For a Variance swap we can split the pnl into a realized part and a "forward going" part. To be more precise:

Assume we enter the trade at t0, and the variance swap has tenor T and a strike $Kvar$. At time $t0< t < T$ we look at the value of variance swap. We will see that: $V (t_0; t) = \lambda ( Var_{realized} - Kvar ) + (1-\lambda)(Knew - Kvar )$ where $\lambda$ is the fraction of time that has passed.

It looks to me that such a decomposition is not possible for correlation swaps but only for covariance swaps. Since the correlation swap depends on realized volatility. And realized volatility can not be decomposed in this fashion.

To be more precise I will define what I mean with correlation swap. The correlation swap is on N stocks. The swap starts at $t_0$ and ends on $T$ assume that between t0 and T there are exactly M trading days. At maturity $T$ the payout will be $\hat{\rho} - K$, where $\hat{\rho}$ is the realized correlation (to be defined below).

For $i=1,..,N$ we have the daily realized return $R_i(t_k) = ln\left(\frac{S_i(t_{k+1})}{S_i(t_k)} \right)$. The realized correlation $\hat{\rho}$ over time interval $[t_0, T]$ is defined to be $\hat{\rho} = \frac{1}{n(n-1)}\sum_{i,j} \frac{COV(i,j)}{\sigma_i \sigma_j}$. In this equation we define $COV(i,j) = \frac{1}{m-1}\sum_{k = 1}^{M} (R_i(t_k) - \bar{R_i}) (R_j(t_k) - \bar{R_j})$. So COV(i,j) measures the realized correlation between stocks i and j over the time interval [t0, T]. The realized vols $\sigma_i$ and $\sigma_j$ are defined similarly in terms of $R_i$.

Therefore the realized correlation measures for each pair i,j for every trading day what the correlation was. The swap pays out the difference between what was realized (as defined above) and the fixed strike.

What would happen with the pnl if we would look at the trade at time $t0 < t < T$? If the swap linear in time $t_k$ then we could split the price up in a piece that was realized up to t and replace the remainder with a new swap.

So my questions are:

  • Do you agree that this decomp can not be done for correlation swap?
  • If one would do this decomp, how bad is the error?
  • Any alternative ways that are better than the decomp discussed above?


  • $\begingroup$ Can you please be more specific, in a mathematical form, on how your correlation swap is defined? $\endgroup$ – Gordon Jan 18 '16 at 14:10
  • $\begingroup$ Hi gordon, thanks for your comments. I added a mathematical description of the payout. $\endgroup$ – mbison Jan 18 '16 at 18:53
  • $\begingroup$ EspressoLover on NuclearPhynance had a really cool proxy: nuclearphynance.com/Show%20Post.aspx?PostIDKey=179529#179553 $\endgroup$ – mbison Jan 19 '16 at 21:10

I assume that this decomposition is possible in the case of a variance swap as variance is decomposable in the sense that $$ V = VAR(R_1+ \cdots + R_N) \approx \frac1N \sum_{i=1}^N R_i^2 . $$ This for any $n \le N$ we can write $$ V \approx \frac1N \sum_{i=1}^n R_i^2 + \frac1N \sum_{i=n+1}^N R_i^2 = \frac n N \frac1n \sum_{i=1}^n R_i^2 + \frac{N-n}{N} \frac{1}{N-n} \sum_{i=n+1}^N R_i^2. $$ In the formulas above you see the weights depending on the time that has passed relative to the total period and the variance terms.

I do not assume that this is possible for a correlation swap. For realized correlation as well as for implied correlation you get not interpretable terms in the denominator.

  • $\begingroup$ Hi Richard, thanks for your response. Given that the correlation swap is not "linear in time". Do you have any idea of how to proxy the size of the error if one would use the linear approach? How bad can it get if we would "pretend" that it is linear in time? Thanks. $\endgroup$ – mbison Jan 18 '16 at 18:55
  • $\begingroup$ I can try, but please helpe me: do you mean implied correlation? as here in formula 3 ? $\endgroup$ – Richard Jan 20 '16 at 6:52
  • $\begingroup$ Hi Richard, in this case it is not like formula 3. Best way to think of the implied correlation in this case is to consider it a constant value that equals the expected value of the realized correlation. It gives a new swap a 0 value. $\endgroup$ – mbison Jan 20 '16 at 22:11

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