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Why a Joint Distribution Cannot Be Recovered from One Strike-Indexed Expectation

Article Quant Q&A · Author: Math Girl

Summary

The document asks whether the joint distribution of two random variables can be recovered when their marginal distributions are known and the expectation of one variable multiplied by the positive part of the other variable minus a threshold is available for every nonnegative threshold. The answers say this information is insufficient to identify the joint distribution: varying the threshold supplies a family of expectations, but does not reveal the full dependence structure between the variables.

The responses frame the issue as an underdetermined inverse problem. Many joint densities can satisfy the available constraints, particularly when the possible dependence structures have more degrees of freedom than the supplied expectations constrain. Identification would require additional information, such as specified distributional forms and dependence assumptions. The possibility of interpreting the expectation as a risk-neutral derivative payoff is mentioned, but no such assumption is given, and no numerical reconstruction method is offered.

Key ideas

  • Knowing the threshold-indexed expectation and both marginal distributions does not generally identify the joint distribution.
  • The expectation data leave the dependence between the random variables underdetermined.
  • Multiple joint densities may be consistent with the supplied constraints.
  • Additional assumptions about distribution families and dependence would be needed for identification.
  • A derivative-pricing interpretation could add constraints, but the document does not assume one.

Tags

Full text
# Joint distribution from expectations


# Joint distribution from expectations












Given two random variables $X$ and $Y$ and let $K$ be a constant value. Assume the expectation $\mathbb{E}[X(Y-K)^{+}]$ is given for all possible values of $K\geq 0$. Is there a way to derive the joint probability distribution of $X$ and $Y$ from this??

The expectation can be written as

$$\mathbb{E}[X(Y-K)^{+}]=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}x(y-K)^{+}dF(x,y)$$ and when density exists $$=\int_{-\infty}^{\infty}\int_{K}^{\infty}x(y-K)f(x,y)dxdy$$

Both marginal distributions $F_{X}$ and $F_{Y}$ are known and densities exists as well. Is there any way I can derive the joint distribution if the expected value is given for all values of $K$?

I have been stuck on this for a while now, even rough approximations would be of much use to me or a collection of properties that can be solved numerically.

Can someone please help me?

## Answer by emcor (score 2)

https://quant.stackexchange.com/a/12857

It is not possible to derive the joint distribution from the expectation under the given information here.

The fact that you have the expectation for all $K$ says nothing about the joint distribution $f(x,y)$ because $K$ just shifts the mean of $Y$ but gives no information on the joint probability for $(x,y)$. You may particularly note if $f(x,y)$ have >1 parameter, it means more degrees of freedom than implied by just given K.

To find the joint distribution from the expectation, we would need at least the actual distribution types of $X$ and $Y$, and information on their dependence.

If the above expression is meant as derivative price by riskneutral expected payoff, we may have some additional no-arbitrage conditions but that presumption isnt given here aswell.

## Answer by Hans (score 0)

https://quant.stackexchange.com/a/12858

Emcor is correct, especially the part regarding the unmatched number of degrees of freedom between that of the known and unknown functions. We can make the problem even clearer. Your problem is essentially finding density $f(x,y)$ given $\int xf(x,y)dx, \forall y$. This obviously has infinitely many solutions.

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.