6
$\begingroup$

I am asked to:

Find a joint probability distribution $P(X_1,\dots, X_n)$ such that $X_i , \, X_j$ are independent for all $i \neq j$, but $(X_1, \dots , X_n)$ are not jointly independent.

I have no idea where to start, please help.

$\endgroup$
8
$\begingroup$

I always remember this clear example from Probability Essentials, chapter 3, by J.Jacod & P.Protter.

Let $\Omega = \{1,2,3,4\}$, and $\mathscr{A} = 2^\Omega$. Let $P(i) = \frac{1}{4}$, where $i = 1,2,3,4$.

Let $A = \{1,2\}$, $B = \{1,3\}$, $C = \{2,3\}$. Then A,B,C are pairwise independent but are not independent.

$\endgroup$
6
$\begingroup$

See this link: http://faculty.washington.edu/fm1/394/Materials/2-3indep.pdf

Throw two fair dice.

Consider the events:

  • $A:=\{\text{the sum of the points is 7}\}$,
  • $B:=\{\text{the first die rolled a 3}\}$,
  • $C:=\{\text{the second die rolled a 4}\}$.

All three events have probability $\displaystyle\frac{1}{6}$.

Moreover, you can check that they are pairwise independent.

However, they are not jointly independent.

For your random variables, just take $1_A, 1_B,$ and $1_C$, the indicator functions of these events.

$\endgroup$
1
$\begingroup$

Hint: Let $n\ge 3$. Toss a fair coin $n-1$ times. For $i=1$ to $n-1$, let $X_i=1$ if the $i$-th toss gave a head, and $X_i=0$ otherwise. Let $X_n=1$ if the sum $\sum_1^{n-1} X_i$ is odd, and $X_n=0$ otherwise.

It is obvious that if $i\ne j$, and they are both less than $n$, then $X_i$ and $X_j$ are independent. It takes a bit more work to show that if $i\le n-1$, then $X_i$ and $X_n$ are independent. And it is clear that $X_1,\dots,X_n$ are not mutually independent, since if we know the values of the $X_1$ from $1$ to $n-1$, then we know the value of $X_n$.

$\endgroup$

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.