I don't like/don't understand the solution to the SOA practice problem #171 so I'm trying to find the marginal density of x in another way without drawing a diamond.

Here is the joint density: $f(x,y)=\begin{cases} {{1\over2}}&0<|x|+|y|<1\\{0}&otherwise\end{cases}$

I thought I could do the following:

$f_x(x)= \int_{-1}^{1-x}{1\over2}dy \,+\,\int_0^{1+x}{1\over2}dy$

Then I tried

$f_x(x)= \int_{0}^{1-x}{1\over2}dy \,+\,\int_{1-x}^{1+x}{1\over2}dy$

but I'm not getting the same density as them with their drawing. If it comes down to it, I can try to learn their method but I would really like to just do it the way I know how. Can anyone tell me where my limits are going wrong?

  • $\begingroup$ Did you sketch a drawing of the support region? $\endgroup$ – leonbloy Apr 1 '17 at 17:13
  • $\begingroup$ @leonbloy it's a diamond, and I thought the 2nd one had the right limits for sure, but it doesn't work. $\endgroup$ – Heavenly96 Apr 1 '17 at 17:15

For a fixed $0<x<1$, the support region (check this) goes from $y=x-1$ to $y=1-x$

Hence, in that range, $$f_x(x)= \int_{x-1}^{1-x}\frac12 dy= 1 - x$$

The range $-1<x<0$ is similar.

  • $\begingroup$ so for $-1<x<0\, f_x(x)=\int_{-x-1}^{x+1}{1\over2}dy=x+1$? $\endgroup$ – Heavenly96 Apr 1 '17 at 18:30
  • 1
    $\begingroup$ yes. you should check that the answer makes sense, for example, that $f_x$ integrates to 1 $\endgroup$ – leonbloy Apr 1 '17 at 19:27

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.