Marginal PDFs and conditional probability

Question

Consider the set \begin{align*} E=\{(x,y) \mid |x| + |y| \le 1\}. \end{align*} Suppose that we choose a point $(X,Y)$ uniformly at random in $E$. That is, the joint PDF of $X$ and $Y$ is given by \begin{align*} f_{XY}(x,y)= \begin{cases} c & (x,y)\in E \\ 0 & \mathrm{otherwise} \end{cases} \end{align*}

I let $A$ be the area of $E$. I know that

\begin{align*} 1&=\iint \limits_{E} f_{XY}(x,y) \mathrm{d}x\mathrm{d}y \\ &= c\iint \limits_{E} \mathrm{d}x\mathrm{d}y \\ &=cA. \end{align*}

$E$ forms a square with the side length $l=\sqrt{1^2+1^2}=\sqrt{2}$, so $A=l^2=2$ og therefore

\begin{align*} 1&=2c \Rightarrow c=\frac{1}{2}. \end{align*}

Determine marginal PDFs

I see that $R_X=R_Y=[-1,1]$. I know that for all $x$ that \begin{align*} f_X(x)&=\int_{-\infty}^{\infty} F_{XY}(x,y)\mathrm{d}y \\ &=\frac{1}{2} \int_{-\infty}^{\infty} \mathrm{d}y \end{align*}

Integration limits: \begin{align*} |x|+|y|\le 1 &\Rightarrow |y| \le 1-|x| \\ &\Rightarrow |x|-1 \le y \le 1-|x|. \end{align*} So for $(x,y) \in E$, I get that \begin{align*} f_X(x)&=\frac{1}{2} \int_{|x|-1}^{1-|x|} \mathrm{d}y \\ &=\frac{1}{2}\left(1-|x|-(|x|-1)\right) \\ &=\frac{1}{2}\left(2-2|x|\right) \\ &=1-|x|. \end{align*} So \begin{align*} f_{X}(x)= \begin{cases} 1-|x| & (x,y)\in E \\ 0 & \mathrm{ellers} \end{cases} \end{align*} Likewise I know that: \begin{align*} f_Y(y)&=\int_{-\infty}^{\infty} F_{XY}(x,y)\mathrm{d}x \\ &=\frac{1}{2} \int_{-\infty}^{\infty} \mathrm{d}x \end{align*} But since $E$ er symmetrical, I conclude that \begin{align*} f_Y(y)= \begin{cases} 1-|y| & (x,y)\in E \\ 0 & \mathrm{ellers} \end{cases} \end{align*}

Determine $f_{X|Y}(x|y)$

I know that \begin{align*} f_{X \mid Y}(x \mid y)&=\frac{f_{XY}(x,y)}{f_{Y}(y)} \\ &=\frac{1}{2(1-|y|}. \end{align*}

From the above I see that $f_{X \mid Y}(x \mid y)$ is not defined for $|y|=1$. So I want to split up into cases. The above result is valid for $|y| < 1$. When $|y|=1$ then $P(X=0 \mid Y=1)=1$ so $X \mid Y$ is a discrete random variable with $R_{X\mid Y}=\{0\}$.

That's as far as I've got here, how do I proceed?

Check if $X$ and $Y$ are independent

If $X$ og $Y$ are independent, then \begin{align*} f_{XY}(x,y)&=f_X(x)f_Y(y) \\ \Rightarrow \frac{1}{2}&=(1-|x|)(1-|y|) \\ &=1-|y|-|x|+|x||y| \end{align*}

That is as far as I've got here, how do I proceed?

I would also like to draw $x \mapsto f_X(x)$ and $x \mapsto f_{X \mid Y} (x \mid y)$ and see where there is highest probability for $X$ and $X \mid Y = y$.

How do I draw $x \mapsto f_{X \mid Y} (x \mid y)$?

Answer 1

Some points:

You conclude that $f_{X}\left(x\right)=\begin{cases} 1-\left|x\right| & \text{if }\left(x,y\right)\in E\\ 0 & \text{otherwise} \end{cases}$

Note however that $f_{X}$ is a function with only one argument. It would be correct to state that: $$f_{X}\left(x\right)=\begin{cases} 1-\left|x\right| & \text{if }\left|x\right|<1\\ 0 & \text{otherwise} \end{cases}$$

A similar story for $f_{Y}$.

Working out $f_{X\mid Y}\left(x\mid y\right)=\frac{f_{X,Y}\left(x,y\right)}{f_{Y}\left(y\right)}$ must take place in the awareness that $y$ is a fixed number with $\left|y\right|<1$ which in this context leads to: $$f_{X\mid Y}\left(x\mid y\right)=\begin{cases} \frac{1}{2\left(1-\left|y\right|\right)} & \text{if }\left|y\right|-1<x<1-\left|y\right|\\ 0 & \text{otherwise} \end{cases}$$

This makes clear how to draw the function prescribed by $x\mapsto f_{X\mid Y}\left(x\mid y\right)$ for a fixed $\left|y\right|<1$. In the case where $\left|y\right|\geq1$ the function is not defined and in the case where $\left|y\right|<1$ it takes two distinct values.

From the fact that $y$ is present in the expression it may be concluded that $X$ and $Y$ are not independent.

The case $\left|Y\right|=1$ is irrelevant because $P\left(\left|Y\right|=1\right)=0$ so that conditional probabilities like $P\left(X\in B\mid Y=1\right)$ and $P\left(X\in B\mid Y=-1\right)$ are not well defined.

It is not correct to state that $\left(X\mid Y\right)$ is a discrete random variable under $Y=1$.

Actually $\left(X\mid Y=y\right)$ does not even denote a random variable but only a distribution.

Answer 2

I'm not sure I'm following all your concerns, since you seem to have things more or less as might be expected, and/or I'm misunderstanding what is being asked, but here are a few thoughts that hopefully might help.

In regards to the independence, perhaps you can show that the equality is not satisfied for particular values of $x$ and $y$, say $x=1/4$ and $y=1/4$. The right hand side would give you $1-1/4-1/4+1/16 \ne 1/2$, therefore the random variables are not independent.

In regards to drawing the functions, it might be useful to start with the basic domain of the valid values of $x$ and $y$ which, I believe, can be described as a rotated square in the x-y plane-looking down from z-direction:

                            y                         f_xy(x|y)
                            |                             |      
                            |                        -----|----- h
                           .|.                       |    |    |
                         .  |  .                     |    |    |
  y=k (fixed)  --------.----|----.-----    -->    ----------------x
                     .      |      .               -(1-k)     1-k
               ----.-----------------.-----x          
                     .      |      . 
                       .    |    .                    h=1/(2*(1-|k|))
                         .  |  .
                           .|.
                            |
                            |

The conditional functions $f_{X|Y}(x|y)$ would be cuts in the horizontal direction, with constant height and limited x range. Perhaps this suggests some directions. I hope it helps.