Let S be the shadowed region:enter image description here

Suppose (X,Y) have a uniform distribution over S, their joint PDF is given by $f_{X,Y}(x,y)=\frac{1}{16}, (x,y) \in S$.

Problem 1: find the marginal PDF $f_X(x)$ of X.

Question 1: I know f_X(x) is the integral of Y=y but how do I represent this in this diagram? $f_X(x)= \int_0^4 \frac{1}{16} dy$?

  • 1
    $\begingroup$ The joint density is not $1/16$ everywhere; it is $0$ outside of $S$. You are correct that $f_X(x) = \int_{-\infty}^\infty f_{X,Y}(x,y) \, dy$, but you need to figure out where the integrand is nonzero; this will depend on the particular value of $x$. $\endgroup$
    – angryavian
    Commented Mar 1, 2020 at 17:13
  • $\begingroup$ @angryavian So how should I fix this? Can you add more detail? $\endgroup$
    – neveryield
    Commented Mar 1, 2020 at 17:16

1 Answer 1


Marginal PDF can be expressed like this: $$f_X(x) = \int_{0}^{2}f(x,y)dy$$ whenever $1 \leq |x| \leq 3$ and $$f_X(x) = \int_{0}^{4}f(x,y)dy$$ whenever $|x|<1$ The first equation is $\frac{1}{16}*2 - \frac{1}{16}*0 = 1/8$ while the second equation is $\frac{1}{16}*4 - \frac{1}{16}*0 = 1/4$. You can check yourself by integrating over $x$: $$2*\int_{1}^{3}\frac{1}{8}dx = 1/2$$ and $$\int_{-1}^{1}\frac{1}{4}dx = 1/2$$, giving $1$ as a total.


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