# Probability of multiplication of RVs into $\pm1$

## Question

Consider the following scenarion:

1. Pick $A_1,A_2$ uniformly and independently from $\left\{ -1,1\right\}$
2. For $i=1,2$, pick $\left(B_i,C_i\right)$ from $\left\{ \left(A_{i},A_{i}\right),\left(-A_{i},A_{i}\right),\left(A_{i},-A_{i}\right)\right\}$ uniformly, that is each assignment happens with probability $1/3$.

I'm trying to find out $Pr\left(A_{1}\cdot A_{2}=B_{1}\cdot B_{2}\vee A_{1}\cdot A_{2}=C_{1}\cdot C_{2}\right)$

## Attempts so far

I know that $$Pr\left(A_{1}\cdot A_{2}=B_{1}\cdot B_{2}\vee A_{1}\cdot A_{2}=C_{1}\cdot C_{2}\right)=Pr\left(A_{1}A_{2}=B_{1}B_{2}\right)+Pr\left(A_{1}A_{2}=C_{1}C_{2}\right)-Pr\left(A_{1}A_{2}=B_{1}B_{2}=C_{1}C_{2}\right)$$ I can prove that $Pr\left(A_{1}A_{2}=B_{1}B_{2}\right)=\frac{5}{9}$ and that $Pr\left(A_{1}A_{2}=B_{1}B_{2}\right)=Pr\left(A_{1}A_{2}=C_{1}C_{2}\right)$, so I'm left with calculating $Pr\left(A_{1}A_{2}=B_{1}B_{2}=C_{1}C_{2}\right)$.

When I try to delve into the different cases I get awfully confused. Any help is appreciated!

• A suggestion for a more informative title would be appreciated. – 8l2s Jan 26 '17 at 19:49

There are only nine possibilities so we can just write them all out:

1. If $(B_1,C_1) = (A_1,A_1)$ then $(B_1B_2, C_1C_2) = (A_1A_2,A_1A_2) , (-A_1A_2,A_1A_2), (A_1A_2,-A_1A_2)$ when $(B_2,C_2)=(A_2,A_2),(-A_2,A_2),(A_2,-A_2)$

2. If $(B_1,C_1) = (-A_1,A_1)$ then $(B_1B_2, C_1C_2) = (-A_1A_2,A_1A_2) , (A_1A_2,A_1A_2), (-A_1A_2,-A_1A_2)$ when $(B_2,C_2)=(A_2,A_2),(-A_2,A_2),(A_2,-A_2)$

3. If $(B_1,C_1) = (A_1,-A_1)$ then $(B_1B_2, C_1C_2) = (A_1A_2,-A_1A_2) , (-A_1A_2,-A_1A_2), (A_1A_2,A_1A_2)$ when $(B_2,C_2)=(A_2,A_2),(-A_2,A_2),(A_2,-A_2)$

The cases of interest are when $B_1B_2 = A_1A_2$ or $C_1C_2=A_1A_2$ which are all the ordered pairs above where both don't have minus signs. I count seven, so the probability is $7/9.$

Pick $(X_i,Y_i)$ from $\{(1,1),(-1,1),(1,-1)\}$ without bias.

Then $(B_i,C_i)$ is $A_i\cdot(X_i,Y_i)$ and you are looking for the probability $\mathsf P(1{=}X_1X_2~\vee~1{=}Y_1Y_2)$

Noting that $X_1X_2{=}1$ is the event that $X_1,X_2$ have the same sign (also likewise for $Y_1Y_2{=}1$).

\begin{align}\mathsf P(1{=}X_1X_2~\vee~ 1{=}Y_1Y_2)~&=~\mathsf P(X_1X_2{=}1)+\mathsf P(Y_1Y_2{=}1)-\mathsf P(X_1X_2{=}1~\wedge~ Y_1Y_2{=}1)\\[1ex] &=~ {\mathsf P(X_1{=}1)\mathsf P(X_2{=}1) +\mathsf P(X_1{=}{-}1)\mathsf P(X_2{=}{-}1)\\+\mathsf P(Y_1{=}1)\mathsf P(Y_2{=}1)+\mathsf P(Y_1{=}{-}1)\mathsf P(Y_2{=}{-}1)\\-\mathsf P((X_1,Y_1){=}(X_2,Y_2))}\\[1ex] & =~ \end{align}