0
$\begingroup$

Totally lost of where to start with this question :

Suppose you roll three distinguishable fair dice and call the resulting numbers a, b, and c. Define events X= “a+b is even", ”Y= “b+c is even", and Z= “a+c is even”. Prove that these three events are pairwise independent but not mutually independent.

Right now, this is how I understand the difference between pairwise independence and mutual independence.

𝐴,𝐵,𝐶 are mutually independent if $𝑃(𝐴∩𝐵∩𝐶)=𝑃(𝐴)𝑃(𝐵)𝑃(𝐶)$

But this is not the case for pairwise independence.

Any help would be greatly appreciated!

$\endgroup$
0
$\begingroup$

Suppose we've been given a collection of events $X_1,X_2,...,X_n.$

The events are mutually independent if $$P\left(\bigcap_{k=1}^n X_k\right)=\prod_{k=1}^n P(X_k),$$ and pairwise independent if for all $j,k$ with $1\le j<k\le n$ we have $$P(X_j\cap X_k)=P(X_j)\cdot P(X_k).$$

So, you'll need to show the following: $$P(A\cap B)=P(A)P(B)\\P(A\cap C)=P(A)P(C)\\P(B\cap C)=P(B)P(C)\\P(A\cap B\cap C)\neq P(A)P(B)P(C)$$

$\endgroup$
0
$\begingroup$

Note that $Z= X\odot Y$ (coincidence operator).

COnsqeuently $P X = P Y = PZ = {1 \over 2}$ and $P[X \cap Y \cap \bar Z| = 0$.

$\endgroup$
0
$\begingroup$

The definition of mutual independence for three events (using $A,B$ to denote $A\cap B$ is: $$ P(A,B)=P(A)P(B), \qquad P(B, C)=P(B)P(C),\qquad P(C,A)=P(C)P(A),\\ P(A,B,C)=P(A)P(B)P(C) $$ So there are four conditions to check, not just the three way one you listed. For pairwise, you only need the first three.

For mutual independence of $n$ events, there are $2^n-n-1$ conditions, one for each subset of the events.


Note that $a+b,a+c,$ and $b+c$ are all even iff and only if $a,b,c$ are all even or $a,b,c$ are all odd. Therefore,

$$ P(X,Y,Z)=P(a,b,c\text{ all even})+P(a,b,c\text{ all odd})=(1/2)^3+(1/2)^3=1/4. $$ However, $$ P(X)=P(a,b\text{ both even})+P(a,b\text{ both odd})=(1/2)^2+(1/2)^2=1/2, $$ so $$ P(X)P(Y)P(Z)=(1/2)^3=1/8. $$

$\endgroup$

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.