# Tail Value at Risk of Normal Distribution

For a random variable $X$, Tail-value-at-risk is denoted as $\operatorname{TVaR}_p(X) = \operatorname E(X \mid X>\pi_p) = \dfrac{ \int_{\pi_p}^\infty xf(x) \, dx}{1-F(\pi_p)}$, where $\pi_p=\operatorname{VaR}_p=$ the value-at-risk $=$ the value such that $P(X>\pi_p)=1-p$.

But I struggle to see how the second result was proved. I'm either not sure how to do the calculus or I don't know what tricks they used to get there, but I can't figure it out. Do you understand how they got there?

• Hi @ereHsaWyhsipS, which book is the above screenshot from? Commented Oct 18, 2022 at 17:47

First Part

Since $$P(X > \pi_p) = 1 - p$$ we have $$P[(X - \mu)/\sigma > (\pi_p - \mu)/\sigma] = 1 - p.$$

Since $$Z = (X - \mu)/\sigma \sim N(0,1)$$, the standard normal distribution function $$\Phi$$ specifies $$\Phi(z) = P(Z \leqslant z)$$ and $$1 - \Phi(z) = P(Z > z).$$

Thus,

$$P[(X - \mu)/\sigma > (\pi_p - \mu)/\sigma] = 1 - p \\ \implies 1- \Phi[(\pi_p - \mu)/\sigma] = 1 - p \\ \implies \Phi[(\pi_p - \mu)/\sigma] = p\\ \implies (\pi_p - \mu)/\sigma = \Phi^{-1}(p)\\ \implies \text{VaR}_p(X) =\pi_p = \mu + \sigma \Phi^{-1}(p)$$

Second Part

Note that

$$\int_{\pi_p}^\infty x f(x) \, dx= \int_{-\infty}^\infty x f(x) \, dx - \int_{-\infty}^{\pi_p} x f(x) \, dx = \mu - \int_{-\infty}^{\pi_p} x f(x) \, dx.$$

Changing variables with $$x = \mu + \sigma z$$ and noting that $$\phi(z) = \sigma f(\mu + \sigma z)$$ and $$dx = \sigma dz$$, we have

\begin{align}\int_{-\infty}^{\pi_p} x f(x) \, dx &= \int_{-\infty}^{(\pi_p - \mu)/\sigma}(\mu + \sigma z) \phi(z) \, dz \\ &= \mu \int_{-\infty}^{(\pi_p - \mu)/\sigma} \phi(z) \, dz + \sigma \int_{-\infty}^{(\pi_p - \mu)/\sigma} z\phi(z) \, dz \\ &= \mu \Phi[(\pi_p - \mu)/\sigma] + \sigma \int_{-\infty}^{(\pi_p - \mu)/\sigma} z\phi(z) \, dz \end{align}

As shown in the first part, $$\Phi[(\pi_p - \mu)/\sigma] = p.$$

The key here is the observation that the standard normal density function satisfies $$z \phi(z) = -\phi'(z)$$.

Hence,

\begin{align}\int_{\pi_p}^{\infty} x f(x) \, dx &= \mu - \mu p + \sigma \int_{-\infty}^{(\pi_p - \mu)/\sigma}\phi'(z) \, dz \\ &= \mu(1-p) + \sigma \phi[(\pi_p - \mu)/\sigma] \\ &= \mu(1-p) + \sigma \phi[\Phi^{-1}(p)],\end{align}

and

$$\text{TVaR}_p(X) = \frac{\mu(1-p) + \sigma \phi[\Phi^{-1}(p)] }{1 - F(\pi_p) } = \frac{\mu(1-p) + \sigma \phi[\Phi^{-1}(p)] }{1 - p } \\ = \mu + \sigma \frac{\phi[\Phi^{-1}(p)] }{1-p}$$

• I think there is a typo in $\int_{-\infty}^{\pi_p} x f(x) \, dx = \int_{-\infty}^{(\pi_p - \mu)/\sigma}(\mu + \sigma z) \phi(z) \, dz \\ = \mu \int_{-\infty}^{(\pi_p - \mu)/\sigma} \phi(z) \, dz - \sigma \int_{-\infty}^{(\pi_p - \mu)/\sigma} z\phi(z) \, dz \\ = \mu \Phi[(\pi_p - \mu)/\sigma] - \sigma \int_{-\infty}^{(\pi_p - \mu)/\sigma} z\phi(z) \, dz$, the "minus" should be plus? Commented Jan 6, 2022 at 12:44
• 2. In $\int_{-\infty}^{\pi_p} x f(x) \, dx = \mu - \mu p + \sigma \int_{-\infty}^{(\pi_p - \mu)/\sigma}\phi'(z) \, dz$, the left hand should be $\int_{\pi_p}^{\infty} x f(x) \, dx$ Commented Jan 6, 2022 at 13:00
• @IronMan: Thank you. I will make those edits.
– RRL
Commented Jan 6, 2022 at 15:07