Estimating P&L Risk from Partial Option Hedges

# Quantifying Costs/Benefits Of Partial Hedging


# Quantifying Costs/Benefits Of Partial Hedging












Say I sold a long-dated European put option and I want to analyze the costs and benefits of partial hedges in a world with stochastic price movements, rate movements, and volatility. For example, let's say I want to hedge 50% of my delta, 75% of my rho, and 25% of my vega.

How can I calculate the expected cost/benefit of my hedge? If all risks were 100% hedged, I'd pay the option premium up front (cost) and nothing more at maturity (benefit). If all risks were 50% hedged, I'd pay 50% of the option premium up front (cost) and only 50% of the payout at maturity (benefit). But if my risks are hedged in varying degrees, what analytical solutions are available for me to estimate these costs and benefits without explicit derivation of Greeks at each rebalancing point in the simulation?

EDIT (Given @Arshdeep's responses):

Maybe it will be better if I express myself and my interpretation of Arshdeep's response formulaically. Suppose my stock doesn't have dividends and follows the Heston model described in Hull Options, Futures, and Other Derivatives, so that in my discrete simulation I have:

$\frac{dS}{S} = r \Delta t + \sqrt {V} dz_s$

$\Delta V_t = a(V_L - V_{t-1}) \Delta t + \xi V^{\alpha} dz_v$

$V_t = V_{t-1} + \Delta V_t$

Where $dz_s$ and $dz_v$ are my correlated standard normal random draws. Then my simulation follows the form:

$S_t = S_{t-1} * e^{(r - V_t / 2) \Delta t + \sqrt{V_t} dz_s}$

According to @Arshdeep, I can independently test the impact of my delta and vega hedges. My interpretation of his response is that I apply the following hedge coverage amounts to my simulation, where X is (1 - % Delta Hedged) and Y is (1 - % Vega Hedged):

$S_t = S_{t-1} * e^{X[(r - V_{t-1} / 2) \Delta t + \sqrt{V_{t-1}} dz_s] + Y [(\sqrt{V_t} - \sqrt{V_{t-1}})dz_s - (dV_t)/2]}$

I don't know what to make of this term that represents your unhedged change in volatility when Y<1 but X=1: $(\sqrt{V_t} - \sqrt{V_{t-1}})dz_s$. If you 100% hedge delta, your equity volatility is 0%, but your change in volatility can be less than 0, resulting in negative volatility. Can your net volatility position be negative? In practice, it just flips the sign of your $dz_s$ term, but I don't know what that intuitively means.

## Answer by Arshdeep (score 2)

https://quant.stackexchange.com/a/79463

One approach I would take is to plot %hedging v/s PnL variance.

A perfect hedge should leak no PnL, and a naked position would have variance of the spot. So you have a downward, convex curve that converges to x-axis as hedge% goes to 100%. Parametrize the curve and run a simulation once for 50% hedge to find the parameter. This should give you an idea of how variance moves as a function of a risk factor being partially hedged.

Edit1: The expected PnL of any strategy in the risk neutral set is 0. So maybe to model the expected PnL in the real world, you have to look at the covariance of RND with the strategy.

Edit2: Here is another thing I would do.

$dS^2=2SdS+dt$ So I know the integral of $SdS$ in closed form.

I approximate delta as $delta(t,S(t))= 2aS+b+c*S^2$ so that I know the PnL in closed form. Then the integral calculation is easier. Now you can look at your PnL in closed form. This will work well where $vol*sqrt(t)$ is low. a b and c can match the current delta and delta at extreme percentiles.

Edit 3:

The difference in C(t,S(t)) of 2 different approaches does not have anything to do with actual PnL of the naked call. Sorry for confusion.

## Answer by Kermittfrog (score 1)

https://quant.stackexchange.com/a/79570

Let try a very simple approximation using a binomial tree where the stock moves from $S$ or to $S\times U$ or $S\times D$ with state factors $U\equiv e^{\sigma\sqrt{\Delta t}}$ and $D$. For simplicity, we assume a one-step model and $\Delta t=1$, furthermore $D=1/U$, the risk free rate is zero, i.e. $R=1$, and the initial asset price is $S_0=1$ for convenience. The option will pay $C^U$ and $C^D$ in the states $S^U=U$ and $S^D=D$.

Under this model, the initial delta (i.e. hedge fraction) and option prices are calculated as:

$$ \Delta = \frac{C^U-C^D}{U-D}\quad\mathrm{and}\quad C_0=\underbrace{\Delta}_{\mathrm{hedge}}+\underbrace{\frac{UC^D-DC^U}{U-D}}_{\mathrm{borrow}} $$

Let's assume that we choose to hedge only partially, i.e. to buy only $\alpha\Delta$ in stock, but we will borrow $\frac{UC^D-DC^U}{U-D}$ in either case to simplify the calculation. This way, when shorting the option, we gain $(1-\alpha)\Delta$.

Incorporating the gain in period 0 into the final payoff states:

$$ \begin{align} \pi^U=(1-\alpha)\Delta-C^U+\alpha U\Delta + \frac{UC^D-DC^U}{U-D}\\ \pi^D=(1-\alpha)\Delta-C^D+\alpha D\Delta + \frac{UC^D-DC^U}{U-D} \end{align} $$

and after some algebra this simplifies to

$$ \begin{align} \pi^U&=(1-\alpha)\Delta(1-U)\\ \pi^D&=(1-\alpha)\Delta(1-D) \end{align} $$

Note that for $0<\alpha<1$ the total payoffs are not hedged. As $pU+(1-p)D=1$ by construction, the expected payoff is:

$$ \mathrm{E}(\pi)=0 $$

Then, remembering that $D=1/U$, we can approximate the variance as:

$$ \mathrm{E}\left(\pi^2\right)=\left(1-\alpha\right)^2\Delta^2\frac{(U-1)^2}{U}\approx \sigma^2 \left(1-\alpha\right)^2\Delta^2 $$