Why Variance Swaps Use Squared Returns Without Subtracting the Mean

# Variance swap : ok for variance, but where's the square expectation?


# Variance swap : ok for variance, but where's the square expectation?












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 ?)

## Answer by Alex C (score 2)

https://quant.stackexchange.com/a/33264

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.

## Answer by zer0hedge (score 0)

https://quant.stackexchange.com/a/33542

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