# Join pmf of 2 dependent random variables

Let $$X_1$$, $$X_2$$, $$X_3$$ be independent Binomial($$3, p$$) random variables.

Let $$Y_1 = X_1 + X_3$$, $$Y_2 = X_2 + X_3$$

Let $$Z_1 = \begin{cases} 1 & \text{if Y_1 = 0} \\ 0 & \text{Otherwise} \end{cases}$$

$$Z_2 = \begin{cases} 1 & \text{if Y_2 = 0} \\ 0 & \text{Otherwise} \end{cases}$$

I'm trying to find the joint pmf of $$(Z_1, Z_2)$$

$$Y_1$$ and $$Y_2$$ are Binimoial($$6, p$$) as the X's are independent

The probabilities of $$Y_1 = 0$$, $$Y_2 = 0$$ are $$(1-p)^6$$ I believe.

The issue is $$Y_1$$ and $$Y_2$$ are not independent as they both contain $$X_3$$, meaning I don't think I can just multiply the pmf of $$Z_1$$ with the pmf of $$Z_2$$. How would I find the joint pmf in this case?

Hint:

Observe that: $$Y_1=Y_2=0\iff X_1=X_2=X_3=0$$

Let $$q:=1-p$$.

$$\{Z_1=1,Z_2=1\}=\{X_1+X_2+X_3=0\}$$ so that:$$P(Z_1=1,Z_2=1)=P(X_1+X_2+X_3=0)=q^9$$

$$\{Z_1=1,Z_2=0\}=\{Y_1>0,Y_2=0\}=\{X_1>0,X_2+X_3=0\}$$ so that:$$P(Z_1=1,Z_2=0)=P(X_1>0,X_2+X_3=0)=P(X_1>0)P(X_2+X_3=0)=$$$$\left(1-P(X_1=0)\right)P(X_2+X_3=0)=(1-q^3)q^6$$

Similarly we find $$P(Z_1=1,Z_2=0)=(1-q^3)q^6$$.

We find $$P(Z_1=0,Z_2=0)$$ on base of:$$P(Z_1=0,Z_2=0)=1-P(Z_1=1,Z_2=0)-P(Z_1=0,Z_2=1)-P(Z_1=1,Z_2=1)$$

Observe that we do not have $$P(Z_1=1,Z_2=1)=P(Z_1=1)P(Z_2=1)$$ so $$Z_1,Z_2$$ are not independent.

• Just to make sure I have understood this correct, I can assume that $Z_1$ and $Z_2$ are independent and so the joint pmf is just $P(Z_1, Z_2) = P(Z_1)*P(Z_2) = ((1-p)^6)^{Z_1 +Z_2} (1-(1-p)^6)^{2-Z_1-Z_2}$, correct? – Jim May 30 at 15:27
• $Z_1$ and $Z_2$ are not independent. Right now I have not the opportunity to explain further. I will do that later. – drhab May 30 at 16:01
• Alright, thanks – Jim May 30 at 16:20
• I thought they were because $X_1 = X_2 = X_3$ are $0$ when $Y_1 = Y_2 = 0$ and when they are $0$, they are independent? – Jim May 31 at 1:38
• I have added something. – drhab May 31 at 7:20