[proofplan] We use the ARCH representation $X_t=\sigma_t Z_t$ and the fact that the volatility $\sigma_t$ is known at time $t-1$, while the innovation $Z_t$ is independent of the past. First we verify the integrability needed for the conditional expectations by using independence to factor products of non-negative random variables. Then we identify the conditional expectations by testing against bounded $\mathcal{F}_{t-1}$-measurable random variables and applying independence. [/proofplan] [step:Verify the integrability of $X_t$ and $X_t^2$] Fix $t \in \mathbb{Z}$. Since $\sigma_t$ is $\mathcal{F}_{t-1}$-measurable and $Z_t$ is independent of $\mathcal{F}_{t-1}$, the random variables $\sigma_t$ and $Z_t$ are independent. Hence $\sigma_t$ and $|Z_t|$ are independent, and Tonelli's theorem for independent non-negative random variables gives \begin{align*} \mathbb{E}[|X_t|] = \mathbb{E}[\sigma_t |Z_t|] = \mathbb{E}[\sigma_t]\mathbb{E}[|Z_t|] < \infty. \end{align*} Thus $X_t \in L^1(\Omega,\mathcal{F},\mathbb{P})$. Similarly, $\sigma_t^2$ and $Z_t^2$ are independent non-negative random variables. Since $\mathbb{E}[Z_t^2]=1$ and $\mathbb{E}[\sigma_t^2]<\infty$, \begin{align*} \mathbb{E}[X_t^2] = \mathbb{E}[\sigma_t^2 Z_t^2] = \mathbb{E}[\sigma_t^2]\mathbb{E}[Z_t^2] = \mathbb{E}[\sigma_t^2] < \infty. \end{align*} Therefore $X_t^2 \in L^1(\Omega,\mathcal{F},\mathbb{P})$. [/step] [step:Compute the conditional mean by testing against past-measurable variables] Let $H: \Omega \to \mathbb{R}$ be a bounded $\mathcal{F}_{t-1}$-measurable [random variable](/page/Random%20Variable). Since $H\sigma_t$ is $\mathcal{F}_{t-1}$-measurable and integrable, and since $Z_t$ is independent of $\mathcal{F}_{t-1}$, the random variables $H\sigma_t$ and $Z_t$ are independent. Therefore \begin{align*} \mathbb{E}[H X_t] = \mathbb{E}[H\sigma_t Z_t] = \mathbb{E}[H\sigma_t]\mathbb{E}[Z_t] = 0. \end{align*} The zero random variable is $\mathcal{F}_{t-1}$-measurable and integrable, and it satisfies \begin{align*} \mathbb{E}[H X_t] = \mathbb{E}[H \cdot 0]. \end{align*} for every bounded $\mathcal{F}_{t-1}$-measurable random variable $H$. By the defining characterization of [conditional expectation](/page/Conditional%20Expectation), \begin{align*} \mathbb{E}[X_t \mid \mathcal{F}_{t-1}] = 0. \end{align*} [guided] The point of conditional expectation is that it is characterized by how it pairs with every bounded random variable measurable with respect to the conditioning $\sigma$-algebra. So we take an arbitrary bounded $\mathcal{F}_{t-1}$-measurable random variable \begin{align*} H: \Omega \to \mathbb{R}. \end{align*} We must show that the candidate conditional expectation, namely the zero random variable, has the same pairings with all such $H$ as $X_t$ does. Using the ARCH representation $X_t=\sigma_t Z_t$, we have \begin{align*} H X_t = H\sigma_t Z_t. \end{align*} Because $H$ and $\sigma_t$ are both $\mathcal{F}_{t-1}$-measurable, their product $H\sigma_t$ is also $\mathcal{F}_{t-1}$-measurable. Since $H$ is bounded and $\mathbb{E}[\sigma_t]<\infty$, the product $H\sigma_t$ is integrable. The innovation $Z_t$ is independent of $\mathcal{F}_{t-1}$, so $Z_t$ is independent of every integrable $\mathcal{F}_{t-1}$-measurable random variable, in particular of $H\sigma_t$. Hence \begin{align*} \mathbb{E}[H X_t] = \mathbb{E}[H\sigma_t Z_t] = \mathbb{E}[H\sigma_t]\mathbb{E}[Z_t]. \end{align*} The standardized innovation assumption gives $\mathbb{E}[Z_t]=0$, so \begin{align*} \mathbb{E}[H X_t] = 0. \end{align*} Since also \begin{align*} \mathbb{E}[H \cdot 0] = 0, \end{align*} we have \begin{align*} \mathbb{E}[H X_t] = \mathbb{E}[H \cdot 0]. \end{align*} for every bounded $\mathcal{F}_{t-1}$-measurable random variable $H$. This is exactly the defining characterization of $\mathbb{E}[X_t \mid \mathcal{F}_{t-1}]$, and therefore \begin{align*} \mathbb{E}[X_t \mid \mathcal{F}_{t-1}] = 0. \end{align*} [/guided] [/step] [step:Compute the conditional second moment by the same independence test] Let $H: \Omega \to \mathbb{R}$ be a bounded $\mathcal{F}_{t-1}$-measurable random variable. Since $H\sigma_t^2$ is $\mathcal{F}_{t-1}$-measurable and integrable, and since $Z_t^2$ is independent of $\mathcal{F}_{t-1}$, \begin{align*} \mathbb{E}[H X_t^2] = \mathbb{E}[H\sigma_t^2 Z_t^2] = \mathbb{E}[H\sigma_t^2]\mathbb{E}[Z_t^2] = \mathbb{E}[H\sigma_t^2]. \end{align*} The random variable $\sigma_t^2$ is $\mathcal{F}_{t-1}$-measurable and integrable. Thus, for every bounded $\mathcal{F}_{t-1}$-measurable random variable $H$, \begin{align*} \mathbb{E}[H X_t^2] = \mathbb{E}[H\sigma_t^2]. \end{align*} By the defining characterization of conditional expectation, \begin{align*} \mathbb{E}[X_t^2 \mid \mathcal{F}_{t-1}] = \sigma_t^2. \end{align*} Together with the first identity, this proves the stated conditional martingale difference and conditional variance properties. [/step]