Question

The fraction x of male runners and the fraction y of female runners who compete in marathon races are described by the joint density function f(x, y) = 8xy, 0 y x 1, 0, elsewhere. Find the covariance of x and y .

Answer 1

Answer:

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Answer 2

The covariance of x and y is $\frac{4}{225}$.

Solution:

Let us first find the marginal density function. They are

$g(x)=\left\{\begin{array}{ll}4 x^{3}, & 0 \leq x \leq 1 \\0, & \text { elsewhere }\end{array}\right.$

and

$h(y)=\left\{\begin{array}{ll}4 y\left(1-y^{2}\right), & 0 \leq y \leq 1 \\0, & \text { elsewhere }\end{array}\right.$

From these marginal density functions, we calculate

$\mu_{x}=E(X)$

    $=\int_{0}^{1} 4 x^{4} d x$

Using integration formula, $\int x^{n} d x=\frac{x^{n+1}}{n+1}+C$

     $=\left[\frac{x^{4+1}}{4+1} \right]^1_0$    

     $=\frac{4}{5}$

$\mu_{Y}=\int_{0}^{1} 4 y^{2}\left(1-y^{2}\right) d y$

     $=\frac{8}{15}$

From the joint density function given above, we have

$E(X Y)=\int_{0}^{1} \int_{y}^{1} 8 x^{2} y^{2} d x d y$

            $=\frac{4}{9}$

$\sigma_{X Y}=E(X Y)-\mu_{X} \mu_{Y}$

       $=\frac{4}{9}-\left(\frac{4}{5}\right)\left(\frac{8}{15}\right)$

       $=\frac{4}{9}-\frac{32}{75}$

       $=\frac{4}{225}$

Hence the covariance of x and y is $\frac{4}{225}$.

To learn more...

1. The covariance between the length and weight of five items is 6 and their standard deviation are 2.45 and 2.61 respectively. find the coefficient of correlation between length and weight​

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2. The covariance between two variables x and y is 72. The variances of x and y are 144 and 84. The

coefficient of correlation is

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