Covarianza de dos variables aleatorias. 92

Por el teorema 4.10

Cov(Y1, Y2) = E(Y1·Y2) - E(Y1)E(Y2)

$$\begin{align}&E(Y_1)=\int_0^1\int_{y_1}^16y_1(1-y_2)dy_2dy_1=\\ &\\ &\int_0^1 6y_1\left[y_2-\frac{y_2^2}{2}  \right]_{y_1}^1 dy_1=\\ &\\ &\int_0^16y_1\left(\frac 12-y_1+\frac{y_1^2}{2}\right)dy_1=\\ &\\ &\int_0^1 (3y_1^3-6y_1^2+3y_1)dy_1=\\ &\\ &\left[ \frac {3y_1^4}{4}-2y_1^3+\frac{3y_1^2}{2} \right]_0^1=\\ &\\ &\frac 34-2+\frac 32 = \frac{3-8+6}{4}= \frac{1}{4}\end{align}$$

Como se nota que hay un limite variable

$$\begin{align}&E(Y_2)=\int_0^1\int_{y_1}^1 (6y_2-6y_2^2)dy_2dy_1=\\ &\\ &\int_0 ^1 \left[3y_2^2-2y_2^3 \right]_{y_1} ^1dy_1 =\\ &\\ &\int_0^1 (1-3y_1^2+ 2y_1^3)dy_1 =\\ &\\ &\left[y_1-y_1^3+\frac 24y_1^4 \right]_0^1= 1-1 +\frac 24 = \frac 12\end{align}$$

Y finalmente

$$\begin{align}&E(Y_1·Y_2)=\int_0^1\int_{y_1}^1 6y_1y_2(1-y_2)dy_2dy_1=\\ &\\ &\int_0^1\int_{y_1}^1 6y_1(y_2-y_2^2)dy_2dy_1=\\ &\\ &\int_0^1 6y_1\left[\frac{y_2^2}{2}-\frac{y_2^3}{3}  \right]_{y_1}^1 dy_1=\\ &\\ &\int_0^1 6y_1\left(\frac 12-\frac 13-\frac{y_1^2}{2}+\frac{y_1^3}{3}\right)dy_1=\\ &\\ &\int_0^1 (y_1-3y_1^3+2y_1^4)dy_1=\\ &\\ &\left[ \frac {y_1^2}{2}-\frac{3y_1^4}{4}+\frac{2y_1^5}{5} \right]_0^1=\\ &\\ &\frac 12-\frac 34+\frac 25=\frac{10-15+8}{20}=\frac{3}{20}\end{align}$$

Cov(Y1, Y2) = 3/20 - (1/4)(1/2) = 3/20 - 1/8 = (24-20)/160 = 4/160 = 1/40

No, no son independientes. Si lo fueran tendría que ser cero la covarianza.

Y eso es todo.