Why EGARCH Uses the Mean Absolute Standard Normal Innovation

Summary

The document explains the constant that appears in the EGARCH(1,1) specification when modeling the magnitude of standardized shocks. If the innovation z is standard normal, its absolute value follows a standard half-normal distribution. The expected absolute value is therefore the half-normal mean, which accounts for the square-root-of-two-over-pi term in the model’s centering expression.

The answer also states how the expectation changes when the normal innovation has standard deviation other than one: it scales linearly with that standard deviation. This explanation depends on normally distributed innovations; the constant should not be carried over unchanged when the assumed shock distribution differs. The exchange gives the distributional identity and expectation but does not discuss EGARCH estimation, alternative innovation distributions, or empirical implications for volatility forecasts.

Key ideas

• The EGARCH centering constant comes from the expectation of the absolute innovation.
• The absolute value of a standard normal variable has a standard half-normal distribution.
• Its expected value is the square root of two over pi.
• For a normal innovation with a different standard deviation, the expected absolute value scales with that deviation.
• The stated constant relies on normally distributed innovations.

Tags

# EGARCH(1,1) mean


# EGARCH(1,1) mean












I'm trying to model an EGARCH(1,1). However, I dont understand why the mean from the general to (1,1) becomes $\sqrt{(\frac{2}{\pi})}$. The following I am refering to is:

## Answer by Pleb (score 5, accepted)

https://quant.stackexchange.com/a/63942

#### This is because $|z_t|$ is a standard half-normal random variable and have expectation $\sqrt{\frac{2}{\pi}}$ .

The expectation, $\mathbb{E}\left[|z_t|\right] = \sqrt{\frac{2}{\pi}}$ is true, when $z_t \overset{iid}{\sim}N(0,1)$. In this case, the absolute value of $z_t$ is called a (standard) half-normal variable that has known expectation. You can verify this from the Wikipedia page. If $z_t \overset{iid}{\sim}N(0,\sigma^2)$ then $\mathbb{E}\left[|z_t|\right]=\sigma\sqrt{\frac{2}{\pi}}$.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.