Given a random variable X with E[X] = 0 and Var(X) = 1, find the expected value and variance of Z = X2.

We are given $Z = X^2$.
The variance of a random variable $X$ is defined as $Var(X) = E[X^2] - (E[X])^2$.
We are given $E[X] = 0$ and $Var(X) = 1$.
Substitute these values into the variance formula:

$$1 = E[X^2] - (0)^2$$ $$1 = E[X^2] - 0$$ $$E[X^2] = 1$$

Since $Z = X^2$, then $E[Z] = E[X^2]$.
Therefore, $E[Z] = 1$.

The variance of $Z$ is defined as $Var(Z) = E[Z^2] - (E[Z])^2$.
We know $Z = X^2$, so $Z^2 = (X^2)^2 = X^4$.
Thus, $Var(Z) = E[X^4] - (E[Z])^2$.
We already found $E[Z] = 1$, so $Var(Z) = E[X^4] - (1)^2 = E[X^4] - 1$.

To determine $E[X^4]$, we need more information about the specific distribution of $X$.
Consider two possible distributions for $X$ that satisfy $E[X]=0$ and $Var(X)=1$:

Case 1: $X$ takes values $-1$ and 1 with probabilities $P(X=-1) = 0.5$ and $P(X=1) = 0.5$.

• $E[X] = (-1)(0.5) + (1)(0.5) = 0$.
• $E[X^2] = (-1)^2(0.5) + (1)^2(0.5) = 1(0.5) + 1(0.5) = 1$.
• $Var(X) = E[X^2] - (E[X])^2 = 1 - 0^2 = 1$. In this case, $Z = X^2$ will always be $1^2 = 1$ or $(-1)^2 = 1$. So $Z$ is always 1. If $Z$ is always 1, then $E[Z] = 1$ and $Var(Z) = 0$.

Case 2: $X$ takes values $-\sqrt{3}$, 0, $\sqrt{3}$ with probabilities $P(X=-\sqrt{3}) = 1/6$, $P(X=0) = 4/6$, $P(X=\sqrt{3}) = 1/6$.

• $E[X] = (-\sqrt{3})(1/6) + (0)(4/6) + (\sqrt{3})(1/6) = 0$.
• $E[X^2] = (-\sqrt{3})^2(1/6) + (0)^2(4/6) + (\sqrt{3})^2(1/6) = 3(1/6) + 0 + 3(1/6) = 1$.
• $Var(X) = E[X^2] - (E[X])^2 = 1 - 0^2 = 1$. In this case, $Z = X^2$ can take values: If $X = -\sqrt{3}$, $Z = (-\sqrt{3})^2 = 3$. If $X = 0$, $Z = 0^2 = 0$. If $X = \sqrt{3}$, $Z = (\sqrt{3})^2 = 3$. So $Z$ can be 0 or 3. $P(Z=0) = P(X=0) = 4/6 = 2/3$. $P(Z=3) = P(X=-\sqrt{3}) + P(X=\sqrt{3}) = 1/6 + 1/6 = 2/6 = 1/3$. Now calculate $E[Z]$ and $Var(Z)$ for this distribution: $E[Z] = (0)(2/3) + (3)(1/3) = 0 + 1 = 1$. (Matches Step 1) $E[Z^2] = (0)^2(2/3) + (3)^2(1/3) = 0 + 9(1/3) = 3$. $Var(Z) = E[Z^2] - (E[Z])^2 = 3 - (1)^2 = 3 - 1 = 2$. In this case, $Var(Z) = 2$, which is positive.

Since $Var(Z)$ can be 0 or 2 (or other positive values depending on the distribution of $X$), $Var(Z)$ can be positive, and its exact value depends on the distribution of $X$.

A. $E[Z] = 1$ but $Var(Z)$ can be positive, depending on the distribution of $X$. This matches our findings.
B. $E[Z] = 1$ and $Var(Z) = 0$ because squaring removes variability when variance is 1. This is incorrect, as $Var(Z)$ can be positive.
C. Neither $E[Z]$ nor $Var(Z)$ can be determined from the given information. This is incorrect, as $E[Z]$ can be determined.
D. $E[Z] = 0$ because $E[X] = 0$ and $Var(Z) = 1$ by definition. This is incorrect, as $E[Z] = 1$.

The correct statement is A.

$$\boxed{\text{A. } E[Z] = 1 \text{ but } Var(Z) \text{ can be positive, depending on the distribution of } X.}$$