Payout of a variance swap at maturity $T$ is proportional to $\left(\frac{252}{N} \sum_{i=0}^{N-1} R_i^2 \right) - \sigma_{\textrm{VS}}^2$ where $R_i \equiv \ln\left( \frac{S_{T_{i+1}}}{S_{T_i}} \right)$ and where strike $\sigma_{\textrm{VS}}^2$ is set such that the payout discounted at inception is equal to $0$. Fine.

Now the variance swap denomination come from the fact that, the normalizing factor $252$ apart, $\sum_{i=0}^{N-1} R_i^2$ is the realized variance of the logarithmic returns (in sampling terms). By definition, I would have rather expected to see a $\frac{1}{N} \sum_{i=0}^{N-1} R_i^2 - \left( \frac{1}{N} \sum_{i=0}^{N-1} R_i \right)^2$, and not seeing the $\left( \frac{1}{N} \sum_{i=0}^{N-1} R_i \right)^2$ bit shows that $\frac{1}{N} \sum_{i=0}^{N-1} R_i$ must have been assumed (for all underlying VSs are written on : liquid stocks, indexes, FX rates etc) equal to $0$.

Why are logarithmic returns supposed to have zero-expectation, i.e. are centered ? Is it really true ? On which time scales is this true if it is ? (Here the VS was daily, but under which other scales do logarithmic returns have zero expectation ?)

  • 2
    $\begingroup$ The assumption that expected returns are zero is pretty common in computing daily variance or standard deviation. That is besause the variance of daily stock returns is about 0.01^2 = 0.0001 and the expectation is about 0.1/252 = 0.004 which when squared is negligible i.e. 0.000016 comapared to the other number. $\endgroup$
    – Alex C
    Mar 26, 2017 at 20:08
  • $\begingroup$ @AlexC Fine by me, modulo the "the variance of daily stock returns is about 0.01^2" bit : where does it come from ? This is observed on the market right ? What about other time scales ? (monthly or weekly, higher frequencies do not exist afaik) $\endgroup$
    – Olórin
    Mar 26, 2017 at 20:15
  • $\begingroup$ I should have said the daily standard deviation of the S&P 500 is about 1% a day; the annual return of S&P500 is about 10% a year. (that's empirical data). For things other than S&P 500 it may be different of course. $\endgroup$
    – Alex C
    Mar 26, 2017 at 21:39
  • 1
    $\begingroup$ I would add that if you adopt the non zero expected value version of the formula, it presents a difficult hedging problem, because your exposure to the variance on any individual day depends in some complex way on the sum of returns so far, whereas in the zero version your exposure to the variance is the same on all the days. $\endgroup$
    – dm63
    Mar 26, 2017 at 21:51
  • $\begingroup$ @AlexC Thx Alex $\endgroup$
    – Olórin
    Mar 26, 2017 at 21:53

2 Answers 2


By market convention, the "variance" in Variance Swaps is computed by the above formula which assumes that the average return is zero.

There are two reasons (at least) why this convention is used:

(1) The expected return for the S&P 500, from historical data, is about 10% per year or a little less, which amounts to about 4 basis points per day. This is small compared to the standard deviation of daily returns which is on the order of 1% per day. So failing to subtract the average return squared introduces a fairly small error in the calculation of variance. It is "close enough" for most purposes.

(2) Perhaps more important, as pointed out by dm63 in a comment above, if the swap had been defined with non zero expected value version of the formula, using actual returns, it would be more difficult to hedge because your exposure to the variance on any individual day depends in some complex way on the sum of returns so far, whereas in the zero version your exposure to the variance is the same on all the days.

  • 2
    $\begingroup$ +1. Another consequence of that choice is that in order to get exposed to the variance which will realise over a future period (forward variance) one can seamlessly trade a calendar spread of standard variance swaps. $\endgroup$
    – Quantuple
    Mar 27, 2017 at 7:41
  • $\begingroup$ Is the sampling expectation somehow replicable ? Or not ? $\endgroup$
    – Olórin
    Mar 28, 2017 at 16:31

As described for example here, realized variance is the sum of squared returns - $\sum_{i=0}^{N-1} R_i^2$ in your notation.

Thus it is wrong to assume that the formula consists of:

  • "normalizing factor" - $252$
  • "variance of the logarithmic returns" in statistical terms - $\frac{1}{N}\sum_{i=0}^{N-1} R_i^2$.

Instead the two pieces of the formula are:

  • realized variance - $\sum_{i=0}^{N-1} R_i^2$
  • reciprocal of duration of time period - $\frac{252}{N}$

And since we are not talking about variance (in statistical sense) of the logarithmic returns, it does not make sense to ask whether and why logarithmic returns are supposed to have zero-expectation!

The realized variance is useful because it provides a relatively accurate measure of volatility of the underlying - $\sigma$.

The actual derivation of the formula is illustraded below.

Let $\alpha$ and $\sigma$ be constants, and define the geometric Brownian motion $$ S(t) = S(0) e^{\sigma W(t)+(\alpha-\frac{1}{2}\sigma^2)t}$$

Let $0 \leq T_1 < T_2$ be given and suppose we observe $S(t)$ for $T_1 \leq t \leq T_2$. Choose some partition of this interval $T_1 = t_0 < t_1 < \cdots < t_m = T_2$ and observe log returns $\log\frac{S_{t_{j+1}}}{S_{t_{j}}}$ over each of subintervals $[t_j, t_{j+1}]$: $$\log\frac{S_{t_{j+1}}}{S_{t_{j}}} = \sigma (W(t_{j+1}) - W(t_j)) + (\alpha - \frac{1}{2}\sigma^2)(t_{j+1}-t_j)$$

The realized volatility $\sum_{j=0}^m\Big(\log\frac{S_{t_{j+1}}}{S_{t_{j}}}\Big)^2$ is:

$$\sum_{j=0}^m\Big(\log\frac{S_{t_{j+1}}}{S_{t_{j}}}\Big)^2 = \sigma^2 \sum_{j=0}^m \big(W(t_{j+1})-W(t_j)\big)^2 + \big(\alpha - \frac{1}{2}\sigma^2\big)^2 \sum_{j=0}^m (t_{j+1}-t_j)^2 + \\2\sigma(\alpha - \frac{1}{2}\sigma^2)\sum_{j=0}^m (W(t_{j+1})-W(t_j))(t_{j+1}-t_j)$$

Let $\|\Pi\| = \max_{j=0,1,\cdots m-1}(t_{j+1}-t_j)$. Then: $$ \lim_{\|\Pi\|\to 0} \sum_{j=0}^m (t_{j+1}-t_j)^2 = 0 \\ \lim_{\|\Pi\|\to 0} \sum_{j=0}^m (W(t_{j+1})-W(t_j))(t_{j+1}-t_j) = 0 \\ \lim_{\|\Pi\|\to 0} \sum_{j=0}^m \big(W(t_{j+1})-W(t_j)\big)^2 = T_2-T_1$$ Thus:

$$\sigma^2 \approx \frac{1}{T_2-T_1}\sum_{j=0}^m\Big(\log\frac{S_{t_{j+1}}}{S_{t_{j}}}\Big)^2$$

Now you can see that $\frac{252}{N} = \frac{1}{T_2-T_1} $ and we don't need the assumption that logarithmic returns have zero-expectation. Instead it is assumed that $S(t)$ follows geometric Browninan motion with constant volatility.

The above derivation has been shamelessly stolen from here (3.4.3 Volatility of Geometric Brownian Motion)

Also have a look at "The Volatility Smile" by E. Derman. Chapter 4 "Variance Swaps" discusses how variance swaps can be replicated

  • $\begingroup$ I don't understand the point of all of this. $\frac{1}{N} \sum_{i=0}^{N-1} R_i^2$ is indeed the realized variance of the logarithmic returns and the $252$ is indeed of normalizing factor as you simply anualize the variance with this factor. The rest of your paragraph is true, so what ? Because it's not an answer to my question. $\endgroup$
    – Olórin
    Apr 7, 2017 at 13:32
  • $\begingroup$ In my opinion, you question is based on the wrong assumptions and in fact I've not answered your question per se but tried to show that the assumptions are wrong! I'll try to elaborate my answer to make it clearer $\endgroup$
    – zer0hedge
    Apr 7, 2017 at 14:01
  • $\begingroup$ No need, just say which are my "assumptions" that you find wrong $\endgroup$
    – Olórin
    Apr 7, 2017 at 14:25
  • $\begingroup$ Done. Have a look :-) $\endgroup$
    – zer0hedge
    Apr 7, 2017 at 14:40
  • $\begingroup$ Added last paragraph about replication of variance swaps $\endgroup$
    – zer0hedge
    Apr 7, 2017 at 14:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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