2
$\begingroup$

So I have this pdf, $f(x)=3x^2$ for $x\in (0,1)$ and I need to find $Cov(2X+7, X^2+3X-12)$. My main concern about how I answer this is, what is the joint pdf for these two distributions? I guess it's $f(x)$ but I wanted to confirm this.

But as for actually calculating the covariance, I first use the fact that constant terms can vanish, so that I instead calculate $Cov(2X, X^2+3X)$. I'll use the definition of covariance which tells me to compute

$$\int\int_{R}(x-\mu_{X})(y-\mu_{Y})f(x,y)dA$$

so I need to know each mean for $2X$ and $X^2+3X$. I know how to calculate those, so to avoid writing too much, let's assume I've found their values. Then I believe the integral I want to compute is

$$\int_{0}^{1}(x-\mu_X)(x-\mu_Y)f(x)dx$$

Is this correct?

$\endgroup$
3
  • 1
    $\begingroup$ It seems you are posting a lot of what looks like school work? If this is school work, please mark your posts with the "self-study" tag. $\endgroup$ Mar 23, 2015 at 0:52
  • 1
    $\begingroup$ @StatsStudent I'm not in the class (so I'm learning this independently) but it is material from a class. I can label it as self-study if that helps. $\endgroup$
    – Addem
    Mar 23, 2015 at 0:56
  • $\begingroup$ You do not need the joint of $(2X+7, X^2+3X-12)$ to compute the covariance. And no, the joint pdf of $(2X+7, X^2+3X-12)$ is not $f(x)$. $\endgroup$
    – Xi'an
    Mar 23, 2015 at 7:14

1 Answer 1

3
$\begingroup$

Use the fact that, $$\text{Cov}(X,X) = \text{Var}(X)$$ and that, $$\text{Cov}(aX+bY,cW+dV) = ac \,\text{Cov}(X,W)+ad\,\text{Cov}(X,V)+bc\,\text{Cov}(Y,W)+ bd\,\text{Cov}(Y,V)$$

The rest should be trivial.

$\endgroup$
5
  • $\begingroup$ OK, I'm pretty sure I can do that. But just for my general knowledge, was there anything wrong in what I did in the OP? Or is it just less efficient/elegant? $\endgroup$
    – Addem
    Mar 23, 2015 at 0:52
  • $\begingroup$ Also, after having done this, I suppose I should remark (to confirm that I'm right, and for the sake of anyone who might read this later) that it seems you still ultimately need to compute $E(X), E(X^2),$ and $E(X^3)$. $\endgroup$
    – Addem
    Mar 23, 2015 at 1:19
  • 1
    $\begingroup$ There is no contradiction between stoched's solution and $$\int\int_{\mathbb{R}^2}(x-\mu_{X})(y-\mu_{Y})f(x,y)\text{d}x\text{d}y$$ since this is the definition of covariance. The second equation $$\int_{0}^{1}(x-\mu_X)(x-\mu_Y)f(x)\text{d}x$$ is wrong. Since $x^2+3x=\psi(x)$ is a function of $x$, it should read $$\int_{0}^{1}(x-\mu_X)(\Psi(x)-\mu_Y)f(x)\text{d}x$$ $\endgroup$
    – Xi'an
    Mar 23, 2015 at 7:18
  • $\begingroup$ I take it $\Psi(x)=\int_0^x(t^2 + 3t)dt$? How does this follow from the definition of covariance? I've tried taking another stab at this to try to make sure I understand all this, and I got: Since we have basically $Cov(V,W)$ where $V=2X+7$ and $W=X^2+3X-12$ so if $V=v=2x+7$ and $W=w=x^2+3x-12$ then $Cov(V,W)=\int_0^1\int_0^1(v-\mu_V)(w-\mu_W)f(x)dx = \int_0^1\int_0^1(2x+7-\mu_V)(x^2+3x-12-\mu_W)f(x)dx$. I'm sure you're right about this, but I'm not understanding the exact justification for what you got or how my result went wrong. $\endgroup$
    – Addem
    Mar 23, 2015 at 20:49
  • $\begingroup$ @Xi'an Actually, I think maybe my mistake was using the same $x$ for both variables when technically they should be distinct? Maybe. I don't know. $\endgroup$
    – Addem
    Mar 23, 2015 at 20:50

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.