Random Variable Distribution Problem Help Needed

Question

Let $Y_1, Y_2, Y_3$ be random variables, each of which take values in {0, 1}

Let: $p(Y_1 = y_1)$=$\frac{y_1+1}{3}$

$p(Y_3 = y_3|Y_1 = y_1)$=$\frac{y_3+y_1+1}{2y_1+3}$

$p(Y_2 = y_2|Y_1 = y_1, Y_3 = y_3)$= $\left\{ \begin{array}{c l} \frac{(y_2+y_1+1)^2}{(y_1+1)^2+(y_1+2)^2} &\mbox{for $y_3=0$} \\ \frac{y_2+1}{3} & \mbox{for $y_3=1$} \end{array}\right.$

Use these distributions to find:

i) $p(Y_3 = 0|Y_1 = 1)$

ii) $p(Y_1 = 0|Y_3 = 1)$

I don't have any problems with i) as you sub $0$ and $1$ into the second distribution and get 2/5 as my answer. But ii) is confusing me so if someone could help me, that will be extremely helpful.

Answer 1

To find $p(Y_1{=}y_1\mid Y_3{=}y_3)$, apply Bayes' Rule (using appropriate events $A,B$)

$$\mathsf P(A\mid B)= \dfrac{p(B\mid A)~p(A)}{p(B)}$$

And using the Law of Total Probability, $p(B)=p(B\mid A)~p(A)+p(B\mid\neg A)~p(\neg A)$ , may also be useful.