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Given this pdf: $$f(y_1,y_2)=\begin{cases} 4y_1y_2 & 0\leq y_1\leq1 \\ 0 & \text{otherwise} \\ \end{cases} $$

The goal is to show that $\text{Cov}(Y_1,Y_2) = 0$.

First I found the marginal pdfs: $$f_1(y_1) = \int_0^1 4y_1y_2\text{ d}y_2 = 2y_1$$ $$f_2(y_2) = \int_0^1 4y_1y_2\text{ d}y_1 = 2y_2$$

Then I found the Expected Values: $$E[Y_1] = \int_0^1 2y_1\text{ d}y_1 = 1$$ $$E[Y_2] = \int_0^1 2y_2\text{ d}y_2 = 1$$ $$E[Y_1Y_2] = \int_0^1 \int_0^1 y_1y_24y_1y_2 \text{ d}y_1\text{d}y_2 = \frac{4}{9}$$

Then trying to find the Covariance I get: $$\text{Cov}(Y_1,Y_2) = E[Y_1Y_2] - E[Y_1]E[Y_2]$$ $$\text{Cov}(Y_1,Y_2) = \frac{4}{9} - 1\cdot1 = -\frac{5}{9}$$

However, I know that the Covariance should be 0 because they are independent (The limits don't depend on each other and the joint pdf can be factored without having the random variables combined).

I'm not sure where I made a mistake.

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$$E[Y_1] = \int_0^1 2y_1^2dy_1 = \frac23$$ $$E[Y_2] = \int_0^1 2y_2^2dy_2 = \frac23$$

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  • $\begingroup$ Why is it $2y_1^2$ and not $2y_1$? Isn't the formula for expected value supposed to be $E[Y] = \int_y y*g(y) dy$ where $g(y)$ is the marginal pdf? $\endgroup$ – user3370201 Nov 22 '15 at 20:54
  • $\begingroup$ @user3370201 Yes, but you got $g(y)=2y$. $\endgroup$ – MoebiusCorzer Nov 22 '15 at 20:54
  • $\begingroup$ I see where I went wrong. I forgot to multiply it by the $y$ in $\int_y y*g(y) dy$ so I get $2y$ and not $2y^2$ $\endgroup$ – user3370201 Nov 22 '15 at 20:57
  • $\begingroup$ @user3370201 Yes, exactly. $\endgroup$ – Chayu Nov 22 '15 at 20:59
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Another way to do it, assuming some knowledge of elementary probabilistic properties. You proved that $f(y_{1},y_{2})=f_{1}(y_{1})f_{2}(y_{2})$, hence $$\mathbb{P}[Y_{1}\in A_{1}, Y_{2}\in A_{2}]=\mathbb{P}[Y_{1}\in A_{1}]\cdot\mathbb{P}[Y_{2}\in A_{2}]$$

Which is the definition of the independence of $Y_{1}$ and $Y_{2}$, which directly implies $\mathbb{E}[Y_{1}Y_{2}]=\mathbb{E}[Y_{1}]\cdot\mathbb{E}[Y_{2}]$ and $\text{Cov}[Y_{1},Y_{2}]=0$

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