0
$\begingroup$

the common probability density function $f(x,y)$ of two random variables $X,Y$ is constant in the rectangle $Q$ and vanishes outside of it.

$f(x,y)=\begin{cases}\frac{1}{2} &\quad, (x,y)\in Q\\0 &\quad, \text{else}\end{cases}$

enter image description here

Now I'm interested in the marginal probability density function for $X$. We have to consider two cases.

Case 1: $-1\leq x \leq 0$

$f_X(x)=\int_{-\infty}^\infty f(x,y) dy = \int_{-1-x}^{1+x} \frac{1}{2} dy = \dots = 1+x$

Case 1: $0\leq x \leq 1$

$f_X(x)=\int_{-\infty}^\infty f(x,y) dy = \int_{-1+x}^{1-x} \frac{1}{2} dy = \dots = 1-x$

So we get

$f_X(x)=\begin{cases} 1+x &\quad, -1\leq x \leq 0\\ 1-x &\quad, 0\leq x \leq 1 \\ 0 &\quad, \text{else}\end{cases}$

Now I kind of have a hard time interpreting $f_X$. Can someone try to explain why the result we got makes sense?

$\endgroup$
  • 1
    $\begingroup$ Your work and answer are correct - what is your question exactly? $f_X$ is a density function, not sure what there is to "interpret" here. $\endgroup$ – Math1000 Aug 1 at 13:13
  • $\begingroup$ Like I think I should be able to look at $f_X$ and think "Yeah, that is a result which does indeed make sense for $f$". E.g. before getting $f_X$ you might think about what kind of solution you get and then you get something and you aren't really surprised. I can't do that here, I don't see why one would expect this kind of $f_X$. $\endgroup$ – handy Aug 3 at 11:35

Your Answer

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

Browse other questions tagged or ask your own question.