Question

Let \((\Omega,\mathcal F,P)\) be a probability space, where for \(A\subset\Omega\), \(A\neq\phi\), \(A\neq\Omega\),
\[ \mathcal F=\{\Omega,\phi,A,A^c\},\quad P(\Omega)=1,\ P(\phi)=0,\ P(A)=\frac{1}{2}=P(A^c). \]
Let \(X\) and \(Y\) be two random variables defined on \(\Omega\) as follows:
\[ X(\omega)=\begin{cases}1 & \text{if }\omega\in A\\ 0 & \text{if }\omega\in A^c,\end{cases} \quad\text{and}\quad Y(\omega)=\begin{cases}1 & \text{if }\omega\in A^c\\ 0 & \text{if }\omega\in A.\end{cases} \]
Then which of the following statements is correct?

• A. \(X\) and \(Y\) have the same distribution function
• B. \(X=Y\) almost everywhere
• C. \(X\) and \(Y\) are independent
• D. \(E(XY)=1\)

Hint:

Both \(X\) and \(Y\) are Bernoulli\((1/2)\) indicator variables with \(Y=1-X\); they share the same distribution but are never equal pointwise, are not independent, and \(XY=0\) always since \(A\cap A^c=\phi\).

Answer 1

The Correct Option is A

Step 1: Build a small probability table over the two atoms.
The sample space effectively splits into two disjoint pieces with $P(A)=1/2$, $P(A^c)=1/2$. On $A$: $X=1$, $Y=0$. On $A^c$: $X=0$, $Y=1$. So the joint pairs $(X,Y)$ only ever take the values $(1,0)$ with probability $1/2$ and $(0,1)$ with probability $1/2$; the pairs $(1,1)$ and $(0,0)$ never happen.

Step 2: Read off the marginal law of $X$ and of $Y$.
Marginal of $X$: $P(X=1)=1/2$, $P(X=0)=1/2$. Marginal of $Y$: $P(Y=1)=1/2$, $P(Y=0)=1/2$. Both marginals are identical, so as distributions $X$ and $Y$ are identical, matching option (A).

Step 3: Use the table to test the other three options.
Every row in the table has $X\neq Y$. So $P(X=Y)=0$, ruling out option (B).

Step 4: Test independence from the table.
Independence would require $P(X=1,Y=1)=1/4$. But the table shows the pair $(1,1)$ never occurs, so $P(X=1,Y=1)=0\neq1/4$. Ruling out option (C).

Step 5: Compute $E(XY)$ straight from the table.
The product $XY$ equals $0$ on $A$ and $0$ on $A^c$. So $E(XY)=0$, not $1$, ruling out option (D).

Only option (A) survives every check. \[ \boxed{X\text{ and }Y\text{ have the same distribution function}} \]