Recall the definition of independence in Chapter 2.
Two events \(A\) and \(B\) are independent if
\[ \text{P}(A \mid B)=\text{P}(A) \]
By the definition of the conditional probability,
\[ \text{P}(A \mid B)=\frac{\text{P}(A \cap B)}{\text{P}(B)} \]
We have
\[ \text{P}(A \cap B) = \text{P}(A) \cdot \text{P}(B), \;\; \text{if $A$ and $B$ are independent.} \]
Two RVs \(X\) and \(Y\) are independent if
\[ p_{X \mid Y}(x \mid y)=p_X(x), \;\;\; \text{for all $x$ and $y$ with $p_Y(y)>0$.} \]
Knowing \(Y=y\) does not affect our believe on how likely \(X=x\) occurs.
By the definition of the conditional PMF,
\[ p_{X \mid Y}(x \mid y)=\frac{p_{X, Y}(x, y)}{p_Y(y)} \]
\(X\) and \(Y\) are independent if
\[ p_{X, Y}(x, y) = p_X(x) \cdot p_{Y}(y), \;\; \text{for all $x$ and $y$.} \]
| \(Y=1\) | \(Y=2\) | \(Y=3\) | \(Y=4\) | \(p_X(x)\) | |
|---|---|---|---|---|---|
| \(X=1\) | \(\frac{1}{16}\) | \(\frac{1}{16}\) | \(\frac{1}{16}\) | \(\frac{1}{16}\) | \(\frac{1}{4}\) |
| \(X=2\) | \(\frac{1}{16}\) | \(\frac{1}{16}\) | \(\frac{1}{16}\) | \(\frac{1}{16}\) | \(\frac{1}{4}\) |
| \(X=3\) | \(\frac{1}{16}\) | \(\frac{1}{16}\) | \(\frac{1}{16}\) | \(\frac{1}{16}\) | \(\frac{1}{4}\) |
| \(X=4\) | \(\frac{1}{16}\) | \(\frac{1}{16}\) | \(\frac{1}{16}\) | \(\frac{1}{16}\) | \(\frac{1}{4}\) |
| \(p_Y(y)\) | \(\frac{1}{4}\) | \(\frac{1}{4}\) | \(\frac{1}{4}\) | \(\frac{1}{4}\) | \(\frac{100}{100}\) |
Knowing the 1st roll doesn’t give us any info about the 2nd roll.
| \(X \setminus T\) | \(2\) | \(3\) | \(4\) | \(5\) | \(6\) | \(7\) | \(8\) | \(p_X(x)\) |
|---|---|---|---|---|---|---|---|---|
| \(1\) | \(\frac{1}{16}\) | \(\frac{1}{16}\) | \(\frac{1}{16}\) | \(\frac{1}{16}\) | \(0\) | \(0\) | \(0\) | \(\frac{1}{4}\) |
| \(2\) | \(0\) | \(\frac{1}{16}\) | \(\frac{1}{16}\) | \(\frac{1}{16}\) | \(\frac{1}{16}\) | \(0\) | \(0\) | \(\frac{1}{4}\) |
| \(3\) | \(0\) | \(0\) | \(\frac{1}{16}\) | \(\frac{1}{16}\) | \(\frac{1}{16}\) | \(\frac{1}{16}\) | \(0\) | \(\frac{1}{4}\) |
| \(4\) | \(0\) | \(0\) | \(0\) | \(\frac{1}{16}\) | \(\frac{1}{16}\) | \(\frac{1}{16}\) | \(\frac{1}{16}\) | \(\frac{1}{4}\) |
| \(p_T(t)\) | \(\frac{1}{16}\) | \(\frac{2}{16}\) | \(\frac{3}{16}\) | \(\frac{4}{16}\) | \(\frac{3}{16}\) | \(\frac{2}{16}\) | \(\frac{1}{16}\) | \(1\) |
Knowing the 1st roll gives us some info about the sum.