Expectation value proof

1. The problem statement, all variables and given/known data



[itex]f(x,y)=6a^{-5}xy^{2}[/itex] [itex]0≤x≤a[/itex] and [itex]0≤y≤a[/itex], [itex]0[/itex] elsewhere



Show that [itex]\overline{xy}=\overline{x}.\overline{y}[/itex]



2. Relevant equations



[itex]\overline{x}=\int^{∞}_{-∞}{x.f(x)dx}[/itex]



3. The attempt at a solution



[itex]\overline{x}=\int^{∞}_{-∞}{x.f(x)dx}[/itex]

[itex]=\int^{a}_{0}{x.6a^{-5}xy^{2}dx}[/itex]

[itex]=6a^{-5}\int^{a}_{0}{x^{2}y^{2}dx}[/itex]

[itex]=6a^{-5}[/itex][itex]\frac{1}{3}a^{3}y^{2}[/itex]

[itex]=2a^{-2}y^{2}[/itex]



Following the same proccess I get [itex]\overline{y}=\frac{3}{2}a^{-1}x^{2}[/itex]



But when it comes to [itex]\overline{xy}[/itex] I'm not really sure how to approach it



I tried

[itex]\overline{xy}=\int^{∞}_{-∞}{x.y.f(x,y)dx}[/itex]

[itex]=\int^{a}_{0}\int^{a}_{0}{x.y.6a^{-5}xy^{2}dx}[/itex]

[itex]=\int^{a}_{0}\int^{a}_{0}{6a^{-5}x^{2}y^{3}dx}[/itex]

[itex]=\frac{1}{2}a^{2}[/itex]

which does not equal [itex]\overline{x}.\overline{y}[/itex]

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