# Pairwise vs. total independence of discrete uniform random deviates

Let $X$ be a discrete uniform random variable on the set $\{000, 011, 101, 110\}$ of four binary integers, and let $X_{i}$ denote the ith digit of $X$, for $i = 1, 2, 3$. Show that $X_{1}, X_{2}, X_{3}$ are independent pairwise, but not totally independent. Can you generalize this example to more than three random variables?

Can anyone help me with this exercise?

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Also posted on math.SE by a different user. –  Dilip Sarwate Apr 30 '12 at 12:09
See here for a full solution. –  Did May 1 '12 at 21:49

You can prove the independence property with brute force calculations. Suppose that $X_{j} = k$ for $k\in\{0,1\}$. Then what is $P(X_{i} = m | X_{j} = k)$? Simply by enumerating the possibilities, you can see that it is the same as the unconditional probability.

However, suppose you're given two of the three bits. Each item in the set of triples is unique, so conditioned on any two bits, you know what the third must be, so it's not totally independent.

For the generalization part, I will just give a hint: Hamming distance. I hope this is for a class on algebraic codes. It's a nice exercise.

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This is way too complicated and brute force calculations are totally unnecessary. All that is needed is to find $P\{X_i = 1, X_j = 1\}$ which can be done by inspection, and compare to the product $P\{X_i = 1\}P\{X_j = 1\}$ which can also be done almost by inspection. Three choices for the pair $(i,j)$. Then, insist that the given information supports the inference that $$P\{X_1=1,X_2=1,X_3=1\}=0\neq P\{X_1=1\}P\{X_2=1\}P\{X_3=1\}$$ since all the terms on the right have value $\frac{1}{2}$. –  Dilip Sarwate Apr 30 '12 at 13:12
My first paragraph, where I say brute force calculation I think that it's clear this means inspection when you're talking about 4 3-bit numbers. I think your criticism is unproductive and you're free to write up your approach in another answer if you believe it is better. I appreciate the constructive criticism, but I do not think it is productive in this instance. –  EMS Apr 30 '12 at 13:55
I didn't down-vote your answer, but it is your answer that is overly complicated and not very useful as an efficient way of solving the problem. Computing $P(X_{i} = m | X_{j} = k)$ for all choices of $m \in \{0,1\}$, as is implied in "Simply by enumerating the possibilities" is unnecessary extra work that is really not needed to verify pairwise independence; checking one possibility is enough. Also, computing conditional probabilities is unnecessary extra work. It is too bad that you feel my comments were unproductive but I don't intend to withdraw them or apologize for making them. –  Dilip Sarwate Apr 30 '12 at 14:27
I know an answer to the generalization question but I have no idea how your indication may lead to it. –  Did May 1 '12 at 21:13
Then why refer to Hamming distance? –  Did May 7 '12 at 12:10