# Variance replication using options

I would like to understand the intuition behind the following question:

Why a certain weighted sum of prices of put and calls is equivalent to the implied variance of an underlying?

A variance swap is replicated with a static basket of calls and puts (I understand that this replicates the implied variances) which are delta-hedged (this replicates the realized variance), but what I find difficult is the intuition behind that.

Let $t_0, t_1, \ldots, t_n$ be observation dates, where $0=t_0 < \cdots < t_n = T$, and $\{S_t \mid t \geq 0\}$ be the equity price process without dividend payments. Then the realized variance is defined by \begin{align*} \frac{252}{n}\sum_{i=1}^n \ln^2 \frac{S_{t_i}}{S_{t_{i-1}}}. \end{align*} Note that, for sufficiently small $x$, \begin{align*} \ln (1+x) \approx x - \frac{1}{2}x^2. \end{align*} Moreover, \begin{align*} \ln^2 (1+x) &\approx x^2\\ &\approx 2x - 2 \ln(1+x). \end{align*} Then \begin{align*} \sum_{i=1}^n \ln^2 \frac{S_{t_i}}{S_{t_{i-1}}} &\approx 2\sum_{i=1}^n \frac{S_{t_{i}}-S_{t_{i-1}}}{S_{t_{i-1}}}-2\ln\frac{S_T}{S_0}. \end{align*} Assuming that the short interest rate $r_t$ is deterministic, \begin{align*} E\bigg(\sum_{i=1}^n \ln^2 \frac{S_{t_i}}{S_{t_{i-1}}}\bigg) &\approx 2\sum_{i=1}^n E\Bigg(E\bigg(\frac{S_{t_{i}}-S_{t_{i-1}}}{S_{t_{i-1}}}\mid S_{t_{i-1}}\bigg)\Bigg)-2E\bigg(\ln\frac{S_T}{S_0}\bigg)\\ &= 2\sum_{i=1}^n E\Bigg(\frac{S_{t_{i-1}}e^{\int_{t_{i-1}}^{t_i}r_s ds}-S_{t_{i-1}}}{S_{t_{i-1}}}\Bigg)-2E\bigg(\ln\frac{S_T}{S_0}\bigg)\\ &= 2\sum_{i=1}^n \Big(e^{\int_{t_{i-1}}^{t_i}r_s ds} -1\Big)-2E\bigg(\ln\frac{S_T}{S_0}\bigg)\\ &\approx 2\sum_{i=1}^n \int_{t_{i-1}}^{t_i}r_s ds - 2E\bigg(\ln\frac{S_T}{S_0}\bigg)\\ &= 2\int_0^T r_s ds - 2E\bigg(\ln\frac{S_T}{S_0}\bigg)\\ &= 2\ln \Big(S_0 e^{\int_0^T r_s ds} \Big) - 2 \ln S_0 - 2E\bigg(\ln\frac{S_T}{S_0}\bigg)\\ &= -2E\bigg(\ln\frac{S_T}{E(S_T)} \bigg). \end{align*} Note that, for any smooth function $f$, $a>0$, and $x>0$, \begin{align*} f(x) = f(a) + f'(a)(x-a) + \int_a^{\infty}(x-k)^+ f''(k)dk + \int_0^a (k-x)^+f''(k)dk. \end{align*} See also How to hedge a derivative that pays the reciprocal of the stock price?.

Consider the function $f(x)=\ln x$ with $x=S_T$ and $a = E(S_T)$. We have that \begin{align*} \ln S_T = \ln E(S_T) + \frac{S_T - E(S_T)}{E(S_T)} - \int_{E(S_T)}^{\infty} \frac{(S_T-k)^+}{k^2} dk - \int_0^{E(S_T)} \frac{(k-S_T)^+}{k^2} dk. \end{align*} Therefore, \begin{align*} E\bigg(\sum_{i=1}^n \ln^2 \frac{S_{t_i}}{S_{t_{i-1}}}\bigg) &\approx -2E\bigg(\ln\frac{S_T}{E(S_T)} \bigg)\\ &=2E\bigg[\int_{E(S_T)}^{\infty} \frac{(S_T-k)^+}{k^2} dk + \int_0^{E(S_T)} \frac{(k-S_T)^+}{k^2} dk\bigg], \end{align*} which is a weighted sum of prices of put and calls.

For an elementary and intuitive explanation, we consider the Black Scholes setting with a geometric Brownian motion. That is, \begin{align*} S_T &= S_0 \exp\big((r-\frac{1}{2}\sigma^2)T+\sigma W_T \big)\\ &= E(S_T) \exp\big(-\frac{1}{2}\sigma^2T+\sigma W_T \big), \end{align*} where $\{W_t\mid t\geq 0\}$ is a standard Brownian motion. Then, we have the variance \begin{align*} \sigma^2 =\frac{2}{T}\Big(\sigma W_T -\ln \frac{S_T}{E(S_T)} \Big). \end{align*} That is, \begin{align*} \sigma^2 = -\frac{2}{T}E\Big(\ln \frac{S_T}{E(S_T)}\Big), \end{align*} which, as demonstrated above, can be approximated by a weighted sum of prices of put and calls.

• Hi Gordon, thanks a lot for your detailed answer. I can follow your reasoning up until "assuming the short interest rate r_t is deterministic" in concrete the assumptions that you are doing in the formula below it. Could you explain that further, please? Besides the expected value of the Taylor expansion is there any other more intuitive way of explaining my question? thank you!! Commented May 28, 2015 at 19:14
• @Escachator: If the interest rate is deterministic, we can the approximate the realized variance by a log function. I added more steps. However, I do not have a more intuitive way to explain. Commented May 28, 2015 at 19:46
• @Escachator: I added an elementary interpretation. Commented May 28, 2015 at 22:34
• Quick one: what did you do in the last step? Did you take expected value on one side of the equality? I have to say that although I understand the process you follow, it is still not so intuitive the relationship between variance and option prices. Now that I think about it, I remember a relationship between the sum of the gammas of a strip of options and the gamma... Maybe that was another intuition? Commented May 28, 2015 at 22:52
• @Escachator: That is correct. we take expectation on both sides. Commented May 28, 2015 at 22:53

The intuition is much simpler in Bachelier setting. The portfolio of being long one each of all out-of-money puts plus all out-of-money calls has a quadratic payoff, $$(S_t-S_0)^2$$ times a constant inversely proportional to the distance between strikes, $$\Delta K$$. Now, this being a parabola with constant convexity, such a portfolio has a constant gamma, and hedging consists of hedging a parabola whereby delta is simply $$(S_t-S_0)$$. The return of such a portfolio is equal exactly to the "simple" variance $$\Sigma_{i=0}^t (S_{i+1}-S_i)^2$$. That's all.