Hi
@theapplecrispguy Yes, but only because occasionally (or rarely) the 95.0% VaR will be represented by an author as the 0.050 VaR, treating them as the same idea. In any realistic use case, we are referring to the losses that happen in the worst 5.0% of times, the losing tail. It would not be realistic/sensible to speak of the losses that occur in the worst 95.0% of times (although mathematically it works of course). But, again, the idea is that we have a cumulative distribution function here, given by F(x) = p = x^3/125. Solving for x, we have x = (p*125)^(1/3). So for example,
- at p = 5%, x = (5%*125)^(1/3) = $1.84
- at p = 10%, x = (10%*125)^(1/3) = $2.32
- at p = 20%, x = (20%*125)^(1/3) = $2.92
- at p = 50%, x = (50%*125)^(1/3) = $3.97; i.e., the median
- at p = 95%, x = (95%*125)^(1/3) = $4.92
It's not a distribution of losses, it's a distribution of future values. We always want to be mindful of the difference, sometimes the distribution is losses. As this is future values, the risk tail is "down" at p = 5%, it is not "up" at p = 95%, so that either a 95% or 0.05 VaR would be realistically referring to the p = 5%. At the same time, there are valid usages of VaR for lower confidence levels, so we could have a
90% VaR (given by the loss of $5.00 minus
$2.32) or a even an
80% VaR (given by the loss of $5.00 minus
$2.92). I hope that helps!
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