Let R be the rectangle $\ \{(x, y); 0 <= x <= 2, 0 <= y<= 1\} $, and let $\ f(x, y) = >k(x^2+ y^2)$ on R and zero elsewhere.

(a) Find the value of k which makes f a joint density function.

(b) If X and Y are random variables with joint density function f(x, y), find

i. the marginal distributions of X and Y ;

ii. the expectations and variances of X and Y ;

iii. the covariance and correlation of X and Y .

For a) I set the double integral equal to 1 and got k = $\ 3\over10 $

for b)I) I got:

$\ f_X(x) = $$\ 3\over 10$$\ (x^2+ $ $1 \over 3$$)$ and

$\ f_Y(y) = $$\ 3\over 10$$\ ($$8 \over 3$$+ 2y^2)$

I'm not sure about the rest. For b) II) I'm thinking it might be:

$E(X) = \int xf_X(x)dx$, $\int E(Y) = yf_Y(y)dy$


$Var(X) = \int (x-E(X))^2f_X(x)dx$, $Var(Y) = \int (y-E(Y))^2f_Y(y)dy$

I've no idea about covariance or correlation yet. I'll cross that bridge when I come to it.

Am I on the right track?


  • $\begingroup$ Yes all good up to here. $\endgroup$ – user88595 Mar 7 '14 at 11:45

It´s all good. However for calculating the variance I do suggest using the following identity, which is easily derived($\mu_x$ is $E(X)$, in case it is not clear): $$\begin{align} var(X) &= E\left((X-\mu_x)^2\right)\\ &= E\left(X^2 - 2X\mu_x + \mu_x^2\right)\\ &= E(X^2) - E(2X\mu_x) + E(\mu_x^2)\\ &= E(X^2) - 2\mu_xE(X) + \mu_x^2\\ &= E(X^2)-2\mu_x^2+\mu_x^2\\ &=E(X^2)-\mu_x^2 \\ \end{align} $$ You already have $\mu_x$, but you still have to solve an integral here for finding $E(X^2)$. However, it is in general easier than directly finding $E((X-\mu_x)^2)$. In any case, it's better if you find the variance both ways a few times, so you convince yourself the identity makes life easier most times.

For covariance and correlation there is not much of a bridge to cross. Covariance is just $E((X-E(X)(Y-E(Y))$ Again in this case, avoid solving the integral directly, as this expected value can be easily split in easier terms and leads to the following identity:

$$\begin{align} cov(X,Y) &= E((X-\mu_x)(Y-\mu_y) \\ &= E(XY - X\mu_y - Y\mu_x + \mu_y\mu_y) \\ &= E(XY) - E(X\mu_y) - E(Y\mu_x) + E(\mu_x\mu_y)\\ &= E(XY) - \mu_yE(X) - \mu_xE(Y) + \mu_x\mu_y\\ &= E(XY) - \mu_x\mu_y - \mu_x\mu_y + \mu_x\mu_y\\ &= E(XY) - \mu_x\mu_y\\ \end{align} $$ If you are taking a course on probability you may want to memorize the previous identity also. Then the correlation is simply: $$ \rho(X,Y)=\frac{cov(X,Y)}{\mu_x\mu_y} $$

  • $\begingroup$ Hi Pedro and welcome to math.SE! I substitued the \hspace expressions in your equations to align. You can edit the question to see how I did it. $\endgroup$ – Daniel R Mar 7 '14 at 14:11
  • $\begingroup$ thanks Daniel, it´s the first time I use LaTeX. The \hspace didn´t feel right to be honest, :) $\endgroup$ – Pedro Mar 7 '14 at 15:19
  • $\begingroup$ I get $var(X) = 2.0564-(7/5)^2 = 0.0964$ Is that right? $\endgroup$ – user123965 Mar 8 '14 at 11:36
  • $\begingroup$ And I get a covariance of $3/4 - (0.55 \times 7/5) = -0.02$ and correlation of -0.026 $\endgroup$ – user123965 Mar 8 '14 at 11:50

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