Independence, part 2

Statement 1: (One die roll vs. the sum of dice)

Note that there is a completely different possible set of sums for each possible roll of a die; for example, if the first die is a 1, then a sum of 12 is impossible, whereas if the first die is a 6, a sum of 2 is impossible. Therefore, the two are not independent.

Statement 2: (One die roll vs. the sum of dice modulo 6)

Each column has the digits 0 through 5, each with a probability of $\frac{1}{6}$ of occuring. Therefore, the first die does not affect the probability of a particular sum of dice modulo 6, and so the probabilities are independent.

Statement 3: (The sum of dice vs. the sum of dice modulo 6)

The sum of dice modulo 6 is completely dependent on the original sum of dice; if rolling a 7, for instance, the only possible sum of dice modulo 6 is 1. This means the probabilities are dependent.

We shall use the following definition of independence:

Two random variables $X$ and $Y$ are independent iff $P(X = x; \; Y=y) = P(X=x)\; P(Y=y)$ for every $x, y$ .

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Claim: $X$ and $Z$ are not independent.

Proof: When $X=1$ , we can never have $Z=12$ , but they are individually possible. Formally,

$P(X=1; Z=12) = 0 \\ P(X=1) > 0 \; P(Z=12) > 0$

The product of two positive numbers can never be zero. $\blacksquare$

Claim: $Z$ and $U$ are not independent.

Proof: When $Z=2$ , we can never have $U=3$ , but they are individually possible. Formally,

$P(Z=2; U=3) = 0 \\ P(Z=2) > 0 \; P(U=3) > 0$

$\blacksquare$

Claim: $X$ and $U$ are independent.

Proof: We shall interpret $x, u$ as remainders modulo 6.

We want to calculate the probability of the event $(X=x; U=u)$ . This is the same as the event $(X=x; Y \equiv u-x)$ .

But we know that $P(X=x; Y \equiv u-x) = P(X=x) \; P(Y \equiv u-x)$ , by independence of $X$ and $Y$ . Hence, it follows that $X$ and $U$ are also independent. $\blacksquare$

Since U = Z (mod 6), then clearly Z and U are dependent.

Z = X + Y, so by simple logic we can say that Z is dependent on Z. To show a bit more rigour E [Z|X] = X + E [Y].

To show U and X are independent, we can demonstrate that U can take any value from 0 to 5 regardless of the value of X. Each of these has the same 1/6 probability so knowledge of X tells us nothing about U -> they are independent.

X is independent because is the first thrown, Y depends on first thrown, Z depends on X and Y. U is independent because is the result from division Z/6