# 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? • 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 $F_{T,T}^P(X) := X_T$