Marie has been taking note of the time she leaves for work and the length of her morning commute. She decided to model the number of hours $X$

Question

Marie has been taking note of the time she leaves for work and the length of her morning commute. She decided to model the number of hours $X$ after 6:30 a.m that she leaves for work with a $uniform(2)$ distribution, ie $X$ has pdf:

$\begin{equation*} f_X(x) = \begin{cases} c & \quad 0<x<2 \\ 0 & \quad\text{otherwise} \end{cases} \end{equation*}$

She notices that, when $0<x<2$ the length of the commute (in hours), $Y$, seems to depend upon the time she leaves for work and decides to model $Y|X$ with a distribution, with pdf:

$\begin{equation*} f_{Y\mid X}(y\mid x) = \begin{cases} 6+8(x-y) & \quad \frac{1}{4}+x<y<\frac{3}{4}+x \\ 0 & \quad\text{otherwise} \end{cases} \end{equation*}$

Note that $f_{Y\mid X}(x,y)$ is defined only for $0<x<2$

$a.$ Determine the value of $c$

$b.$ Find the joint pdf of $X$ and $Y$, $f_X,_Y(x,y).$ Remember to include the range of the values of $x$ and $y$ for which the result holds.

$c.$ Find that probability that Marie left home before 7:00 am and her commute is less than $45$ minutes.

$d.$ Copy and complete following to rewrite $f_X,_Y(x,y).$ with $y$ as leading variable.

$\begin{equation*} f_{Y,X}(y,x) = \begin{cases} 3+4(x-y) & \quad ...<y<...\max(...,0)<x<\min(2,...) \\ 0 & \quad\text{otherwise} \end{cases} \end{equation*}$

MY WORKING

$a.$ By integrating the given pdf $f_X(x)$ over its domain I find that $c=\frac{1}{2}$

$b.$ Since $f_{X,Y}(x,y)=f_{Y\mid X}(x\mid y)\cdot f_X(x)$. Since we know values of $f_{Y\mid X}(x\mid y)$ & $f_X(x)$ already. By putting them in the formulae I get:

$\begin{equation*} f_{Y,X}(y, x) = \begin{cases} \frac{1}{2}\cdot [6+8(x-y)] & \quad 0<x<2, 0.25<y<2.75\\ 0 & \quad\text{ otherwise} \end{cases} \end{equation*} $

I am having issue in the other two parts.

In part $c$; I am unable to write mathematically what the question asks. As a result I can't proceed further.

As for part $d$ I don't understand what does the phrase "rewrite $f_{X,Y}(x,y).$ with $y$ as leading variable. " even mean.

Any help and guidance will be appreciated. Thanks

Answer 1

Indeed, $c=1/2$.

$\qquad\begin{align} f_{X}(x)&=\begin{cases} 1/2 &:& 0<x<2\\0&:&\text{otherwise}\end{cases} \\[1ex] f_{Y\mid X}(y\mid x) &= \begin{cases} 6+8(x-y) &:& \frac{1}{4}+x<y<\frac{3}{4}+x \\ 0 &:&\text{otherwise} \end{cases} \end{align}$

You are correct that the joint probability mass function is the product of these two functions. However, the support of the joint pmf is the intersection of their supports. This can be presented in various ways, for example:

$\hspace{10ex}\begin{align}f_{X,Y}(x,y) &=\begin{cases} 3+4(x-y) &:& 0<x<2~,~ x+1/4<y<x+3/4 \\ 0 &:&\text{otherwise} \end{cases}\\[2ex]&=\begin{cases} 3+4(x-y) &:& 1/4<x+1/4<y<x+3/4<2+3/4 \\ 0 &:&\text{otherwise} \end{cases} \end{align}$

Part (d) just requests you to rewrite that support in the given format.