# P&L when hedging with realized volatility

Quoting "The Volatility Smile" of EMANUEL DERMAN MICHAEL B. MILLER, pag.95-96:

The hedged portfolio at any time $$t$$ is given by $$\Pi(I,R)=V_{I}-\Delta_{R}X$$, where $$\sigma_{R}$$ is the computed realized volatility, $$V_{I}$$ is the value of the option computed with the implied volatility $$\sigma_{I}$$, and $$\Delta_{R}$$ is the hedge ratio valued with $$\sigma_{R}$$. The increment of the portfolio hedged with the realized volatility is $$dP\&L(I,R)=dV_{I}-\Delta_{R}dX-\Delta_{R}XDdt-(V_{I}-\Delta_{R}X)rdt$$, which can be rewritten as $$dP\&L(I,R)=dV_{I}-rV_{I}dt-\Delta_{R}[dX-(r-D)Xdt]$$, where $$D$$ is the dividend yield. Had we valued at $$\sigma_R$$ and hedged at $$\sigma_R$$, the hedging startegy would have been the riskless one that leads to the BSM equation. Hence, $$dP\&L(R,R)=0=dV_{R}-V_{R}rdt-\Delta_{R}[dX-(r-D)Xdt] \to \Delta_{R}[dX-(r-D)Xdt] = dV_{R}-V_{R}rdt \to dP\&L(I,R)=dV_{I}-dV_{R}-(V_{I}-V_{R})rdt.$$

Now the author does something that I don't get:

Using the product rule to take the derivative of $$e^{-rt}(V_{I}-V_{R})$$ with respect to $$t$$, we obtain $$dP\&L(I,R)=e^{rt}d[e^{-rt}(V_{I}-V_{R})]$$, expressing the incremental P&L in terms of a complete differential, which will make it easier to calculate the total P&L over the life of the option.

What is exactly doing the author in the last step and which are the mathematical steps to go from $$e^{rt}\partial/\partial t(V_{I}-V_{R})+(V_{I}-V_{R})\partial/\partial t (e^{rt})$$ to $$e^{rt}d[e^{-rt}(V_{I}-V_{R})]$$? Many thanks!

Without diving too much into the paper, the derivations can be verified as follows.

### Derivation:

Let us start from the end result: Using the product rule with respect to $$t$$ on $$e^{-rt} \left(V_I(t) - V_R(t)\right)$$:

\begin{align*} \frac{d}{dt}\left[e^{-rt} \left(V_I(t) - V_R(t)\right)\right] &= -re^{-rt} \left(V_I(t) - V_R(t)\right) + e^{-rt} \left[\frac{d}{dt}\left(V_I(t) - V_R(t)\right)\right]\\ \end{align*} Now multiply with $$e^{rt}$$ on both sides: \begin{align*} e^{rt}\frac{d}{dt}\left[e^{-rt} \left(V_I(t) - V_R(t)\right)\right] &= -r \left(V_I(t) - V_R(t)\right) + \frac{d}{dt}\left(V_I(t) - V_R(t)\right)\\ \end{align*} and lastly we multiply with $$dt$$ on both sides (I have removed the subscript $$t$$ to follow authors notation): \begin{align*} e^{rt}d\left[e^{-rt} \left(V_I - V_R\right)\right] &= -r \left(V_I - V_R\right)dt + d\left(V_I - V_R\right)\\ &=-r \left(V_I - V_R\right)dt + dV_I - dV_R\\ &= dP\&L(I,R) \end{align*}

Hence, it holds that we can re-write the P&L equation as the complete differential provided by the authors.

### Alternative derivation:

An alternative derivation can be provided using the product rule for Itô processes.

Let $$f = X \cdot Y$$, where $$X = e^{-rt}$$ and $$Y=(V_I - V_R)$$, then following from the product rule for Itô processes:

\begin{align} df &= d(X\cdot Y)\\ &= Y dX + X dY + d[X,Y]\\ &= (V_I - V_R)d\left(e^{-rt}\right) + e^{-rt} d\left(V_I - V_R\right) + 0\\ &= (V_I - V_R) \left(-re^{-rt} dt\right) + e^{-rt} d\left(V_I - V_R\right), \end{align} where we have used Itô's lemma on $$d(e^{-rt}) = -re^{-rt} dt$$ and the fact that the quadratic variation of a finite variation process is zero (ie. $$d[X,Y] = 0$$, since $$e^{-rt}$$ is a continuous differentiable function).

Now, multiplying with $$e^{rt}$$ on both sides gives you the desired result: \begin{align} e^{rt} df&= e^{rt} d(X\cdot Y)\\ &= e^{rt}d\left(e^{-rt} \left(V_I - V_R\right)\right)\\ &= -r (V_I - V_R) dt + dV_I - dV_R\\ &= dP\&L(I,R). \end{align}

• Thanks! Just a small comment for the notation that I mentioned on the question above. It is not correct to write $\partial / \partial t (e^{-rt}(V_{I}(t)-V_{R}(t)))$ since stochastic paths are not differentiable by $t$. Correct me if I'm wrong Commented May 30 at 13:11
• @DivertingPie I implicitly assumed that $V_I(t)$ and $V_R(t)$ are adapted processes and hence deterministic at time $t$. You can also derive the same results following the product rule for Ito processes. It might be more along the lines of the book/paper. I will provide another alternative derivation.
– Pleb
Commented May 30 at 19:15
• amazing, thanks! Commented May 30 at 20:02