Marking a Variance Swap Using Realized and Forward Variance
Summary
The document derives an approximate mark-to-market value for a variance swap partway through its life. It splits total realized variance into the portion already observed and the expected variance over the remaining period, then weights each by its number of observations. The worked example uses 40 elapsed days with 20% realized volatility, a 22% break-even strike for the remaining term, and a 25% strike for the original swap. It reports a payoff of about $154 million for the stated notional example.
The derivation relies on the variance-swap payoff convention shown, annualization by 252, and the assumption that the remaining variance expectation can be inferred from the shorter-dated swap's break-even strike. The document flags that its example corresponds to a very large vega notional. Its indexing and day-count notation are not fully consistent, so the calculation should be checked against the contract's precise conventions before use.
Key ideas
- Expected swap value can be calculated by taking the expectation of its payoff.
- Split the squared-return sum into realized observations and future observations.
- Annualized variance over each segment contributes in proportion to that segment's observation count.
- A remaining-term break-even variance strike can inform the expected variance for the unobserved period.
- Payoff scaling and contract conventions materially affect the monetary mark.
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# Structured question on mark-to-market value of a variance swap
# Structured question on mark-to-market value of a variance swap
anyone can provide solution or some idea to the following question? thanks
## Answer by will (score 4, accepted)
https://quant.stackexchange.com/a/54700
The (undiscounted) value of any derivative is the expected value of the payoff.
So the (undiscounted) value of a varswawp is:
$$\mathbb{E}\left[ \mathrm{Notional} \cdot 10000 \cdot \left( K^2 - \frac{252}{N} \sum_{i=1}^N \left(\ln \frac{S_i}{S_{i-1}} \right)^2\right) \right]$$
Where, we can move everything that's static outside of the expectation, and then separate the terms of the sum too:
$$\mathrm{Notional} \cdot 10000 \cdot \left( K^2 - \frac{252}{N} \sum_{i=1}^N \mathbb{E}\left[ \left(\ln \frac{S_i}{S_{i-1}} \right)^2 \right] \right)$$
which we can then farther serparate:
$$\mathrm{Notional} \cdot 10000 \cdot \left( K^2 - \frac{252}{N} \left( \sum_{i=1}^m \mathbb{E}\left[ \left(\ln \frac{S_i}{S_{i-1}} \right)^2 \right] + \sum_{i=m+1}^N \mathbb{E}\left[ \left(\ln \frac{S_i}{S_{i-1}} \right)^2 \right] \right) \right)$$
And then we can move the expectations back outside each of the sums (it will be clear why later):
$$\mathrm{Notional} \cdot 10000 \cdot \left( K^2 - \frac{252}{N} \left( \mathbb{E}\left[\sum_{i=1}^m \left(\ln \frac{S_i}{S_{i-1}} \right)^2 \right] + \mathbb{E}\left[\sum_{i=m+1}^N \left(\ln \frac{S_i}{S_{i-1}} \right)^2 \right] \right) \right)$$
where $m$ is some point in time part way through the life of the varswap. If we set $m$ such that it's 40 (i.e. 2 months have 40 days), and given that we have the realised voolatility for those 40 days (20%):
$$ \sqrt{\frac{252}{40} \sum_{i=1}^{40}\left(\ln\frac{S_i}{S_{i-1}} \right)^2} = 20\%$$ $$ \sum_{i=1}^{40}\left(\ln\frac{S_i}{S_{i-1}} \right)^2 = 20\%^2 \cdot \frac{40}{252}$$
and we know that the striek of the 10m varswap is 22% - where the break even strike of varswap is the strike such that it has a value of zero (keeping the values of $i$ and $N$ to mean the same as in the original 1y varswap):
$$\mathbb{E}\left[ \mathrm{Notional} \cdot 10000 \cdot \left( K^2 - \frac{252}{N-41} \sum_{i=41}^{N} \left(\ln \frac{S_i}{S_{i-1}} \right)^2\right) \right] = 0$$
so again, take out the static parts, and split the expectation, then move the strike over the equals sign:
$$ \mathbb{E}\left[ \sum_{i=41}^{N} \left(\ln \frac{S_i}{S_{i-1}} \right)^2 \right] = K^2 \cdot \frac{N-41}{252} = 22\%^2 \cdot \frac{N-41}{252}$$
Then sub these back in to the broken up payoff we created above: $$\mathrm{Notional} \cdot 10000 \cdot \left( K^2 - \frac{252}{N} \left( \mathbb{E}\left[\sum_{i=1}^m \left(\ln \frac{S_i}{S_{i-1}} \right)^2 \right] + \mathbb{E}\left[\sum_{i=m+1}^N \left(\ln \frac{S_i}{S_{i-1}} \right)^2 \right] \right) \right)$$
$$\mathrm{Notional} \cdot 10000 \cdot \left( K^2 - \frac{252}{N} \left( 20\%^2 \cdot \frac{40}{252} + 22\%^2 \cdot \frac{N-41}{252} \right) \right)$$
multiply out all of the factors you have: $$\mathrm{Notional} \cdot 10000 \cdot \left( K^2 - \frac{40 \cdot 20\%^2 + (N-41) \cdot 22\%^2 }{N} \right)$$
Where, you can quite easily see here that the expected realised variance is the time weighted sum of the realised variance so far, and the expected realised variance in the future.
putting all the numbers in:
$$\mathrm{$1m} \cdot 10000 \cdot \left( 25\%^2 - \frac{40 \cdot 20\%^2 + (N-41) \cdot 22\%^2 }{252} \right) = \mathrm{$1m} \cdot 10000 \cdot \left( 25\%^2 - 21.69\%^2 \right)$$
which gives a payoff of ~\$154m. This example though is a varswap with a vega notional of \$50m, which is huge.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.