a continuous random variable and $Q_x$ is the associated quantile function. Show that expected shortfall $ES_X[p]$ at the confidence level $p$ which is defined as
$$ES_X[p]=\Bbb E[X|X\leq Q_x(1-p)]$$ has the representation $$ES_X[p]=\frac{1}{1-p}\int_0^{1-p} Q_x(a)da.$$

Can some one give me a hint for this?

I know the definition of quantile function $Q_X(p)=\inf\{x: F_x \geq p\}$, I can think of it on an intuitive level, but want some thoughts to get started mathematically

Answer 1

Gordon's answer is spot on. Another way to see it though, would be using Bayes formula and a change of variable.

\begin{align*} ES_X(p) &=E\left(X \mid X\le Q_X(1-p)\right)\\ &=\int_{-\infty}^{\infty} x\, \phi\left(x \mid x\le Q_X(1-p)\right) dx \\ &=\int_{-\infty}^{\infty} x\, \frac{\phi\left(x\le Q_X(1-p) \mid x \right)\phi(x)}{\int_{-\infty}^{\infty} \phi\left(u\le Q_X(1-p) \mid u \right)\phi(u) du } dx \end{align*} Now simply note that $$\phi\left(x \le Q_X(1-p) \mid x \right) = 1\left\{ x \le Q_X(1-p) \right\} $$ to be able to rewrite the integral as in the second line in Gordon's answer \begin{align*} ES_X(p) &= \frac{E\left(X\, 1_{X\le Q_X(1-p)} \right)}{P(X\le Q_X(1-p))} \\ &= \frac{1}{1-p} \int_{-\infty}^{Q_X(1-p)} x\, \phi(x) dx \end{align*} a change of variable $x \to y=\Phi(x)$ (the cdf of $X$) finishes the demonstration.

Answer 2

Note that $Q_X$ is the pseudo-inverse of the distribution function $F$, and for any uniform random variable $U$ over $[0, 1]$, the random variable $Q_X(U)$ has the same distribution as $X$. Moreover, since $X$ is continuous, $Q_X$ is strictly increasing. Proofs of these facts are purely mathematical, and can be discussed some other questions.

Here, we assume these facts are true. Then \begin{align*} ES_X(p) &=E\left(X \mid X\le Q_X(1-p)\right)\\ &=\frac{E\left(X\, 1_{X\le Q_X(1-p)} \right)}{P(X\le Q_X(1-p))}\\ &=\frac{E\left(Q_X(U)\, 1_{Q_X(U)\le Q_X(1-p)} \right)}{P(Q_X(U)\le Q_X(1-p))}\\ &=\frac{E\left(Q_X(U)\, 1_{U\le 1-p} \right)}{P(U\le 1-p)}\\ &=\frac{1}{1-p}\int_0^{1-p} Q_X(a) da. \end{align*}