Why Partial Moments Cannot Determine an Arbitrary Payoff Expectation
Summary
Knowing the mean of a random variable and the expectations of its positive and negative parts does not, in general, determine the expectation of an arbitrary twice differentiable payoff. For a positive lognormal variable, the positive-part expectation equals the mean and the negative-part expectation is zero, yet many distributions share a mean while producing different expectations for nonlinear functions.
The answer notes that a mean can suffice when the distribution belongs to a family characterized by one parameter, and that Jensen’s inequality or a Taylor approximation may offer limited bounds or estimates. For a broader valuation, a static replication identity expresses a payoff using a linear position and integrals of calls and puts across strikes. Thus a continuum of option prices can recover the expectation under the pricing framework, while only a few partial moments cannot; assumptions and the availability of those prices matter.
Key ideas
- The mean and positive and negative partial moments do not identify a general distribution.
- A positive random variable can have the same mean while producing different expectations for nonlinear payoffs.
- A mean may be sufficient for distributions characterized by a single parameter.
- Jensen’s inequality and Taylor expansion can provide limited information about a payoff expectation.
- Static replication uses option prices across a continuum of strikes to value general payoffs.
Tags
Full text
# Given $\mathbb{E}[X]$, $\mathbb{E}[\max(0,X)]$, and $\mathbb{E}[\min(0,X)]$, what is $\mathbb{E}[f(X)]$
# Given $\mathbb{E}[X]$, $\mathbb{E}[\max(0,X)]$, and $\mathbb{E}[\min(0,X)]$, what is $\mathbb{E}[f(X)]$
Let $X$ be any random variable with any distribution. Given that we know $\mathbb{E}[X]$, $\mathbb{E}[\max(0,X)]$, and $\mathbb{E}[\min(0,X)]$, can you write a formula for $\mathbb{E}[f(X)]$ where $f$ is any twice differentiable function?
I am guessing the solution might have something to do with the arbitrage free pricing of $f(X)$ given price of $X$, price of call option on $X$, and price of put option on $X$, but I am not an expert in pricing.
Any ideas?
## Answer by Kevin (score 7, accepted)
https://quant.stackexchange.com/a/73442
I don't think one can answer your question. Suppose $X=e^{\mu+\sigma Z}$ is log-normal, i.e. positive. Thus, $\mathbb{E}[\max\{0,X\}]=\mathbb{E}[X] $ and $\mathbb{E}[\min\{0,X\}]=0$. From just knowing $\mathbb{E}[X]$, you cannot conclude what $\mathbb{E}[f(X)]$ is for an arbitrary function $f$. An implication of your statement is that the mean fully characterises the distribution of positive random variables, which is not true in general.
It would work for distributions that only have one parameter (are characterised by their mean) like the Poisson distribution or exponential distribution. Depending on your function, you might get some information about $\mathbb{E}[f(X)]$ from Jensen's inequality or a first-order Taylor expansion.
Famously, Carr and Madan derived a static replication formula, see @Gordon's answer here:
\begin{align*} f(x) &= f(a) + f'(a) (x-a) + \int_a^{\infty}(x-u)^+f''(u)\text du + \int_{0}^a(u - x)^+f''(u)\text du. \end{align*} This allows you to calculate expectations of the form $\mathbb{E}[f(X)]$ - but you do need a continuum of option prices (at all positive strikes). This finding echoes the result from Breeden-Litzenberger (1978): A complete list of option prices fully characterises the distribution of the underlying asset at maturity. However, you need to know much more than just the mean of the positive and negative part of $X$.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.