how to write a joint distribution when using one random variable as a selector

Question

Suppose there are 3 coins, A, B, and C, toss A first. If it is a head, toss B, otherwise toss C. Can I write this as a joint distribution of 3 random variables? Is this correct? $$ P(A,B,C)=P(A)(P(B))^{I(A=1)}(P(C))^{I(A=0)} $$ , where $I(...)$ is an indicator function.
If it can be written as a joint distribution, how many entries are the joint distribution has? Is it 8 entries (2x2x2) or is it 4 entries (2 entries for A=1 and B, and 2 entries for A=0 and C).
What is the right formula to exactly describe this scenario?

Answer 1

Suppose there are 3 coins, A, B, and C, toss A first. If it is a head, toss B, otherwise toss C. Can I write this as a joint distribution of 3 random variables? Is this correct?

From the comments:

I suppose the value of B is undefined or I don't care the value of B if A is T. In the joint distribution I would imagine, P(A,B=H,C) always equal to P(A,B=T,C) if A is T.

What you have is a variable (call it $D$) who takes on four enumerated values, $H_B, H_C, T_B, T_C$ (or possibly just two values, $H, T$) whose value is the result of whichever coin, $B$ or $C$, is flipped.

Of course, unless these coins are biased these results are equally probable.

$$\text{Let }p:=\Pr(A=H), q:=\Pr(B=H), r:=\Pr(C=H)$$

$$\begin{align}\text{So we have:} \\ \Pr(D=H_B) & = \Pr(A=H)\Pr(B=H) \\ & = pq, \\ \Pr(D=T_B) & = \Pr(A=H)\Pr(B=T) \\ & = p(1-q), \\ \Pr(D=H_C) & = \Pr(A=T)\Pr(C=H) \\ & = (1-p)r, \\ \Pr(D=T_C) & =\Pr(A=T)\Pr(C=T) \\ & = (1-p)(1-r). \\ \text{or else:} \\ \Pr(D=H) & = \Pr(A=H)\Pr(B=H)+\Pr(A=T)\Pr(C=H) \\ & = pq+(1-p)r, \\ \Pr(D=T) & =\Pr(A=H)\Pr(B=T)+\Pr(A=T)\Pr(C=T) \\ & = p(1-q)+(1-p)(1-r).\end{align}$$

Answer 2

I doubt that it can be written as a joint distribution because in a discrete case we have to make sense for the expression like $P\{A=0,B=0,C=0\}$ (0 stands for head).

And even this simple expression cannot have sense because if $A=0$, then $B$ is either $0$ or $1$ with known probabilites, but we know nothing about $C$.

Answer 3

Edit:

May be we can write the joint distribution as below. It is called the conditional probability mass function.

P(B = b, C = c/ A = a) = P(A).P(B/A) + P(A).P(C/A).

For example, we want to evaluate P(B=H,C=T/A=H) it will be $ = \frac{1}{2}\cdot\frac{1}{2}+\frac{1}{2}\cdot0 = 1/4$

Similarly, we want to evaluate the P(B=T,C=H/A=T) = $ = \frac{1}{2}\cdot0+\frac{1}{2}\cdot\frac{1}{2} = 1/4$

It will boil down to four outcomes.

Thanks Satish