Conditional Probability Calculations

Question

The random variable $X$ and $Y$ have the joint PDF

$f_{X,Y}(x,y)= \begin{cases} 2, & \text{if } x > 0\ \text{and} \ y>0\ \text{and} \ x+y\leq 1 \\ 0, & \text{otherwise} \end{cases} \quad $

Let $A$ be the event ${Y ≤ 0.5}$ and let $B$ be the event ${Y >X}$.

I Solved part A but am really stuggling with part B

(a) Calculate $P(B|A)$

I found that $P(B|A)=P(B,A)/P(A)$

$P(A)=P(Y\leq 0.5)=0.75$

$P(B,A)=0.5$

So

$P(B|A)=2/3$

(b) Calculate $f_{X|Y}(x|0.5).$ Calculate also the conditional expectation and the conditional variance of X, given that $Y=0.5$

(c) Calculate $f_{X|B}(x)$

(d) Calculate $E[XY]$

(e) Calculate the PDF of $Y/X$

Could I please get help with part(a) I am struggling with this part, which will allow me to solve the rest of the problem!

Answer 1

In this figure, we have that:

1. the blue area is where the probability is not null
2. the red area corresponds to the event A.
3. the yellow area corresponds to the event B.
4. the green area corresponds to the event "A and B" (hereafter called "A,B").

The red area is equal to: $$\frac{(1+0.5)0.5}{2} = 0.375.$$

The yellow area is equal to: $$\frac{1\cdot 0.5}{2} = 0.25.$$

The green area is equal to: $$\frac{0.5 \cdot 0.5}{2} = 0.125.$$

As a consequence, $P(A) = 0.375 \cdot 2 = 0.75$, $P(B) = 0.25\cdot 2 = 0.5$ and $P(A,B) = 0.125 \cdot 2 = 0.25$ (i.e. we multiply the area by the value of the pdf, 2, which is constant all over the blue set).

Then:

$$P(B|A) = \frac{P(A,B)}{P(A)} = \frac{0.25}{0.75} = \frac{1}{3}.$$