# 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.

• An expression with only 1 integral would be $E(Z=X(Y-K)^+) = \int_0^\infty \left( 1 - F_Z(x) \right) \,\mathrm{d}x$ – emcor Jun 30 '14 at 15:26
• Could you clarify the application to Quant Finance? Pure math questions belong to Mathematics. – SRKX Jun 30 '14 at 15:34
• You may note $(y-K)^+=0$ when $y\leq K$, so the second integration goes only from $K$ to $\infty$. – emcor Jun 30 '14 at 16:15

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.
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.
• Precisely we are looking for $f(x,y):E(g(x,y,K)|K)=\int\int g(x,y|K)f(x,y|K)dxdy$ (the integral goes dxdy) – emcor Jul 1 '14 at 8:56