Where do these conditional probabilities come from?

Question

For questions b.ii and b.iii of this problem 3 of this PDF, the solution given states that:

I would like to ask where do these conditional probabilities come from all of a sudden? This solution does not feel like a solution at all.

Answer 1

These are not conditional probabilities but PDF's.

Under condition $X=0.5$ random variable $Y$ takes values in interval $[0,0.5]$ and the fact that the PDF of $(X,Y)$ is constant there indicates that the distribution involved is uniform.

A uniform distribution on $[0,0.5]$ corresponds with a PDF taking value $2$ on this interval and value $0$ elsewher.

In mathematical notation: $$f_{Y\mid X}\left(y\right)=\begin{cases} 2 & \text{if }y\in\left[0,0.5]\right]\\ 0 & \text{otherwise} \end{cases}$$

For $f_{X\mid Y}(x\mid 0.5)$ a similar story except that in that case there is not uniformity but something that quite looks like that. There the original PDF takes two distinct values and these cases must be discerned.

Answer 2

Hint

\begin{eqnarray} f(x,y)=\left\{ \begin{array}{cc} \frac{1}{2} & x<1 \hspace{1cm} y<x \\ \frac{3}{2} & x>1 \hspace{1cm} y+x<2 \end{array} \right. \end{eqnarray}

\begin{eqnarray} f_X(x)=\left\{ \begin{array}{cc} \int_0^x \frac{1}{2} dy & x<1 \\ \int_0^{2-x} \frac{3}{2} dy & x>1 \end{array} \right. \end{eqnarray}

\begin{eqnarray} f_{Y|X}(y|x)=\frac{f(x,y)}{f_X(x)}=\left\{ \begin{array}{cc} \frac{\frac{1}{2}}{\frac{x}{2}} =\frac{1}{x} & x<1 \hspace{1cm} y<x\\ ? & x>1 \hspace{1cm} y+x<2 \end{array} \right. \end{eqnarray}

so $f(y|x=.5)=\frac{1}{.5}=2$

Calculating $F_Y(a)=P(Y\leq a)$

$$F_Y(a)=P(Y\leq a)=(s_1 +s_2)\times \frac{1}{2} +(s_3 +s_4)\times \frac{3}{2}$$ $$=(\frac{a^2}{2} +(1-a)*a)\times \frac{1}{2} +((1-a)*a +\frac{a^2}{2})\times \frac{3}{2}$$ $$=(a-\frac{a^2}{2} )\times \frac{1}{2} +(a-\frac{a^2}{2})\times \frac{3}{2}=2a-a^2$$

now $f_Y(a)=\frac{d}{da} F_Y(a)=2-2a$ for $a\in(0,1)$