Deriving Hedged Option P&L from Implied and Realized Volatility

# P&L when hedging with realized volatility


# 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!

## Answer by Pleb (score 5, accepted)

https://quant.stackexchange.com/a/79551

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}