Determine marginal densities and distributions from joint density

[psie]

Homework Statement

Let $X$ and $Y$ have joint density $$f(x,y)=\begin{cases} 1& \text{for }0\leq x\leq 2,\max(0,x-1)\leq y\leq\min(1,x) \\ 0 &\text{otherwise}.\end{cases}$$ Find the marginal density functions and the joint and marginal distribution functions.

Relevant Equations

The marginal distribution of $X$ given the joint density $f_{X,Y}$ is given by $f_X(x)=\int_\mathbb{R} f_{X,Y}(x,y) \,dy$ and similar for $Y$.

This is the follow-up problem to my previous problem.

"Integrating out" the $y$-variable and $x$-variable separately, we see that $f_Y(y)=2$ and $f_X(x)=\min(1,x)-\max(0,x-1)$. From my previous post, we see that $X$ is the sum of two independent $U(0,1)$-distributed r.v.s. What is the distribution of $Y$ though? It looks to me that if $0\leq x\leq 2$, i.e. if $0\leq x\leq 1$ or $1<x\leq 2$, then $0\leq y\leq x$ or $x-1\leq y\leq 1$ respectively. So $y$ ranges from $0$ to $1$, which doesn't make sense since then the pdf $f_Y(y)=2$ does not integrate to $1$. However, currently I don't see the error.

 

[Orodruin]

What makes you think that this

$$f(x,y)=\begin{cases} 1& \text{for }0\leq x\leq 2,\max(0,x-1)\leq y\leq\min(1,x) \\ 0 &\text{otherwise}.\end{cases}$$

Is the density function of the independent RVs $X$ and $Y$ from your previous post?

If $X$ and $Y$ are uniform independent RVs then the density function is a constant on the product of the supports of each variable.

 

[Orodruin]

Additionally, even if the problem is not supposed to be the independent RVs from your previous post, the correct computation of $f_Y(y)$ has the result $f_Y(y) = 1$, not 2.

 

[psie]

What makes you think that this is the density function of the independent RVs $X$ and $Y$ from your previous post?

I am not saying that $f$ given here is the joint density of two $U(0,1)$-distributed, independent RVs. What I'm claiming is that $X$ in this post is the sum of two independent RVs, both of which are $U(0,1)$, since in my previous post, we found that the density of the sum of two independent RVs which are $U(0,1)$ is $f_X$ as specified here. Am I making sense?

Additionally, even if the problem is not supposed to be the independent RVs from your previous post, the correct computation of $f_Y(y)$ has the result $f_Y(y) = 1$, not 2.

How did you obtain $f_Y(y) = 1$? I just don't see it.

 

[Orodruin]

How did you obtain $f_Y(y) = 1$? I just don't see it.

Did you try drawing the support of the joint distribution function? If you do it should be pretty clear.

 

[Orodruin]

Also note that writing down the support region is significantly simpler if you change the order. In other words, start writing down the domain for $y$ and then write the restriction on $x$ given $y$.

 

[psie]

Did you try drawing the support of the joint distribution function? If you do it should be pretty clear.

Ok, I will try. What is wrong about calculating the marginal density by integrating out the $x$-variable though? In other words, $$f_Y(y)=\int_\mathbb{R} f(x,y) \,dx=\int_0^2 1\, dx=2?$$

 

[Orodruin]

Ok, I will try. What is wrong about calculating the marginal density by integrating out the $x$-variable though? In other words, $$f_Y(y)=\int_\mathbb{R} f(x,y) \,dx=\int_0^2 1\, dx=2?$$

$$f_Y(y)=\int_\mathbb{R} f(x,y) \,dx \neq \int_0^2 1\, dx=2?$$

Draw the support region for $f$ and you will see it.