Given a GARCH(1,1) model

\begin{aligned} X_{t} &= \sigma_{t}Z_{t} \\ \sigma^{2}_{t} &= \omega + \alpha_{1}X^{2}_{t-1} + \beta_{1}\sigma^{2}_{t-1} \end{aligned}

with $Z_{t} \sim i.i.d(0,1)$:

\begin{aligned} X_{t}^2 &= \sigma_{t}^2 Z_{t}^2 \\ &= (\omega + \alpha_{1}X^{2}_{t-1} + \beta_{1}\sigma^{2}_{t-1}) Z_t^2. \end{aligned}

How does one write out the conditional expectation $$E[X_t^2|X_{t-1}^2, X_{t-2}^2, ...]$$

as a summation of alpha, expectation of alpha, beta, etc.?

  • $\begingroup$ Related question "Expectation of $X_t^2$ in ARCH(1) model". $\endgroup$ Commented Apr 8, 2018 at 15:37
  • $\begingroup$ @RichardHardy, what I mean was that you can calculate the variance from, what's known at $t-1$, it's not stochastic at this point. need to clean my language $\endgroup$
    – Aksakal
    Commented Apr 11, 2018 at 13:57

2 Answers 2


GARCH was made to address varying volatility in financial and economic return time series. Note, that the returns on these series usually have very small or no drift at all. Hence, $E[X_t]\approx 0$ and subsequently $\operatorname{Var}\left[X_t\right]\approx X_t^2$ is a good estimate of the observed volatility.

This will explain the form of the second equation in GARCH. Compare it to a typical exponential smoothing: $$s_t=\lambda x_t+(1-\lambda)s_{t-1}$$

So, the idea was to construct a process where the previous volatility $\sigma_{t-1}$ for time $t-1$ gets constantly updated by the new observed volatility estimate $X_{t-1}^2$ that defines the volatility for the next time step $\sigma_{t}$.

Note, that at time $t-1$ all terms on the right hand side of your second equation of GARCH are already observed, which means that $\sigma_t^2$ is deterministic as time $t-1$. The random draw of the return $X_t$ happens in the first equation using this known (at time $t-1$) volatility $\sigma_t$.

When you understand the background of GARCH it should be clear that the answer to your question must be $\sigma_t^2$


As you have already written:

$ E[X^2_t | X^2_{t-1},...] = E[\sigma^2_t Z^2_t| X^2_{t-1},...] = E[(w+\alpha_1 X^2_{t-1}+\beta_1 \sigma^2_{t-1})Z^2_t| X^2_{t-1},...] $

Now, by linearity of the expectation and knowing that $Z_t$ is independent from $X_{t-1},...$ you have:

$ = w E[Z^2_t] + \alpha_1 X^2_{t-1} E[Z^2_t] + \beta_1 \sigma^2_{t-1} E[Z^2_t] = \sigma^2_t $

since $E[Z^2_t] = \operatorname{Var}(Z_t) = 1$.

The reason is that once you condition on the past history of $X^2_{t-1},...$ you are treating both $X^2_{t-1}$ and $\sigma^2_{t-1}$ as constant.

  • 1
    $\begingroup$ How is $Z_t$ independent of $X_t$ if $X_t=\sigma_t Z_t$? $\endgroup$ Commented Apr 11, 2018 at 8:51
  • $\begingroup$ Thank you for the comment @RichardHardy. I edited the answer. $\endgroup$
    – Ale
    Commented Apr 11, 2018 at 12:21

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