$E_1, E_2$ 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 $P(XY=X^2Y^2)=P(XY(1-XY)=0)$



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

This is where I'm lost. Please help!

  • 1
    $\begingroup$ $xy = 0$ 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 $1 = \Pr\{X = X^2, Y = Y^2\} \leq \Pr\{XY = X^2Y^2\}$. $\endgroup$ – BGM Sep 28 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.


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