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 $\int_0^1f(x,y)dx$?

  • 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


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.


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