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In a paper I encountered the following notation

$$P(Z\leq z,u\leq Y\leq v)=C(F_{Z}(z),F_{Y}(v)-F_{Y}(u))$$

However I don't see why this holds in relation to uniform random variables. Usually $$P(Z\leq,Y\leq v)=P(F_{Z}(Z)\leq F_{Z}(z),F_{Y}(Y)\leq F_{Y}(v))=P(U_{1}\leq F_{Z}(z), U_{2}\leq F_{Y}(v))=C(F_{Z}(z),F_{Y}(v))$$

But probability above i would write $$P(Z\leq z,u\leq Y\leq v)=P(F_{Z}(Z)\leq F_{Z}(z),F_{Y}(Y)\leq F_{Y}(v))-P(F_{Z}(Z)\leq F_{Z}(z),F_{Y}(Y)\leq F_{Y}(u))$$ and then use copulas.

Can anyone explain to me where the copula $C(F_{Z}(z),F_{Y}(v)-F_{Y}(u))$ comes from in terms of uniform random variables?

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if you agree that the marginal probability $P(u\le Y\le v)=F_Y(v)-F_Y(u)$, then your formula follows immediately, because next you simply plug the marginals into the copula.

your 3rd equation for the joint probabilities is incorrect for $P(Z\le z,u\le Y\le v)$, I'm not sure where you got it from

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  • $\begingroup$ My problem is that the copula for me is defined as the distribution of uniform random variables and when i work this out with $$C(F_{Z}(z),F_{Y}(v)-_{Y}(u))=P(U_{1}\leq F_{Z}(z),U_{2}\leq F_{Y}(v)-_{Y}(u))$$ I don't see how this reduces to the orginal probability. How does it work when using the definition of copula as a distribution of uniform random variables? $\endgroup$
    – Math Girl
    Mar 28, 2014 at 21:11
  • $\begingroup$ where did you get this equation from? specifically, the right hand side of it. please, refer to some text so we can see where you're wrong $\endgroup$
    – Aksakal almost surely binary
    Mar 28, 2014 at 21:18
  • $\begingroup$ The right hand side in my previous comment I computed myself using the definition of a copula as the cdf of uniform random variables. When looking at Sklars theorem and it's proof, a copula is the distribition function of uniform random variables . And i am trying to use this definition. $\endgroup$
    – Math Girl
    Mar 29, 2014 at 7:02

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