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?


1 Answer 1



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, 2019 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, 2019 at 16:01
  • $\begingroup$ Alright, thanks $\endgroup$
    – Jim
    May 30, 2019 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, 2019 at 1:38
  • $\begingroup$ I have added something. $\endgroup$
    – drhab
    May 31, 2019 at 7:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.