# How to notate joint conditional probability

Given $$X = \{ x_1, x_2, \dots, \}$$ and $$Y = \{ y_1, y_2, \dots \}$$ let $$P(X,Y)$$ be their joint probability. Conditioning $$P(X, Y)$$ on $$y \in Y$$ corresponds to looking at the distribution of the elements of $$X$$ when we disregard all elements of $$Y$$ other than $$y$$. This is achieved by normalizing $$P(X,Y)$$ by $$P(Y = y) := \sum_{x \in X} P(X=x, Y=y)$$. The resulting quantity is usually notated with $$P(X \vert Y = y)$$.

Now imagine that we only want to consider two elements $$y_{j_1}$$ and $$y_{j_2}$$ of $$Y$$. The joint distribution between $$X$$ and $$Y$$ that we obtain by disregarding all elements of $$Y$$ other than $$y_{j_1}$$ and $$y_{j_2}$$. I would like to notate this as follows $$P(X=x,Y=y \ \vert \ \{ y_{j_1}, y_{j_2} \}) := \begin{cases} \frac{P(X=x,Y=y)}{P(Y=y_{j_1}) + P(Y=y_{j_2})} & \text{if } y = y_{j_1}, y_{j_2} \\ 0, & \text{otherwise} \end{cases}$$

Question 1: Is there a standard way to notate this quantity?

Question 2: according to wikipedia "the conditional probability $$P(X \mid Y)$$ is a funciton of $$Y$$: e.g., if the function $$g$$ is defined as $$g(y) = P(X \mid Y = y)$$ then $$P(X \mid Y) = g \circ Y$$". If we introduce a variable $$Z$$ defined as $$Z = \{ z_1, z_2, \dots \} := \{ \{ y_{j_1}, y_{j_2} \}: y_{j_1}, y_{j_2} \in Y, y_{j_1} \ne y_{j_2} \}$$ and denote $$g(z) = P(X,Y \mid z)$$ according to the definition of $$P(X,Y \mid \{ y_{j_1}, y_{j_2} \})$$ above, would it make sense to denote $$P(X,Y | Z) = g \circ Z$$ even though $$Z$$ is not properly a random variable?

Question 3: would the notation $$P(X \vert \{y_1, y_2 \}) := \sum_{y = y_1, y_2}$$ be acceptable? And its "extended" version $$P(X \vert Z)$$?

Among the set of all possible values for $$Y$$, that is $$\{y_1,y_2,\dots\}$$ you want to condition on a subset of $$\{y_1,y_2\}$$, so use set notation i.e.

$$P(X,Y|Y\in \{y_1,y_2\})$$

but this is not equal to $$\frac{P(X,Y)}{P(Y=y_1) + P(Y=y_2)}$$, since in $$P(X,Y)$$ you need to consider only such cases where $$Y$$ is either $$y_1$$ or $$y_2$$ if you condition on those values.

This may be a confusing case, since you first need to notice that the notation means in fact

$$P(X,Y|Z) = \frac{P(X,Y,Z)}{P(Z)}$$

where $$Z$$ is an event for $$Y \in \{y_1, y_2\}$$.

To simplify the example, let's ignore $$X$$, and focus on $$P(Y|Z)$$. Moreover, instead of asking about $$P(Y=y_i|Z)$$, let's make it more general and ask about $$P(Y \notin \{y_1, y_2\}|Z) = P(Z^c|Z)$$. By definition $$P(Z^c|Z) = \frac{P(Z^c, Z)}{P(Z)}$$, and we know that $$P(Z^c, Z) = 0$$, since the common subset of something and the opposite of this thing, is empty $$Z^c \cap Z = \varnothing$$ and that $$P(\varnothing)=0$$.

To give example, let's say that we're talking about throwing a dice $$Y \in \{$$ ⚀, ⚁, ⚂, ⚃, ⚄, ⚅ $$\}$$. For any of the outcomes, the probability is the same, say $$P(Y=\;$$$$) = 1/6$$. The probability of throwing one, or two pips is $$P(Y\in\{$$ ⚀, ⚁$$\}) = 2/6$$. If I told you that I've threw ⚀, or ⚁, and you are to guess the result, then you know for sure, that the probability for any other outcome then ⚀, or ⚁, is zero. You are left with only two possible outcomes, that have equal probabilities $$P($$$$) / [P($$$$) + P($$$$)] = P($$$$) / [P($$$$) + P($$$$)] = \tfrac{1}{6} / \tfrac{2}{6} = \tfrac{1}{2}$$. The other outcomes are excluded by the conditioning.

• @Cesare yes, it is a standard notation in probability theory. google.com/search?q=probability+set+notation
– Tim
Apr 21 '20 at 20:05
• Do you think it would also then be possible to define $P(X,Y \vert YY)$ where $YY$ is the set of all possible pairs of elements of $Y$? Just like one generalizes from $P(X \vert Y=y)$ to $P(X \vert Y)$. Apr 21 '20 at 20:14
• @Cesare this would be nonsense. It is like you asked someone "What is your name, and your name?". A value of $Y$ cannot be equal to "pair of values". The above notation means that $Y$ can be either $y_1$ or $y_2$.
– Tim
Apr 21 '20 at 20:17
• Wait, you changed your answer. Now I am confused. Apr 21 '20 at 20:21
• Yes, $P(X,Y \vert Y \in \{ y_1, y_2 \})$ would only be defined for $Y$ equal to $y_1$ or $Y´y_2$. And is that case the normalization should hold. Or am I missing something? Apr 21 '20 at 20:26