Tail Value at Risk of Normal Distribution

Solution 1:

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

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