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I do not understand the notation of Spearman's correlation in the article of B Schweizer, EF Wolff - The annals of statistics, 1981 - JSTOR given by $\rho = \frac{1}{\sigma_X \sigma_Y}\int_{R^2}( F_{X,Y}(x,y) - F_X(x)F_Y(y)) dx dy$

Whn I use analytical formula of the correlation $\rho = E(XY) - E(X) E(Y)$ gives $\rho = \int_{R^2}( f_{X,Y}(x,y) - f_X(x) f_Y(y) )dx dy$

Should I developp the above two expresions to find an equivalence or is there any probability property to apply to get directly the result?

Many thanks in advance

HammerPower
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  • The integral in your second paragraph is equal to $0$, which is probably not what you meant to write. Maybe the formula that puzzles you is related to the result described in https://en.wikipedia.org/wiki/Expected_value#General_case_2 . – kimchi lover Nov 23 '17 at 21:45
  • Hi, Thanks for the article that is pretty useful,

    I may believe that my second integral is equal to 0 only in the case of independance $f_{X,Y}(x,y)=f_X(x)f_Y(y)$ if and only if $X$ and $Y$ are independant random variables, Additional example is the independant copula is given by $C(u,v)=uv$ were according to Sklar theorem $F_{X,Y}(x,y)=C(u,v)$, the joint distribution function accepts one and one copula only

     – HammerPower Nov 25 '17 at 16:16
  • As to your second integral: $\int_{R^2}f_{X,Y}(x,y)dxdy=1$ and $\int_Rf_X(x)dx=1$ and $\int_Rf_Y(y)dy=1$, so the whole thing equals $1-1\times1=0$. – kimchi lover Nov 25 '17 at 16:32
  • True I meant to write $\int_{R^2}[xyf_{X,Y}(x,y)-xf_X(x)yf_Y(y)]dxdy$ – HammerPower Nov 28 '17 at 06:16

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