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: Recognize \(X\) and \(Y\) as indicator variables.
By definition, \(X=\mathbf{1}_A\), the indicator of the event \(A\), and \(Y=\mathbf{1}_{A^c}\), the indicator of \(A^c\). Since every \(\omega\) is in exactly one of \(A\) or \(A^c\), we always have
\[ Y=1-X. \]

Step 2: Check statement (A), same distribution function.
\(X\) takes the value \(1\) with probability \(P(A)=1/2\) and the value \(0\) with probability \(P(A^c)=1/2\); so \(X\) is Bernoulli\((1/2)\). Likewise \(Y\) takes the value \(1\) with probability \(P(A^c)=1/2\) and the value \(0\) with probability \(P(A)=1/2\); so \(Y\) is also Bernoulli\((1/2)\). Both have the same probability mass function, so their distribution functions agree at every point. Statement (A) is TRUE.

Step 3: Check statement (B), \(X=Y\) almost everywhere.
For \(\omega\in A\): \(X(\omega)=1\) and \(Y(\omega)=0\), so \(X\neq Y\) there. For \(\omega\in A^c\): \(X(\omega)=0\) and \(Y(\omega)=1\), so \(X\neq Y\) there too. So \(X\neq Y\) on the whole of \(\Omega\), meaning \(P(X=Y)=0\), not \(1\). Statement (B) is FALSE.

Step 4: Check statement (C), independence.
If \(X\) and \(Y\) were independent, we would need \(P(X=1,Y=1)=P(X=1)P(Y=1)\). But \(X=1\) and \(Y=1\) together would require \(\omega\in A\) and \(\omega\in A^c\) at once, impossible since \(A\cap A^c=\phi\). So \(P(X=1,Y=1)=0\), while \(P(X=1)P(Y=1)=1/4\neq0\). So \(X\) and \(Y\) are not independent. Statement (C) is FALSE.

Step 5: Check statement (D), \(E(XY)=1\).
Since \(Y=1-X\), we get \(XY=X(1-X)=X-X^2\). Because \(X\) only takes the values \(0\) or \(1\), \(X^2=X\) always, so \(XY=X-X=0\) for every outcome. Hence \(E(XY)=0\), not \(1\). Statement (D) is FALSE.

Final Answer:
\(X\) and \(Y\) have the same distribution function. \[ \boxed{X\text{ and }Y\text{ have the same distribution function}} \]