I need to solve the following question. I apologize in advance not quite sure how to note piece wise functions.

$$ f(x,y)=\begin{cases}72x^2y(1-x)(1-y),& 0 \leq x\leq1, 0\leq y \leq 1\\ 0,& \mathrm{otherwise.} \end{cases} $$

Now I know that the $\operatorname{Cov}(X)=E[XY]-E[X]E[Y]$ and I have attempted to solve by using the same equation for the above continuous function, however, I'm not sure if I am doing this right. By taking the expected values of $x$ and $y$ seperately, there will be variables left and it won't give an exact constant as an answer. For example:


I'm not sure if I'm doing this right. Also, the next question is:

Determine $P({X>Y})$. Which I don't know how to solve

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    $\begingroup$ $E[X]=\int_0^1\int_0^1x\times72x^2y(1-x)(1-y)dxdy$ $\endgroup$ – Guangliang Mar 23 '17 at 23:25
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    $\begingroup$ $P(X>Y) = \int_0^1dy\int_y^1dx f(x,y)$ $\endgroup$ – Guangliang Mar 23 '17 at 23:27
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    $\begingroup$ and $\mathsf P(X>Y) = \int_0^1\int_0^x f_{X,Y}(x,y)\operatorname dy\operatorname dx$ $\endgroup$ – Graham Kemp Mar 23 '17 at 23:28
  • $\begingroup$ Excellent that's where my mistake was. Also, the two equations for the second question you both gave me are different, which one is correct? $\endgroup$ – JKawa Mar 23 '17 at 23:54
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    $\begingroup$ Both They are equivalent. (Use Fubini's Theorem to change operator order.) $\endgroup$ – Graham Kemp Mar 23 '17 at 23:59

You have the joint probability density function, not the marginal, we have to use that.

In general $\mathsf E(g(X,Y)) = \int_0^1\int_0^1 g(s,t)~f_{X,Y}(s,t) \operatorname d s\operatorname d t$


$$\begin{align}\mathsf E(X)&=\int_0^1\int_0^1 x f(x,y)\operatorname d y\operatorname d x\\ &=\int_0^1\int_0^1 x\cdot 72 x^2y(1-y)(1-x)\operatorname d y\operatorname d x\\ &= \int_0^1 x\cdot 12x^2(1-x)\operatorname d x\cdot\int_0^1 6y(1-y)\operatorname d y\end{align}$$


$$\begin{align}\mathsf E(Y)& =\int_0^1\int_0^1 y f(x,y)\operatorname d y\operatorname d x\\ &= \int_0^1 12x^2(1-x)\operatorname d x\cdot\int_0^1 y\cdot6y(1-y)\operatorname d y\\[2ex]\mathsf E(XY)&=\int_0^1\int_0^1 xy f(x,y)\operatorname d y\operatorname d x\\ &= \int_0^1 x\cdot12x^2(1-x)\operatorname d x\cdot\int_0^1 y\cdot6y(1-y)\operatorname d y\end{align}$$

You may now find $~\mathsf E(XY)-\mathsf E(X)\mathsf E(Y)~$.   It is easier than it looks at first glance.

Then because $\mathsf P(X>Y) =\mathsf E(\mathbf 1_{X>Y})$, we have:

$$\begin{align}\mathsf P(X>Y) &= \int_0^1 12x^2(1-x) \int_0^x 6y(1-y)\operatorname dy\operatorname d x\\[2ex] &= \int_0^1 6y(1-y) \int_y^1 12x^2(1-x)\operatorname dx\operatorname d y\end{align}$$

  • $\begingroup$ Excellent explanation. Thanks! $\endgroup$ – JKawa Mar 24 '17 at 0:14

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