# About uniform distribution and marginal PDF.

Let S be the shadowed region: 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? $$f_X(x)= \int_0^4 \frac{1}{16} dy$$?

• 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 $x$. Mar 1, 2020 at 17:13
• @angryavian So how should I fix this? Can you add more detail? Mar 1, 2020 at 17:16

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.