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 $P(X \vert Z)$?


1 Answer 1


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.

  • $\begingroup$ @Cesare yes, it is a standard notation in probability theory. google.com/search?q=probability+set+notation $\endgroup$
    – Tim
    Apr 21, 2020 at 20:05
  • $\begingroup$ 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 $P(X \vert Y)$. $\endgroup$
    – Cesare
    Apr 21, 2020 at 20:14
  • $\begingroup$ @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 $y_2$. $\endgroup$
    – Tim
    Apr 21, 2020 at 20:17
  • $\begingroup$ Wait, you changed your answer. Now I am confused. $\endgroup$
    – Cesare
    Apr 21, 2020 at 20:21
  • $\begingroup$ 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? $\endgroup$
    – Cesare
    Apr 21, 2020 at 20:26

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