How to compute $VaR_\lambda(X_n)$?

Question

I have the following problem. Let $X\in L^\infty$ and $\lambda\in(0,1)$, we define $$VaR_\lambda(X)=\inf\{m\in\mathbb{R} : \mathbb{P}(X+m<0)\leq\lambda\}.$$ I have to prove that $VaR_\lambda(X_n)=0$, where $X_n:=1_{[ \lambda+\frac{1}{n},1]}.$ I calculated $\mathbb{P}(X_n+m<0)$ in this way: $$\mathbb{P}(X_n+m<0)=\int_\mathbb{R} 1_{[\lambda+\frac{1}{n},1]} 1_{(-\infty,-m)} \frac{1}{1-\lambda-\frac{1}{n}} \, dx=\left\{\begin{array}{ll} \frac{-m-\lambda-\frac{1}{n}}{1-\lambda-\frac{1}{n}} &\ \text{ se}\quad -1<m\leq -\lambda-\frac{1}{n}\\ 1&\ \text{ se}\quad m\leq -1 \\ 0&\ \text{ se} \quad m> -\lambda-\frac{1}{n}. \end{array}\right.$$ Now, $\frac{-m-\lambda-\frac{1}{n}}{1-\lambda-\frac{1}{n}}\leq \lambda$ if and only if $m\geq \lambda^2+\lambda(\frac{1}{n}-2)-\frac{1}{n}$, the minimum of this parabola is attained in $\lambda=1-\frac{1}{2n}$, then we found $m\geq-1-\frac{1}{4n^2}$ but we have the condition $-1<m\leq -\lambda-\frac{1}{n}$ too, so it must be $-1<m\leq -\lambda-\frac{1}{n}$. At the same time, if $m> -\lambda-\frac{1}{n}$, $\mathbb{P}(X_n+m<0)=0\leq\lambda$, so I will say that $$VaR_\lambda(X_n)=\inf\{m>-1\}=-1.$$ I don't know what is wrong... maybe I have to found the maximum of the parabola?

Answer 1

First, I assume $n>0$ and that you mean $X_n:=1_{u\in [\lambda+1/n,1]}=1_{u \geq \lambda+1/n}$ for $u\sim \text{Unif}(0,1).$ If so, then note that

$$P(X_{n}<-m)=\begin{cases} 0, & m\geq0\\ 1, & m<-1\\ \lambda+1/n & m\in[-1,0) \end{cases}.$$

It's clear $P(X_{n}<-m)\leq \lambda$ for $m\geq 0$, so the infimum of the set of such $m$ is zero, as desired.