probabilityprobability distributions
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 conditional density of $Y$ given $X=x$ is $f_{Y|X}(y|x)=\frac{1}{x+1}1_{0<y<x+1}$, hence the joint density of $X$ and $Y$ is $$ f(x,y)=f_{Y|X}(y|x)f_X(x)=\frac{1}{x+1}1_{0<x<1,0<y<x+1}$$
The marginal density can then be obtained by "integrating out" the $x$-variable: $$ f_Y(y)=\int f(x,y)\;dx$$
The joint density looks like this:
So, the marginal distribution of $Y$ is given by the following integral
$$\int_y^1 2 \ dx=2(1-y)$$
if $0\leq y\leq 1$.
Best Answer
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.