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}
$$Z_2 =
\begin{cases}
1 & \text{if $ Y_2 = 0$} \\
0 & \text{Otherwise}
\end{cases}
$
I'm trying to find the joint pmf of $(Z_1, Z_2)$
$Y_1$ and $Y_2$ are Binimoial($6, p$) as the X's are independent
The probabilities of $Y_1 = 0$, $Y_2 = 0$ are $(1-p)^6$ I believe.
The issue is $Y_1$ and $Y_2$ are not independent as they both contain $X_3$, meaning I don't think I can just multiply the pmf of $Z_1$ with the pmf of $Z_2$. How would I find the joint pmf in this case?
Best Answer
Hint:
Observe that: $$Y_1=Y_2=0\iff X_1=X_2=X_3=0$$
addendum:
Let $q:=1-p$.
$\{Z_1=1,Z_2=1\}=\{X_1+X_2+X_3=0\}$ so that:$$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.