Why a Basket of Calls Can Exceed a Call on the Basket

Summary

The document compares a portfolio of call options on individual assets with a call option written on a weighted portfolio of those assets. For nonnegative weights summing to one, the payoff of a call is a convex function of its underlying value. Convexity implies that the call payoff on the weighted basket is no greater than the weighted sum of the individual call payoffs, state by state.

Taking discounted expectations under the pricing measure preserves that inequality, establishing the corresponding option-value relationship. The argument relies on matched strikes and maturities, nonnegative weights that form a normalized basket, and comparable pricing assumptions for the assets. The displayed question initially allows general nonnegative weights, while the proof explicitly treats weights summing to one; it does not establish the same statement for arbitrary unnormalized weights. The result is a payoff-convexity argument, not a numerical valuation or empirical study.

Key ideas

• A call payoff is convex in the value of its underlying asset.
• For weights summing to one, the call payoff on a basket is bounded by the weighted sum of calls on the individual assets.
• Discounting and taking expectations carries the payoff inequality into an option-price inequality.
• The stated proof assumes normalized nonnegative weights and matched strikes and maturities.

Tags

# Call options and portfolio of the same options worth less?


# Call options and portfolio of the same options worth less?












A portfolio of long positions in call options with the same maturity and strikes on different assets is worth more than a call option on a portfolio of the same assets with the same weight; i.e. $$\sum_{i=1}^{n}\lambda_i C(T,K,S^{(t)},t) \geq C(T,K,\hat{S},t),$$ where $\lambda_i\geq 0$ and $S^{(i)}$ for $i = 1,\ldots,n$ are assets and $\hat{S} = \sum_{i=1}^{n}\lambda_i S^{(i)}$ is a value of a portfolio which has $\lambda_i$ units of asset $S^{(i)}$ for each $i = 1,\ldots,n$.

This seems like an application of the triangle inequality, but I am not sure how to formally write it. Any suggestions is greatly appreciated.

## Answer by Gordon (score 2, accepted)

https://quant.stackexchange.com/a/23082

For the case where $\sum_{i=1}^n \lambda_i =1$, you need only note that the payoff \begin{align*} (x-K)^+ \end{align*} is a convex function in $x$. That is, \begin{align*} \Big(\sum_{i=1}^n \lambda_i S_i -K\Big)^+ \le \sum_{i=1}^n\lambda_i(S_i-K)^+. \end{align*} Then \begin{align*} e^{-rT}E\left(\Big(\sum_{i=1}^n \lambda_i S_i -K\Big)^+\right) \le \sum_{i=1}^n\lambda_ie^{-rT}E\left((S_i-K)^+\right). \end{align*}

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.