[proofplan] The [random variable](/page/Random%20Variable) $\varepsilon$ takes only the two values $1$ and $-1$, each with probability $1/2$. Therefore each moment is computed by the finite distribution formula for expectation. The parity of $k$ determines whether $(-1)^k$ equals $-1$ or $1$, giving the two moment values. The mean and variance then follow from the cases $k=1$ and $k=2$ together with the definition of variance. [/proofplan] [step:Compute the $k$th moment from the two point distribution] Fix $k\in\mathbb N$. Since the map \begin{align*} x\mapsto x^k \end{align*} is finite on the set $\{-1,1\}$ and $\varepsilon$ takes values in $\{-1,1\}$ almost surely, the random variable $\varepsilon^k$ is bounded and hence integrable. Using the finite distribution formula for expectation, \begin{align*} \mathbb E[\varepsilon^k]=1^k\mathbb P(\varepsilon=1)+(-1)^k\mathbb P(\varepsilon=-1). \end{align*} By the Rademacher distribution of $\varepsilon$, \begin{align*} \mathbb E[\varepsilon^k]=1^k\frac{1}{2}+(-1)^k\frac{1}{2}. \end{align*} Since $1^k=1$, this becomes \begin{align*} \mathbb E[\varepsilon^k]=\frac{1+(-1)^k}{2}. \end{align*} [guided] Fix $k\in\mathbb N$. The point of the computation is that no limiting argument is needed: a [Rademacher random variable](/page/Rademacher%20Random%20Variable) has only two possible values. Since $\varepsilon$ takes values in $\{-1,1\}$ almost surely, the random variable $\varepsilon^k$ also takes values in $\{-1,1\}$ almost surely. In particular, $\varepsilon^k$ is bounded, so the expectation $\mathbb E[\varepsilon^k]$ is well-defined. The finite distribution formula for expectation says that, for a random variable supported on finitely many values, its expectation is the sum of each value multiplied by its probability. Applying this to the random variable $\varepsilon^k$, the value corresponding to $\varepsilon=1$ is $1^k$, and the value corresponding to $\varepsilon=-1$ is $(-1)^k$. Hence \begin{align*} \mathbb E[\varepsilon^k]=1^k\mathbb P(\varepsilon=1)+(-1)^k\mathbb P(\varepsilon=-1). \end{align*} Now we insert the defining probabilities of a Rademacher random variable: \begin{align*} \mathbb P(\varepsilon=1)=\frac{1}{2} \end{align*} and \begin{align*} \mathbb P(\varepsilon=-1)=\frac{1}{2}. \end{align*} Therefore \begin{align*} \mathbb E[\varepsilon^k]=1^k\frac{1}{2}+(-1)^k\frac{1}{2}. \end{align*} Since $1^k=1$ for every $k\in\mathbb N$, this simplifies to \begin{align*} \mathbb E[\varepsilon^k]=\frac{1+(-1)^k}{2}. \end{align*} This formula is the whole moment computation; the remaining distinction is only the parity of $k$. [/guided] [/step] [step:Separate the formula according to the parity of $k$] If $k$ is odd, then $(-1)^k=-1$, and therefore \begin{align*} \mathbb E[\varepsilon^k]=\frac{1+(-1)}{2}=0. \end{align*} If $k$ is even, then $(-1)^k=1$, and therefore \begin{align*} \mathbb E[\varepsilon^k]=\frac{1+1}{2}=1. \end{align*} This proves the stated formula for all $k\in\mathbb N$. [/step] [step:Apply the first two moment formulas to compute the mean and variance] Taking $k=1$, which is odd, gives \begin{align*} \mathbb E[\varepsilon]=0. \end{align*} Taking $k=2$, which is even, gives \begin{align*} \mathbb E[\varepsilon^2]=1. \end{align*} By the definition of variance, \begin{align*} \operatorname{Var}(\varepsilon)=\mathbb E[\varepsilon^2]-(\mathbb E[\varepsilon])^2. \end{align*} Substituting the two moment values, \begin{align*} \operatorname{Var}(\varepsilon)=1-0^2=1. \end{align*} Thus $\mathbb E[\varepsilon]=0$ and $\operatorname{Var}(\varepsilon)=1$. [/step]