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?



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.

  • $\begingroup$ 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? $\endgroup$ – Jim May 30 '19 at 15:27
  • 2
    $\begingroup$ $Z_1$ and $Z_2$ are not independent. Right now I have not the opportunity to explain further. I will do that later. $\endgroup$ – drhab May 30 '19 at 16:01
  • $\begingroup$ Alright, thanks $\endgroup$ – Jim May 30 '19 at 16:20
  • $\begingroup$ 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? $\endgroup$ – Jim May 31 '19 at 1:38
  • $\begingroup$ I have added something. $\endgroup$ – drhab May 31 '19 at 7:20

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