# Is Value-at-Risk translation invariant?

Let: $X=V_1-V_0R_0$ where $R_0$ is the interest rate. Then, is it so that this risk measure is Translation Invariant as:

$\textit{VaR}_{\alpha}(X)=\textit{VaR}_{\alpha}(V_1-V_0R_0)=V_0+\textit{VaR}_{\alpha}(V_1)=V_0+g(F^{-1}_{V_{1}}(\alpha))$?

Appreciate for anyone clarifying this. Thanks

• What is $g()$ above? – Richard Jan 13 '16 at 10:09
• We have: $F^{-1}_{g(X)}(1-p)=g(F^{-1}(1-p))$ – Elekko Jan 13 '16 at 12:33
• Still: how is $g()$ defined? What is it supposed to be. Is $g(x) =x^2$ ? $g$ just pops up in the last term, why? – Richard Jan 13 '16 at 12:38

Translation invariance of a risk measure $\rho$ is defined as $$\rho(X+k) = \rho(X)-k,$$ where $X$ is a random variable such that $\rho(X)$ exists and $k$ is a constant. The meaning is that if I add an amount $k$ to my risky positions then the risk is reduced by this amount.
For VaR we consider the case that $X$ has a continuous distribution and that it is a profit and loss random varibale. Then $$VaR_\alpha(X) = -F^{-1}_{1-\alpha}(X)$$ and $$P[-VaR_\alpha(X) \le X] = 1-\alpha.$$ Note that $VaR$ is a posive number and e.g. for $\alpha=99\%$ the quantile $F^{-1}_{1-\alpha}(X)$ is a negative number.
It also holds that $$P[-VaR_\alpha(X)+k \le X+k] = P[-VaR_\alpha(X) \le X] = 1-\alpha,$$ and thus $VaR_\alpha(X+k) = VaR_\alpha(X)-k$.