Calculating European Option Payoff Variance in Black–Scholes
Article Quant Q&A · Author: Pxx
Summary
The document asks how to calculate the variance of a discounted European call or put payoff under the Black–Scholes assumptions. Its answer outlines a second-moment approach: variance equals the expected squared payoff minus the square of its expected value. For a call, the squared positive-part payoff is expanded over the event that the terminal stock price exceeds the strike, producing terms involving the truncated first and second moments of the stock price and the probability of exercise.
The familiar Black–Scholes derivation supplies the latter two quantities, while the first can be evaluated using a related lognormal calculation with a changed distribution. This gives a route to the call payoff variance, rather than a fully worked formula. The response does not spell out the final expression or separately derive the put case, and it does not discuss variance of the discounted payoff beyond the question’s setup.
Key ideas
- Payoff variance can be found from the second moment minus the squared first moment.
- For a call, square the positive payoff and expand it on the event that the stock finishes above the strike.
- The exercise probability and truncated first moment are available from the standard pricing derivation.
- The truncated second moment can be computed with a related lognormal calculation.
- The document sketches the method but does not provide a final formula or a separate put derivation.
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# Expected value and variance of the stock log-returns under Local Volatility framework
# Expected value and variance of the stock log-returns under Local Volatility framework
I want to calculate the expected value and the variance of the stock process log-returns in the Local Volatility setting (and the realized/terminal correlation but let us begin in the one-dimentional setting). First some intro and my progress so far.
Consider the standard local volatility SDE: $$ d S_t = \mu(t) S_t\, dt + \sigma(t,S_t) S_t dW_t $$ with deterministic drift $\mu(t)$ and some deterministic local volatility function $\sigma(t,x)$. Using Ito's lemma we can derive the following SDE for the $\ln S_t$: $$ d \ln S_t = \ln S_0 + \left( \mu(t) - \frac{1}{2}\sigma^2(t,S_t)\right) dt + \sigma(t, S_t)dW_t $$ which can be written in the integral form as $$ \ln S_t = \ln S_0 + \int_0^t \left( \mu(u) - \frac{1}{2}\sigma^2(u,S_u)\right) du + \int_0^t \sigma(u,S_u) dW_u $$
I am interested in the log-returns over some (small) time interval $\Delta t$, thus, from the above we get $$ \ln\left(\frac{S_{t+\Delta t}}{S_t}\right) = \ln(S_{t+\Delta t}) - \ln(S_{t}) = \\ = \int_t^{t+\Delta t} \left( \mu(u) - \frac{1}{2}\sigma^2(u,S_u)\right) du + \int_t^{t+\Delta t} \!\! \sigma(u,S_u) dW_u $$
Proceeding to the quantities that I want to calculate, namely, the expected value and variance of the log-returns, we have $$ \mathbb{E}\left[ \ln\left(\frac{S_{t+\Delta t}}{S_t}\right) \right] =\\ = \mathbb{E}\left[ \int_t^{t+\Delta t} \left( \mu(u) - \frac{1}{2}\sigma^2(u,S_u)\right) du + \int_t^{t+\Delta t} \!\! \sigma(u,S_u) dW_u \right] = \\ = \int_t^{t+\Delta t} \mu(u) du - \frac{1}{2} \, \mathbb{E} \left[ \int_t^{t+\Delta t} \sigma^2(u,S_u) du\right] + \mathbb{E} \left[ \int_t^{t+\Delta t} \!\! \sigma(u,S_u) dW_u \right] $$ Since the last inegral is zero, we arrive at the following \begin{eqnarray}\label{ev} \mathbb{E}\left[ \ln\left(\frac{S_{t+\Delta t}}{S_t}\right) \right] = \int_t^{t+\Delta t} \mu(u) du - \frac{1}{2} \, \int_t^{t+\Delta t} \mathbb{E} \left[ \sigma^2(u,S_u) \right] du = \quad ... \,\, ??? \end{eqnarray} Here is the place where I get stuck. Question 1: Is there a way to further simplify the second integral? It comes back in the variance derivation below, where I continue with this form of the expected value.
Further, the variance calculation goes as follows: $$ \textrm{var} \left[ \ln\left(\frac{S_{t+\Delta t}}{S_t}\right) \right] = \mathbb{E}\left[ \ln\left(\frac{S_{t+\Delta t}}{S_t}\right) - \mathbb{E}\left[ \ln\left(\frac{S_{t+\Delta t}}{S_t}\right) \right] \right]^2 = \\ = \mathbb{E}\left[ \int_t^{t+\Delta t} \left( \mu(u) - \frac{1}{2}\sigma^2(u,S_u)\right) du + \int_t^{t+\Delta t} \!\! \sigma(u,S_u) dW_u \\ - \int_t^{t+\Delta t} \mu(u) du + \frac{1}{2} \, \int_t^{t+\Delta t} \mathbb{E} \left[ \sigma^2(u,S_u) \right] du \right]^2 = \\ = \mathbb{E}\left[ \int_t^{t+\Delta t} \!\! \sigma(u,S_u) dW_u -\frac{1}{2} \int_t^{t+\Delta t} \left( \sigma^2(u,S_u) - \mathbb{E} \left[ \sigma^2(u,S_u) \right] \right) du \right]^2 = ... $$ Now, we take the square and get the three components $$ ... = \mathbb{E} \left[ \int_t^{t+\Delta t} \!\! \sigma(u,S_u) dW_u \right]^2 \\ - \mathbb{E} \left[ \int_t^{t+\Delta t} \!\! \sigma(u,S_u) dW_u \times \int_t^{t+\Delta t} \left( \sigma^2(u,S_u) - \mathbb{E} \left[ \sigma^2(u,S_u)\right]\right) du \right] \\ + \frac{1}{4} \, \mathbb{E} \left[ \int_t^{t+\Delta t} \left( \sigma^2(u,S_u) - \mathbb{E} \left[ \sigma^2(u,S_u)\right]\right) du \right]^2 $$
By virtue of the Ito's isometry the first component equals $$ \mathbb{E} \left[ \int_t^{t+\Delta t} \!\! \sigma(u,S_u) dW_u \right]^2 = \int_t^{t+\Delta t} \mathbb{E} \left[ \sigma^2(u,S_u) \right] du $$ which is the integral we have already seen in the expected value formula. Thus, the whole variance of the log-return is (so far) given by $$ \textrm{var} \left[ \ln\left(\frac{S_{t+\Delta t}}{S_t}\right) \right] = \int_t^{t+\Delta t} \mathbb{E} \left[ \sigma^2(u,S_u) \right] du \\ - \mathbb{E} \left[ \int_t^{t+\Delta t} \!\! \sigma(u,S_u) dW_u \times \int_t^{t+\Delta t} \left( \sigma^2(u,S_u) - \mathbb{E} \left[ \sigma^2(u,S_u)\right]\right) du \right] \\ + \frac{1}{4} \, \mathbb{E} \left[ \int_t^{t+\Delta t} \left( \sigma^2(u,S_u) - \mathbb{E} \left[ \sigma^2(u,S_u)\right]\right) du \right]^2 $$
Question 2: In general, how to proceed further? Does the expected value of the product of the integrals simplify to something? Third component resembles the variance of the squared local volatility function - can we use it somehow?
Note that in case of constant volatility the variance of the log-return simplifies to $\sigma^2 \Delta t $, which is exactly what we define as volatility in the Black-Scholes setting. But I want to know what is the (empirical/realized) volatility in case of the Local Volatility model setting (as defined by the time scaled variance of the log-returns of the stock process).
Any advice/hint (also partial) would be appreciated!Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.