I would like to know how to calculate the CDF $F_y(y)$ of the following (if possible):
$$Y \sim U[1.4,\ x], \text{ where } X \sim U[1.4,\ 2].$$
I have tried to calculate $f_Y(y)$, taking the following steps:
$f_X(x) = \left \{ \begin{array}{l}
\frac{1}{2-1.4}\quad\text{for $x \in [1.4,\ 2],$}\\
0\quad\text{otherwise}.
\end{array}\right.$
$f_{Y|X}(y|x) = \left \{ \begin{array}{l}
\frac{1}{x-1.4}\quad\text{for $y \in [1.4,\ x]$ and $x \in [1.4,\ 2]$,}\\
0\quad\text{otherwise}.
\end{array}\right.$
Therefore, as $f_{Y|X}(y|x) = \frac{f(x,\ y)}{f_X(x)}$,
$f(x,\ y) = \left \{ \begin{array}{l}
\frac{5}{3}\frac{1}{x-1.4}\quad\text{for $y \in [1.4,\ x]$ and $x \in [1.4,\ 2]$,}\\
0\quad\text{otherwise}.
\end{array}\right.$
However, trying to integrate this to either find $F(x,\ y)$ or $f_Y(y)$ seems impossible to me, because there is a $\ln|x-1.4|$ term, which evaluates to $\ln|0|$.
Any help would be very much appreciated.
Best Answer
I came up with a general solution:
First, consider the case where:
$$X \sim U[0, a] \text{ and } Y \sim U[0, x].$$
So, $$ f_X(x) = \left \{ \begin{array}{l} \frac{1}{a}\quad\text{$x \in [0,\ a]$,}\\ 0\quad\text{otherwise}. \end{array}\right. $$
and
$$ f_{Y|X}(y|x) = \left \{ \begin{array}{l} \frac{1}{x}\quad\text{$x \in [0,\ a]$ and $y \in [0,\ x]$,}\\ 0\quad\text{otherwise}. \end{array}\right. $$
Therefore,
$$ f(x, y) = \left \{ \begin{array}{l} \frac{1}{ax}\quad\text{$x \in [0,\ a]$ and $y \in [0,\ x]$,}\\ 0\quad\text{otherwise}. \end{array}\right. $$
We can then integrate this to find $f_Y$:
$$ f_Y(y) = \frac{1}{a}\int_{-\infty}^{\infty}\frac{1}{x}\,dx = \frac{1}{a}\left[\ln{x}\right]_{y}^{a} = \frac{1}{a}\left[\ln{\frac{a}{y}}\right] \quad\text{(If $y \in [0,\ a])$.}$$
Again, we can integrate to find $F_Y(y)$:
$$\int_{-\infty}^{y}f_Y(y)\,dy = \frac{1}{a}\left[y\ln{\frac{a}{y}} + y\right]_0^y = \frac{y}{a}\left[\ln{\frac{a}{y}} + 1\right].$$
Therefore, $$ F_{Y}(y) = \left \{ \begin{array}{l} 1\quad\text{$y \ge a$},\\ \frac{y}{a}\left[\ln{\frac{a}{y}} + 1\right]\quad\text{$y \in (0,\ a)$,}\\ 0\quad\text{$y \le 0$}. \end{array}\right. $$
We can now consider cases where we transform the distributions.
If $X \sim U[a, b]$ and $Y \sim U[a, x]$,
$$ F_{Y}(y) = \left \{ \begin{array}{l} 1\quad\text{$y \ge b$},\\ \frac{y - a}{b - a}\left[\ln{\frac{b - a}{y - a}} + 1\right]\quad\text{$y \in (a,\ b)$,}\\ 0\quad\text{$y \le a$}. \end{array}\right. $$