Why Prepaid Forward Volatility Matches Underlying Asset Volatility
Summary
The note explains why an asset’s volatility over a fixed horizon matches the volatility of a prepaid forward contract on that asset with the same expiry. Both are measured as the standard deviation of a log return over the period. At the start, the asset value and the prepaid forward price are known, so they contribute only constants to the log return and do not affect its variance.
At expiry, the prepaid forward’s value equals the spot asset price under the no-arbitrage relation. Thus the random terminal value in both returns is the same. The reasoning supports equivalence for this horizon-based volatility definition; it assumes a forward contract signed at the start and held to expiry, and does not claim that every other measure of contract risk is identical to the asset’s volatility.
Key ideas
- Volatility over a fixed horizon can be defined as the standard deviation of the asset’s log return.
- A known initial value contributes only a constant to a log return, leaving its variance unchanged.
- At expiry, a prepaid forward’s value equals the underlying spot price under no arbitrage.
- The equivalence applies to horizon log-return volatility for a forward held from inception to expiry.
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# Answer by Quantuple (score 3, accepted)
# Why is the statement "the volatility of a $T - t$-month prepaid forward on asset X is $\sigma$" the same as "the volatility of asset X is $\sigma$"?
I'm self studying and I'm having trouble with understanding the equivalent formulations of the volatility $\sigma$ of an asset $X$, as in the below problem.
In the below the problem (and the first part of the solution that I posted), what is highlighted in red implies that the statement "the volatility of a $6$-month prepaid forward on $X$ is $0.3$" is equivalent to stating "the volatility of $X$ is $0.3$.
I'm trying to convince myself of why that is true.
The volatility of an asset $X$ means the standard deviation of the return, or $\sqrt{\text{Var}(\ln X_t / X_0)} = \sqrt{\text{Var}(\ln{X_t})}$.
The standard deviation of the return on a $T - t$-month prepaid forward on $X$, $F^p_{t, T}(X)$, would be $\sqrt{\text{Var}(\ln X_T / F^p_{t, T}(X))} = \sqrt{\text{Var}(\ln{X_t})}$, since $F^p_{t, T}(X)$ is a known constant.
Hence, the two formulations for volatility are equivalent. Is this reasoning correct?
## Answer by Quantuple (score 3, accepted)
https://quant.stackexchange.com/a/32199
Almost, indeed
- The volatility of an asset over a horizon $[t,T]$ indeed refers to the standard deviation of the log-return observed over that period: $$ \sqrt{\text{Var}\left(\ln\left(\frac{X_T}{X_t}\right)\right)} = \sqrt{\text{Var}\left( \ln X_T - \ln X_t \right)} = \sqrt{\text{Var}\left( \ln X_T \right)} $$ because $X_t$ is a known constant at $t$
- The standard deviation of the log-return of a forward contract on $X$ signed at $t$ and expiring at $T$ would be: $$ \sqrt{\text{Var}\left(\ln\left(\frac{F_{T,T}^P(X)}{F_{t,T}^P(X)}\right)\right)} = \sqrt{\text{Var}\left( \ln F_{T,T}^P(X) \right)} = \sqrt{\text{Var}\left( \ln X_T \right)} $$ because $F_{t,T}^P(X)$ is a known constant at $t$ but most of all because by absence of arbitrage opportunity $F_{T,T}^P(X) := X_T$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.