For what values of $\alpha <\frac 12$ occurs $CVaR_{\alpha}(X) \ge VaR_{\alpha} (X)$

Question

The payoff function for an investment is modeled with a random variable $X$ with a density function $$f(x)=\frac 12 e^{-|x|}$$ For what values of $\alpha <\frac 12$ occurs $$CVaR_{\alpha}(X) \ge VaR_{\alpha} (X)$$

My problems:

1. I'm not sure if I got the absolute value right - when calculating the distribution function I rejected $x<0$ because $VaR$ is defined for $\alpha \in (0,1)$ (here $\alpha <\frac 12$ )
   
2. Assuming my approach to the absolute value was correct, I arrived at an inequality from which it is difficult to derive the relation to $\alpha$
   

Below is my solution:
$VaR_{\alpha}(X)=-F_X^{-1}(\alpha)$
$CVaR_{\alpha}(X)=-\mathbb E(X | X \le -VaR_{\alpha}(X))=-\mathbb E(X | X \le F_X^{-1}(\alpha))$

For $t> 0$ (other cases are not needed for this task) the distribution function is represented by the formula: $$F_X(t)=\int_{0}^t \frac{1}{2}e^{-x} dx=-\frac 12 e^{-x} \bigg|_0^t =-\frac 12 e^{-t}+\frac 12$$ $F_X^{-1}$: $$-\frac{1}{2}e^{-t}+\frac 12=y \Rightarrow \frac{1}{2}e^{-t}=\frac 12 -y \Rightarrow e^{-t}=1-2y \Rightarrow -t=\ln (1-2y) \Rightarrow t=-\ln(1-2y), y\in (0,\frac 12)$$ So: $$VaR_{\alpha}(X)=-F_X^{-1}(\alpha)=\ln(1-2\alpha)$$ Calculate $CVAR_{\alpha}$: \begin{align} CVaR_{\alpha}(X) &= - \frac{1}{\mathbb P(X \le F_X^{-1}(\alpha ))} \int_{-\infty}^{F_X^{-1}(\alpha)} x \cdot f(x) dx \\ &= - \frac{1}{F_X \left( F_X^{-1}(\alpha ) \right)} \int_{-\infty}^{-\ln (1-2\alpha) } x \cdot \frac 12 e^{-|x|} dx \\ &= -\frac 1{2\alpha} \left( \int_{-\infty}^0 x e^x dx+ \int_0^{-\ln(1-2\alpha)} x e^{-x} dx\right) \\ &= -\frac{1}{2\alpha} \left( xe^x \bigg|_{-\infty}^0-\int _{-\infty}^0 e^xdx + (-xe^{-x}) \bigg|_0^{\ln(1-2\alpha)}-\int_0^{\ln(1-2\alpha)} (-e^{-x})dx \right) \\ &= -\frac{1}{2\alpha} \left( (0-0)-(1-0) + (-\ln(1-2\alpha)e^{-\ln(1-2\alpha)}-0) -(e^{-\ln(1-2\alpha)}-1) \right) \\ &= \frac 1{2\alpha} e^{-\ln(1-2\alpha)}(1+\ln(1-2\alpha)) \\ &=\frac{1}{2\alpha (1-2\alpha)} (1+\ln(1-2\alpha)) \end{align}

Inequality: $$CVaR_{\alpha}(X) \ge VaR_{\alpha} (X)$$ $$\frac{1}{2\alpha (1-2\alpha)} (1+\ln(1-2\alpha)) \ge \ln(1-2\alpha)$$ $$ 1+\ln(1-2\alpha) \ge 2\alpha (1-2\alpha) \ln(1-2\alpha)$$ $$(1-2\alpha +4\alpha^2)\ln(1-2\alpha)\ge -1$$

Answer 1

I will use the definitions you provided of VaR and CVaR. We can see that $$F(c)=\begin{cases} \frac{1}{2}\int_{-\infty}^ce^{x}dx=e^c/2&c\leq 0\\ \frac{1}{2}+\frac{1}{2}\int_{0}^ce^{-x}dx=\frac{1}{2}+\frac{1}{2}(1-e^{-c})=1-e^{-c}/2&c>0 \end{cases}$$ Now, the exact inverse exists and for $u \in (0,1)$ $$F^{-1}(u)=\begin{cases} \ln(2u)&u\leq 1/2\\ -\ln(2(u-1))&u>1/2 \end{cases}\implies \textrm{VaR}(u):=-F^{-1}(u)=\begin{cases} -\ln(2u)&u\leq 1/2\\ \ln(2(u-1))&u>1/2 \end{cases}$$ Now let $u<1/2$. Then $F^{-1}(u)<0$. We then get $$\textrm{CVaR}(u):=-\frac{\int_{-\infty}^{F^{-1}(u)}xe^xdx/2}{P(X\leq F^{-1}(u))}=-\frac{\int_{-\infty}^{\ln(2u)}xe^xdx}{2u}=1-\ln(2u)$$ We conclude that $\textrm{CVaR}(u)>\textrm{VaR}(u),\,\forall u <1/2$.