# $E_1, E_2$ are independent events and $X,Y$ are random variables associated with them, then which of the following is true.

$$E_1, E_2$$ are independent events such that $$P(E_1)=\frac14, P(E_2| E_1)=\frac12, P(E_1|E_2)=\frac14$$. Define random variables $$X$$ and $$Y$$ by

$$X=\begin{cases} & 1 & \text{if } E_1 \text{ occurs} \\ & 0 & \text{if } E_1^C \text{ occurs} \\ \end{cases}$$

$$Y=\begin{cases} & 1 & \text{if } E_2 \text{ occurs} \\ & 0 & \text{if } E_2^C \text{ occurs} \\ \end{cases}$$

Consider the following statements:

$$\alpha: X$$ is uniformly distributed on the set [0,1]

$$\beta: X,Y$$ are identically distributed.

$$\gamma: P(X^2+Y^2=1)=\frac12$$

$$\delta: P(XY=X^2Y^2)=1$$

Choose the correct combination:

1. $$(\alpha,\beta)$$ 2. $$(\alpha,\gamma)$$ 3. $$(\beta,\gamma)$$ 4. $$(\gamma, \delta)$$

Given answer is 4.

My attempt:

$$P(E_2| E_1)=\frac12 \implies P(E_1 \cap E_2) = \frac18$$

$$P(E_1|E_2)=\frac14 \implies P(E_2) = \frac12$$

So that $$P(X=1)=\frac14, P(X=0)=\frac34, P(Y=1)=\frac12, P(Y=0)=\frac12$$

Now $$P(XY=X^2Y^2)=P(XY(1-XY)=0)$$

$$=P(XY=0)+P(XY=1)$$

$$=P(X=0)P(Y=0)+P(X=1)P(Y=1)$$

$$=\frac34 \frac12+ \frac14 \frac12=\frac12$$

• $xy = 0$ if and only if $x = 0$ or $y = 0$, not and. You have already apply this on the second step, but forget in the third step. Another way to view this is that $1 = \Pr\{X = X^2, Y = Y^2\} \leq \Pr\{XY = X^2Y^2\}$.
– BGM
Sep 28, 2019 at 9:12

• $$X$$ takes only two values from $$\{0,1\}$$, it is not a continuous random variable. $$\alpha$$ can't be correct.
• We are told that $$E_1$$ and $$E_2$$ are independent, $$P(E_2|E_1)=P(E_2)=\frac12$$ and $$P(E_1)=\frac14$$, hence $$X$$ and $$Y$$ can't be identically distributed. Hence $$\beta$$ can't be correct.
• Notice that $$X$$ and $$Y$$ are binary, \begin{align}P(X^2+Y^2=1)&=P(X+Y=1)\\&=P(X=1)P(Y=0)+P(X=0)P(Y=1)\\ &=\frac14 \cdot \frac12+ \frac34 \cdot \frac12\\ &=\frac12\end{align}
Hence $$\gamma$$ is true.
Notice that $$XY$$ is binary, hence $$XY=X^2Y^2$$. $$\delta$$ is true.