The intervals for x and y is (-1,1). X and Y are independent if "E(XY)=E(X)\u00d7E(Y)"
"E(X)=\\iint(\\frac{X}{4}(1+XY))dx dy X,Y\\isin(-1,1)"
"=\\intop(\\frac{X^2}{8}+\\frac{X^3Y}{12})dy X,Y\\isin(-1,1)"
"=\\frac{X^2Y}{8}+\\frac{X^3Y^2}{24}. X,Y\\isin(-1,1)"
"=\\frac{Y}{8}+\\frac{Y^2}{24}-\\frac{Y}{8}+\\frac{Y^2}{24},\nX\\isin(-1,1)"
"=\\frac{1}{24}+\\frac{1}{24}-\\frac{1}{24}-\\frac{1}{24}=0"
Since both X and Y have same domain and are at same position on the distribution, "E(X)=E(Y) =0"
"E(XY)=\\iint(\\frac{XY}{4}(1+XY))dx dy, X,Y\\isin(-1,1)"
"=\\intop(\\frac{X^2Y}{8}+\\frac{X^3Y^2}{12})dy ,X,Y\\isin(-1,1)"
"=\\frac{X^2Y^2}{16}+\\frac{X^3Y^3}{36}X,Y\\isin(-1,1)"
"=\\frac{Y^2}{16}+\\frac{Y^3}{36}-\\frac{Y^2}{16}+\\frac{Y^3}{36} , Y\\isin(-1,1)"
"=\\frac{1}{36}+\\frac{1}{36}+\\frac{1}{36}+\\frac{1}{36}=\\frac{1}{9}"
Since E(XY) is not equal to the product of E(X) and E(Y), X and Y are not independent.
"E(X^2)=\\iint(\\frac{X^2}{4}(1+XY))dx dy X,Y\\isin(-1,1)"
"=\\intop(\\frac{X^3}{12}+\\frac{X^4Y}{16})dy X,Y\\isin(-1,1)"
"=\\frac{X^3Y}{12}+\\frac{X^4Y^2}{32}X,Y\\isin(-1,1)"
"=\\frac{Y}{12}+\\frac{Y^2}{32}+\\frac{Y}{12}-\\frac{Y^2}{32},Y\\isin(-1,1)"
"=\\frac{1}{12}+\\frac{1}{12}+\\frac{1}{12}+\\frac{1}{12}=\\frac{1}{3}"
Similarly, since both X and Y have same domain and are at same position on the distribution, "E(X^2)=E(Y^2)=\\frac{1}{3}"
"E(X^2Y^2)=\\iint(\\frac{X^2Y^2}{4}(1+XY))dx dy, X,Y\\isin(-1,1)"
"=\\intop(\\frac{X^3Y^2}{12}+\\frac{X^4Y^3}{16})dy ,X,Y\\isin(-1,1)"
"=\\frac{X^3Y^3}{36}+\\frac{X^4Y^4}{64} ,X,Y\\isin(-1,1)"
"=\\frac{Y^3}{36}+\\frac{Y^4}{64}+\\frac{Y^3}{36}-\\frac{Y^4}{64}\n ,Y\\isin(-1,1)"
"=\\frac{1}{36}+\\frac{1}{36}+\\frac{1}{36}+\\frac{1}{36}=\\frac{1}{9}"
"E(X^2)\u00d7E(Y^2)=\\frac{1}{3}\u00d7\\frac{1}{3}=\\frac{1}{9}"
Since the product of "E(X^2)" and "E(Y^2)" is equal to "E(X^2Y^2)," "X^2" and "Y^2" are independent.
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