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Why Expected Option Strategy Payoffs Add Linearly

Article Quant Q&A · Author: unsupervised
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Summary

The document asks whether the expected payoff of a multi-leg option position, such as a spread or condor, equals the sum of the expected payoffs of its individual options. It represents each leg’s payoff as a function of the same market value at expiration and illustrates the expectation operation with powers of a random variable.

The answer applies linearity of expectation: the expected value of a sum is the sum of the expected values. This holds whether or not the component payoffs are independent or correlated, so expected payoff for a linear combination of derivatives can be calculated by adding the legs’ expected payoffs with their position weights. The discussion concerns expected payoffs under a chosen probability distribution; it does not establish that the expectation under a real-world distribution equals an option’s market price, nor does it address financing, execution, or risk. Those distinctions matter when applying the algebra to actual option valuation or strategy returns.

Key ideas

  • The expected payoff of a sum of option legs equals the sum of their expected payoffs.
  • Linearity of expectation does not require the component payoffs to be independent.
  • Each leg can be represented as a payoff function of the same underlying price at expiration.
  • The result concerns expectation under a probability distribution and does not by itself determine market price.

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OptionsStatisticsDerivatives pricing
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# is price of multiple option strategy linear under expectation?


# is price of multiple option strategy linear under expectation?












I wonder if someone can confirm (or refute) that the expected payoff of several option (in a strategy such as a spread, condor, etc) behaves as "expection of a sum is sum of expectations".

Here is my thinking: First denote $E_m$ = expectation of market at expiration. (I think is log-normal to a first approximation, but that does not matter for the question).

Then a single option's payout is $$ E_m[ f(m) ] $$ where $f()$ is the function describing a call (or put) payout.

if two options $f_1, f_2$ are added to make a strategy such as a vertical spread, notate the payout $$ E_m [ f_1(m) + f_2(m) ] $$ where $f_1,f_2$ are both functions of the single random variable $m$, the market value at expiration.

Now the question: Is it true that $$ E_m [ f_1(m) + f_2(m) ] = E_m f_1(m) + E_m f_2(m) $$

Here is an analogy outside of finance (from a reddit discussion), take $f_1(x)=x^2$, $f_2(x)=x^4$. These functions of a random variable $x$ are clearly are correlated. however, $$ E_x[ x^2 + x^4 ] = \int p(x) (x^2 + x^4) dx = E_x[x^2] + E_x[x^4] $$ so they do separate under the expectation!

## Answer by will (score 3, accepted)

https://quant.stackexchange.com/a/59039

$$\mathbf{E}\left[X+Y\right] = \mathbf{E}\left[X\right] + \mathbf{E}\left[Y\right]$$

This is just a property of random variables (see here).It doesn't matter that $X$ and $Y$ are not independent, or correlated.

So yes, when you have a linear combination of derivatives, the value is the linear sum of the values of the individual derivatives.

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

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