Interpreting VaR in a Rare-Event Coin-Toss Example
Summary
The document explains a sign convention in value at risk using a bet that pays $100 if seven coin flips are not all heads and loses $12,700 if they are. Under the quoted loss convention, gains are negative losses, so the 99% VaR is −$100: the common outcome is a $100 gain, while the large loss falls in the less-than-1% tail. VaR describes a quantile threshold, not the size of the rare loss beyond that threshold.
The answer also clarifies the probability statement by comparing losses with the threshold and with zero. Since the common outcome is a gain, its loss is negative; the example's wording about losses below zero can therefore be confusing when read alongside the −$100 threshold. The explanation is limited to this discrete illustration and does not address broader choices of VaR convention or the shortcomings of VaR as a tail-risk measure.
Key ideas
- VaR can be expressed using losses, making gains negative values.
- In the coin-toss example, the common outcome is a gain and the rare outcome is a large loss.
- A VaR threshold identifies a quantile and does not measure the size of losses beyond it.
- The sign convention is essential for interpreting the example's probability statements.
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# Understanding example on VaR
# Understanding example on VaR
Trying to understand the example below on VaR in Wikipedia. I don't really understand how the 1% VaR is being defined here. Firstly, shouldn't it be 1% Var is 100 since its the amount her looses? And secondly, I dont understand this statement: "the probability that he loses less than zero (which is impossible) is less than 99%.". Given 1% Var = -100 as per their definition shouldnt it be "the probability that he loses less than -100 is less than 99%." and not 0?
> More formally, p VaR is defined such that the probability of a loss greater than VaR is less than or equal to p while the probability of a loss less than VaR is less than or equal to 1−p. For instance, assume someone makes a bet that flipping a coin seven times will not give seven heads. The terms are that he gains 100 if it doesn't happen (with probability 127/128) but loses $12,700 if it does (with probability 1/128). The 1% VaR is then -100, because the probability that he loses more than that is less than 1% while the probability that he loses less than zero (which is impossible) is less than 99%.
Thanks
## Answer by AK88 (score 2)
https://quant.stackexchange.com/a/39795
The way I understood it is as follows:
- If you DO NOT get 7 heads out of seven flips, then you make 100\$;
- If you DO get 7 heads out of seven flips, then you lose 12 700\$;
The probability of NOT getting seven heads is $\frac{127}{128} = 99,21875\%$ and $1-p = 0.78125\%$. Therefore, with 99% confidence level VaR you are still making 100\$ (I think that's why there is a minus sign). Since VaR ignores tail events, you are not losing at 99% VaR. Hence the statement:
> ... the probability that you lose MORE than -100\$ is less than 1%, while the probability that he loses LESS than zero (which is impossible) is less than 99%.
Maybe taking a look at the original statement by Einhorn may be helpful:Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.