0
$\begingroup$

anyone can provide solution or some idea to the following question? thanks Picture

$\endgroup$
4
$\begingroup$

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.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.