Let $X$ be a set of cardinality $n$ and let $R$ be the set of all relations over $X$. Consider the probability space $(R,P)$ with uniform distribution, that is, $P[\{R_1\}]] = P[\{R_2\}]$ for all $ R_1,R_2 \in R$. Define the relation ”equiprobable” on the set of all events of $(R, P)$ as follows: Events $A$ and $B$ are equiprobable iff $P[A] = P[B]$.

a. Prove that ”equiprobable” is an equivalence relation.

b. How many equivalence classes does this relation define?

Here is my attempt at a, can anyone help me check?

  • Reflexive: $R_1 ,\in R$, then $P[\{R_1\}] = P[\{R_1\}]$, which is equiprobable $\Rightarrow$ $R_1\,R\, R_1$ $\Rightarrow$ reflexive

  • Transitive: $R_1,R_2,R_3 \in R$, suppose $P[\{R_1\}] = P[\{R_2\}\,] \& \,P[\{R_2\}] = P[\{R_3\}] \,\Rightarrow\, P[\{R_1\}] = P[\{R_3\}] \,\Rightarrow\, R_1 \,R \,R_3 \,\Rightarrow $transitive

  • Symmetric: $R_1, R_2 \in R$, $ P[\{R_1\}] = P[\{R_2\}]$ $\Rightarrow$ $P[\{R_2\}] = P[\{R_1\}]$ $\Rightarrow$ $R_2\, R\, R_1$ $\Rightarrow$ symmetric

I don't know how to solve b, and my teacher gave me a hint but I'm still a lot confused about equivalence classes

If $R_1$ and $R_2$ are equiprobable, is there any relation between $|R_1|$ and $|R_2|$?

This is just my guess: $|R_1| = |R_2|$ because they are equivalence relations, so the size of them are equal? Please help me understand this

  • $\begingroup$ How many elements does $R$ have? So how many values can $P$ take? $\endgroup$ Jun 3, 2019 at 5:55
  • $\begingroup$ @user10354138 erm I'm not sure.. the question doesn't state, does it? Can you give me some more hints? $\endgroup$
    – amV
    Jun 3, 2019 at 6:03
  • $\begingroup$ Please use Mathjax when writing math :) $\endgroup$ Jun 3, 2019 at 6:04
  • $\begingroup$ @RiccardoSvenRisuleo I will, thanks $\endgroup$
    – amV
    Jun 3, 2019 at 6:05
  • $\begingroup$ The equivalence class of an event $A\subseteq R$ is the set of all events $B\subseteq R$ which are equiprobable to $A$. Under the uniform distribution, the only condition to be satisfied is that $A$ and $B$ contain the same number of outcomes. You might as well label each equivalence class by the size of its elements! $\endgroup$
    – M. Nestor
    Jun 3, 2019 at 6:17

1 Answer 1


In general if $f:A\to B$ is some function then a relation $S$ over $A$ that is prescribed by: $$xSy\iff f(x)=f(y)$$can "immediately" be recognized as an equivalence relation on base of the observations that $f(x)=f(x)$ for every $x\in A$, $f(x)=f(y)\implies f(y)=f(x)$ for all $x,y\in A$ and $f(x)=f(y)\wedge f(y)=f(z)\implies f(x)=f(z)$ for all $x,y,z\in A$.

You did that in your attempt with function $P$.

Further the equivalence class represent by $a\in A$ is the set $\{x\in A\mid f(x)=f(a)\}$.

The cardinality of the set of equivalence classes equals the cardinality of the image of the function, i.e. the set $\{f(x)\mid x\in A\}$. This because there is a one-to-one relation between the image and the set of equivalence classes: every element $y$ of this image corresponds with equivalence class $\{x\mid f(x)=y\}$.

Application in your situation:

The set of outcomes is $R=\wp\left(X\times X\right)$. From the fact that $X$ has $n$ elements we conclude that $X\times X$ has $n^{2}$ elements, and secondly that $R=\wp\left(X\times X\right)$ has $2^{n^{2}}$ elements.

Further $P$ is an additive function $\wp\left(R\right)\to\mathbb{R}$ determined by $P\left(\left\{ r\right\} \right)=2^{-n^{2}}$ for every $r\in R$.

That implies that for event $A\subseteq\wp\left(R\right)$ we have $P\left(A\right)=\sum_{r\in A}P\left(\left\{ r\right\} \right)=\left|A\right|2^{-n^{2}}$ where $\left|A\right|$ can take the values $\left\{ 0,1,2,\dots,2^{n^{2}}\right\} $.

We conclude that the cardinality of the image of function $P$ is: $$2^{n^{2}}+1$$ and as shown above this is also the number of equivalence classes.

  • $\begingroup$ can you explain a little bit about the set of the image of P? why is it {0, 1/n^2, ... (n^2-1)/n^2, 1}? Sorry I'm not quite good at this $\endgroup$
    – amV
    Jun 3, 2019 at 7:00
  • $\begingroup$ I have repaired my answer. $\endgroup$
    – drhab
    Jun 3, 2019 at 8:25
  • $\begingroup$ Where does the "+ 1" come from for the cardinality of the relation? I'm a little unclear on that part $\endgroup$ Dec 15, 2020 at 17:11
  • $\begingroup$ @quantumferret Do you agree that set $\{0,1,2,\dots m\}$ has $m+1$ elements? Here $m=2^{n^2}$. $\endgroup$
    – drhab
    Dec 15, 2020 at 18:37
  • $\begingroup$ @drhab would it be incorrect for me to think of it in terms of a probability space (Ω, F, P)? Ω would be |X x X| = n^2 and F = 2^(n^2), so F = m? $\endgroup$ Dec 16, 2020 at 13:06

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