Best-of Puts Versus Worst-of Calls Under Perfect Negative Correlation

# Which option more expensive?


# Which option more expensive?












What's more expensive between an ATM call WO or an ATM Put BO on two underlyings S1, S2 assuming correlation between S1 and S2 = -1? Suppose r=q=repo=0, vol is constant.

A colleague of mine got asked this question during an interview and failed to answer.

## Answer by Kermittfrog (score 4)

https://quant.stackexchange.com/a/82421

Below, I argue that the best-of-put should be more expensive under a bivariate geometric Brownian motion model.

#### Definitions

We define our basket product to be written on some characteristic of the relative returns of some assets $S^{(1)},S^{(2)}$

$$ R \equiv\left\{R_T^{(1)},R_T^{(2)}\right\} = \left\{\frac{S^{(1)}_T}{S_0^{(1)}},\frac{S^{(2)}_T}{S_0^{(2)}}\right\} $$ and the best-off ("BO") and worst-off ("WO") to be defined as $$ \begin{align} P_{\mathrm{BO}}&\equiv\max{R}\\ P_{\mathrm{WO}}&\equiv\min{R} \end{align} $$

#### Assumptions

We assume the two assets to jointly follow a bivariate geometric Brownian motion (GBM), specifically, that their return innovations be correlated with parameter $\rho$. We also assume both assets to have identical volatility, $\sigma_1 = \sigma_2 = \sigma$:

$$ \begin{align} S_T^{(1)}&\sim S_0^{(1)}e^{-\frac{1}{2}\sigma^2T+\sigma\sqrt{T}x}\\ S_T^{(2)}&\sim S_0^{(2)}e^{-\frac{1}{2}\sigma^2T+\sigma\sqrt{T}y}\\ &\sim S_0^{(2)}e^{-\frac{1}{2}\sigma^2T+\sigma\sqrt{T}\left(\rho x+\sqrt{1-\rho^2}z\right)} \end{align} $$ where $x,y$ are bivariate (standard) normally distributed with correlation $\rho$, and $z$ is another standard normal. As $\rho = -1$,

$$\begin{align} S_T^{(2)}&\sim S_0^{(2)}e^{-\frac{1}{2}\sigma^2T-\sigma\sqrt{T} x} \end{align}$$

This assumption results in

$$ \begin{align} R&=\left\{e^{-\frac{1}{2}\sigma^2 T+\sigma\sqrt{T}z},e^{-\frac{1}{2}\sigma^2 T-\sigma\sqrt{T}z}\right\}\\ &=e^{-\frac{1}{2}\sigma^2 T}\left\{e^{+\sigma\sqrt{T}z},e^{-\sigma\sqrt{T}z}\right\} \end{align} $$ (with some sloppiness in the notation). Ultimately, as we have zero dividends / interest rates, $\mathrm{E}\left(S_T^{(i)}\right)=S_0^{(i)}$, allowing us to assume at the money to mean a strike of $100\%$, i.e. $K=1$.

### Result

Thus, the (stochastic) payoff of a worst of ATM call becomes: $$ \begin{align} \Pi_{\mathrm{WO-ATM-C}}&=\max\left(0,P_{\mathrm{WO}}-K\right) \\ &=\max\left(0,\min\left\{R_T^{(1)},R_T^{(2)}\right\}-1\right)\\ &=\max\left(0,e^{-\frac{1}{2}\sigma^2T}\min(e^{\sigma\sqrt{T}x},e^{-\sigma\sqrt{T}x})-1\right) \end{align} $$ Clearly, the payoff is zero in all states of the world, the option premium is zero.

For the best of ATM put, on the other hand: $$ \begin{align} \Pi_{\mathrm{BO-ATM-P}}&= \max\left(0,K-P_\mathrm{BO}\right)\\ &= \max\left(0,1-\max\left\{R_T^{(1)},R_T^{(2)}\right\}\right)\\ &=\max\left(0,1-e^{-\frac{1}{2}\sigma^2T}\max(e^{\sigma\sqrt{T}x},e^{-\sigma\sqrt{T}x})\right) \end{align} $$ This payoff is positive whenever $e^{-\frac{1}{2}\sigma^2T+\sigma\sqrt{T}x}<1$, i.e. $|x|<\frac{1}{2}\sigma\sqrt{T}$, which is attainable. Hence, the best-of put commands a positive price.