> [!quote] > > A pair of [[Random Variable|random variables]] $(X, Y)$ is **non-degenerate** for all $1 \le j \le n$ (assuming $n \le m$), there exists $1 \le k \le m$, such that the [[Joint Distribution|joint probability]] $p_{jk} > 0$. If $n \le m$, then swap the roles of $j$ and $k$. > [!quote] My Reaction > > What? What the what? What? This one small paragraph in the middle of nowhere took a few hours to digest, so what does *non-degenerate* mean? Let's start with the table of two coin flips, $X$ and $Y$, with their joint probabilities calculated in the middle. | Joint Probability | Tail | Head | | ----------------- | ---- | ---- | | Tail | 0.25 | 0.25 | | Head | 0.25 | 0.25 | Here we can see that for all values of $n$ (in every column), there exists at least one entry where the joint probability is greater than 0. This means that two independent coin flips are *non-degenerate*. To make some progress, I think we should look at the other side of the coin, what does *degenerate* mean? Flipping the definition: *A pair of random variables $(X, Y)$ is **degenerate** if for **any**[^1] $1 \le j \le n$ (assuming $n \le m$) and for **all**[^1] $1 \le k \le m$ the joint probability $p_{jk} = 0$.* Let's try to alter that coin flip table such that it is *degenerate*. Note that since we are modifying the probabilities, they may no longer make sense as coin flips, but that's fine, we're just playing. | Joint Probability | Tail | Head | | ----------------- | ---- | ---- | | Tail | 0.5 | 0 | | Head | 0.5 | 0 | Here we have changed one of the columns to $0$, meaning that the first coin can never flip heads. This seems to give us an answer, a pair of random variables are *degenerate* if looking at their joint probabilities tells us that one of the variables' [[Sample Space|sample space]] contains an impossible outcome. In other words, that outcome has been *degenerated*. What if we measure it in terms of rows? None of the rows are empty, therefore from the textbook definition, the pair is non-degenerate, right? There is an answer to this question, but let's wait a bit before answering it. However, consider this table, where we make it such that the second coin will always flip a head if the first coin also flips a head. While a *possibility* has been eliminated (Head + Tail), this is *not* a case of a pair of *degenerate* random variables, because the column is not entirely empty. | Joint Probability | Tail | Head | | ----------------- | ---- | ---- | | Tail | 0.25 | 0 | | Head | 0.25 | 0.5 | If the pair is only degenerate if the entire column is empty, then the outcome of $X$ that the column represents is impossible *independent* of the value of the other variable, making it entirely redundant, *degenerating* it from the sample space. Remember the earlier question? Despite the fact that there are no empty rows, the empty column has been eliminated. Since we consider the act of eliminating a possible outcome of one of the random variables as to degenerate, it does not matter if it eliminates a row or a column. Both an empty row and an empty column make the pair degenerate. [^1]: Since only one exception is needed to break the rule, we swap all for any and vice versa.