What is VaR of an indicator function

Question

$\operatorname{VaR}_{\lambda}(X):=\inf\{m \in \mathbb R: P(X+m<0)\leq \lambda\}$ and $\lambda \in [0,1]$

Let $A\in \mathcal{F}$ where $(\Omega,\mathcal{F},P)$ is the probability space.

$1.$ Determine $ \operatorname{VaR}_{\lambda}(-1_{A})$

$2.$ Determine $ \operatorname{VaR}_{\lambda}(-X)$ where $X = \exp(\sigma Z + \mu)$ and $Z$ ~ $\mathcal{N}(0,1)$ and $\mu,\sigma \in \mathbb R$.

My attempt:

$1.$ for any $m \in \mathbb R$ we have: $P(-1_{A}+m<0)=P(m<-1_{A})=1-P(1_{A}\leq m)$

now for $\lambda$, we have $1-P(1_{A}\leq m)\leq \lambda \iff 1- \lambda \leq P(1_{A} \leq m)$.

Note that if $ \lambda = 0$ we have $1\leq P(-1_{A} \leq m)$ and if follows that $m \geq 0$ and hence $m = 0$ would be the infimum in this case.

Next if $0<\lambda <1$: I am not sure about this case. Note that if $\lambda > P(A)$ then we could choose the $m=0$. Otherwise if $\lambda \leq P(A)$ , then choose $m = 1$. Is this correct?

$2.$ $P(-X+m<0)=1-P(X\leq m)=1- P(\exp(\sigma Z+\mu)\leq m))=1-P(Z\leq \frac{\log(m)-\mu}{\sigma}) $

and $1-P(Z\leq \frac{\log(m)-\mu}{\sigma})= 1-\Phi(\frac{\log(m)-\mu}{\sigma})\leq\lambda\iff 1-\lambda \leq \Phi(\frac{\log(m)-\mu}{\sigma})$

How should I now calculate the $m$ that fullfills this?

Answer 1

1. Simply write out the probability: $$P(-1_A+m<0)= \begin{cases} 0 & m\ge1\\ P(A) & 0\le m<1\\ 1 & m<0 \end{cases}$$ so the value at risk is $$\text{VaR}_\lambda(-1_A)= \begin{cases} 1 & \lambda<P(A)\\ 0 & \lambda \ge P(A) \end{cases}$$
2. You are almost there, you just need to remember that you are looking for the smallest m which fulfills this. And then use monotonicity of the functions involved. And then you just compute for m. Spoiler

$$1-\lambda=\Phi(\frac{\ln(m)-\mu}{\sigma}) \implies m=\exp(\sigma\Phi^{-1}(1-\lambda)+\mu)$$