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The loss $L$ has the $\lambda_1^2$ distribution, i.e. the distribution of the random variable $X^2$, where $X$ has a standard normal distribution. Proof that $VaR_c(L)=(\Phi^{-1}(\frac{c+1}2))^2$, where $\Phi$ is the cumulative distribution function of the normal distribution.

Edit: definition $VaR_c(L)$ is here: https://en.wikipedia.org/wiki/Value_at_risk#Mathematical_definition

My try: $$VaR_c(L)=F_L^{-1}(c)$$ $$F_L(t)=\mathbb P(L\le t)=\mathbb P(X^2 \le t) =\mathbb P(-\sqrt t \le X \le \sqrt t) = F_X(\sqrt t)-F_X(-\sqrt t)=\Phi(\sqrt t)-\Phi(-\sqrt t)$$

However I don't know what I can do to finish my task.

john1235
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1 Answers1

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Note $\Phi(-x)=1-\Phi(x)$ so $F(x)=2\Phi(\sqrt{x})-1$. We conclude: $2\Phi(\sqrt{x})-1=u\implies x=(\Phi^{-1}((1+u)/2))^2$.

Snoop
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