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Let $X$ be a Loss random variable (Positive values of X represents Losses) and let $p \in (0,1)$. I know that the Value at Risk at level $p$ of $X$ is defined as:

$$VaR_p(X) = inf{\{x \in \mathbb{R} : F(x) \ge p \}}= inf{\{x \in \mathbb{R} : P[X \gt x] \le 1- p \}}$$

(Also this infimum is equal to the minimum value because $F(VaR_p(X))\ge p$). My problem is the interepretation of this quantity:

  1. In some books (for example: Loss Models) $Var_p(X)$ is interpreted as the minimum capital required such that the probability of being insolvent is at most $1-p$: that is : $P[X \gt VaR_p(X)] \le 1-p$. This interpretation is fine with me.

  2. In some other references and books (for example Wikipedia) $VaR_p(X)$ is interpreted as the maximum possible loss (at the level $p$) such that the probability of loss being less than $VaR_p(X)$ is at least $p$: that is: $P[X \le VaR_p(X)] \ge p$

This second definition of maximum possible loss doesn't make sense to me because formally the definition of $VaR_p(X)$ is with an infimum (which coincides with the minimum)

I know that the value at risk is also equal to:

$$VaR_p(X) = sup{\{x \in \mathbb{R} : F(x) \lt p \}}= sup{\{x \in \mathbb{R} : P[X \gt x] \gt 1- p \}}$$

But if we try to intepret the $VaR_p(X)$ using this definition as a maximum possible loss it would be: The maximum possible Loss such that the probability of having a loss $X$ less than $VaR_p(X)$ is less than $p$ but again it still doesn't make sense to me.

I would really appreciate if someone can help me understanding this concept.

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    $\begingroup$ I have trouble following your reasoning. But isn't $P[X \gt VaR_p(X)] \le 1-p$ formally equivalent to $P[X \le VaR_p(X)] \ge p$ ? In English saying "If I have 1000 in capital, there is a 10% chance I will go bankrupt tomorrow", is pretty much the same as saying "If I have 1000 in capital there is a 90% I will not go bankrupt tomorrow" ( i made these numbers up, of course). What substantive difference do do you see between interpretation 1 and 2? $\endgroup$ – noob2 Dec 26 '20 at 13:18
  • $\begingroup$ My problem is when we think of $VaR_p(X)$ as a maximum possible loss, not minimum capital. For instance if we take $VaR_p(X) + 1$, this is a greater number such that: $P[X \gt VaR_p(X)+1] \le 1-p$ and $P[X \le VaR_p(X)+1] \ge p$. So $VaR_p(X)$ would not be a maximum number that has this properties. $\endgroup$ – user128422 Dec 26 '20 at 15:34

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