Determine the random event X from the equation AX = AB

Determine the random event $X$ from the equation $AX = AB$.

Step 1: What the equation says

In set notation the equation reads

\[ A\cap X = A\cap B. \]

In words: "$A$ and $X$ occur together" is the same event as "$A$ and $B$ occur together". Informally: whenever $A$ occurs, $X$ occurs exactly when $B$ occurs. What happens when $A$ does not occur is not restricted at all.

Note that $X$ is an unknown event (a set of outcomes), not a number, so we cannot "divide by $A$". For example, $X=B$ is obviously a solution, but in general it is not the only one.

Step 2: Split every event into its part inside $A$ and outside $A$

Every outcome is either in $A$ or in $\overline{A}$, and never in both. So for any event $X$

\[ X = XA + X\overline{A}, \qquad (XA)(X\overline{A}) = \varnothing. \]

• $XA$ = the part of $X$ inside $A$. The equation says $XA = AB$, so this part is fixed.
• $X\overline{A}$ = the part of $X$ outside $A$. It does not appear in the equation, so it can be any part of $\overline{A}$.

The diagram shows the three regions of $\Omega$ that matter:

Step 3: The general solution

Putting the two parts together: $X$ must contain all of $AB$, nothing from $A\overline{B}$, and an arbitrary piece of $\overline{A}$. An arbitrary piece of $\overline{A}$ can be written as $\overline{A}D$ where $D$ is any event. Therefore

\[ \boxed{X = AB + \overline{A}D, \qquad D \text{ an arbitrary event.}} \]

Step 4: Check that every such $X$ is a solution

Multiply $X = AB + \overline{A}D$ by $A$ and use the distributive law:

\[ AX = A(AB) + A(\overline{A}D) = (AA)B + (A\overline{A})D = AB + \varnothing\cdot D = AB. \]

Here we used $AA = A$ (an event combined with itself does not change) and $A\overline{A} = \varnothing$ (an event and its opposite cannot occur together). ✓

Step 5: Check that there are no other solutions

Let $X$ be any solution, i.e. $AX = AB$. Using Step 2,

\[ X = XA + X\overline{A} = AB + \overline{A}X. \]

This is exactly the form from Step 3 with $D = X$. So every solution has that form. $\blacksquare$

Step 6: The smallest and the largest solution

• Smallest: take $D=\varnothing$ (add nothing outside $A$): $\;X_{\min} = AB$.
• Largest: take $D=\Omega$ (add all of $\overline{A}$): $\;X_{\max} = AB + \overline{A}$.

The largest solution can be simplified. Since $B = AB + \overline{A}B$ and $\overline{A}B \subseteq \overline{A}$,

\[ B + \overline{A} = AB + \overline{A}B + \overline{A} = AB + \overline{A}, \qquad\text{so}\qquad X_{\max} = B + \overline{A}. \]

An equivalent description of all solutions is therefore

\[ AB \subseteq X \subseteq B + \overline{A}. \]

(If $X$ lies between these two events, then $AB\subseteq AX \subseteq A(B+\overline{A}) = AB$, so $AX = AB$.)

Step 7: Example with a die

Roll a die: $\Omega = \{1,2,3,4,5,6\}$. Let $A = \{1,2,3,4\}$ and $B = \{2,4,6\}$. Then

\[ AB = \{2,4\}, \qquad A\overline{B} = \{1,3\}, \qquad \overline{A} = \{5,6\}. \]

So $X$ must contain $2$ and $4$, must not contain $1$ and $3$, and may contain any of $5$, $6$. There are $2^2 = 4$ solutions:

Step 8: Special cases

• If $A = \Omega$ (the certain event), then $\overline{A} = \varnothing$ and the only solution is $X = B$.
• If $A = \varnothing$ (the impossible event), both sides are $\varnothing$ and every event $X$ is a solution.
• If $A \subseteq B$, then $AB = A$ and the equation becomes $AX = A$, i.e. $A \subseteq X$.