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Let$\ H(u_1,u_2)$ and$\ M(u_1,u_2)$ be bivarate CDF with standard uniform margins. What probability will be defined by the following integral (if any): $$ \int\limits_{[0;1]^2} H(u_1,u_2)\;dM(u_1,u_2) = P(\cdots)\ ?$$

Thanks in advance.

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Why do you think that this integral should define any probability distribution ? It's a number. – Sasha Feb 6 '12 at 16:48
The result of your double integral will be a number, not a probability distribution, right? If not, could you give more details about what exactly the integral is, even if it is for the specific case $M(u_1, u_2) = u_1u_2$ for $0 \leq u_1, u_2, \leq 1$ which corresponds to independent $U[0,1]$ random variables? – Dilip Sarwate Feb 6 '12 at 16:50
Could it be that you meant $[0,1]\times[0,y]$ instead of $[0,1]^2$? BTW, you had lots of nested {{braces}} that weren't needed, which I deleted. – Michael Hardy Feb 6 '12 at 17:38
Sorry for confusion. Yes, this integral is a number. I modified the question. What I actually wanted to know if this number actually denotes some meaningfull probablity. – learningmath Feb 6 '12 at 17:56
@MichaelHardy No. Thanks for your edit. – learningmath Feb 6 '12 at 17:57
up vote 1 down vote accepted

Assume that $H$ is the CDF of $(X_1,X_2)$, that $M$ is the CDF of $(Y_1,Y_2)$, and that $(X_1,X_2)$ and $(Y_1,Y_2)$ are independent. Then, $$ \int H(u_1,u_2)\mathrm dM(u_1,u_2)=\mathrm E(H(Y_1,Y_2))=\mathrm P(X_1\leqslant Y_1,X_2\leqslant Y_2). $$

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