0
$\begingroup$

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$$f(x,y) = \begin{cases} 8xy & 0 \le x \le y \le1 \\ 0 & \mbox{elsewhere,} \end{cases}$$

Find the covariance of $X$ and $Y$ .

I know the formula as $σ_{xy}=E(XY ) − μ_Xμ_Y $

And the given solution is as follows

We first compute the marginal density functions. They are

$$g(x) = \begin{cases} 4x^3 & 0 \le x \le1 \\ 0 & \mbox{elsewhere.} \end{cases}$$

and $$h(y) = \begin{cases} 4y(1 − y^2) & 0 \le y \le1 \\ 0 & \mbox{elsewhere.} \end{cases}$$

My Question:

How did they get $g(x)$ and $h(y)$?

Did they use $\int_0^1 f(x,y)$ dy and $\int_0^1f(x,y)dx$?

$\endgroup$
2
  • 4
    $\begingroup$ Please add the self-study tag and expand on the difficulties you have with notions like marginal densities and covariance of continuous variables. In particular explain why you did not try to compute $\int_0^1 f(x,y)\text{d}x = h(y)$ $\endgroup$
    – Xi'an
    Nov 8, 2018 at 18:44
  • 1
    $\begingroup$ See this answer for some details about the marginal density can be derived from the joint density. $\endgroup$ Nov 8, 2018 at 19:13

1 Answer 1

1
$\begingroup$

Since your joint is non-zero when $x\leq y$, $h(y)=\int_0^{y}{f(x,y)dx}$. And, it appears that your $h(y)$ is not true, PDF of Y should be $f_Y(y)=g(y)$. Or the initial condition should be $y\leq x$. Anyway, after finding marginals, you calculate the means. And for $E[XY]$, you'll just perform a joint integration by respecting the limits.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.