Consider the following GARCH(1,1) model for the volatility of asset returns
𝑟
𝑡
:
𝑟
𝑡
=
𝛼
+
𝛽
𝑟
𝑡
−
1
+
𝜀
𝑡
𝜀
𝑡
=
ℎ
𝑡
𝑢
𝑡
ℎ
𝑡
=
𝜇
+
𝛿
ℎ
𝑡
−
1
+
𝜙
𝜀
𝑡
−
1
2
𝔼
𝑡
−
1
(
𝑢
𝑡
)
=
0
𝔼
𝑡
−
1
(
𝑢
𝑡
2
)
=
1
You estimated the following values for the parameters

Estimates
Parameters
𝛼

𝛽

𝜇

𝛿

𝜙

Estimates
0.111

0.8122

0.0011

0.9321

0.0511

Assume that the last 2 observations of the return process are
𝑟
𝑇
=
0.27
and
𝑟
𝑇
−
1
=
0.02
, and the value of the conditional variance in the last period of your sample is
ℎ
𝑇
=
0.75
. Then what is the predicted value of the conditional variance
ℎ
𝑇
+
1
in period
𝑇
+
1
?

Question

Consider the following GARCH(1,1) model for the volatility of asset returns 
𝑟
𝑡
:
𝑟
𝑡
=
𝛼
+
𝛽
𝑟
𝑡
−
1
+
𝜀
𝑡
𝜀
𝑡
=
ℎ
𝑡
𝑢
𝑡
ℎ
𝑡
=
𝜇
+
𝛿
ℎ
𝑡
−
1
+
𝜙
𝜀
𝑡
−
1
2
𝔼
𝑡
−
1
(
𝑢
𝑡
)
=
0
𝔼
𝑡
−
1
(
𝑢
𝑡
2
)
=
1
You estimated the following values for the parameters

Estimates					
Parameters	
𝛼
	
𝛽
	
𝜇
	
𝛿
	
𝜙

Estimates	
0.111
	
0.8122
	
0.0011
	
0.9321
	
0.0511

Assume that the last 2 observations of the return process are 
𝑟
𝑇
=
0.27
 and 
𝑟
𝑇
−
1
=
0.02
, and the value of the conditional variance in the last period of your sample is 
ℎ
𝑇
=
0.75
. Then what is the predicted value of the conditional variance 
ℎ
𝑇
+
1
 in period 
𝑇
+
1
?

BlackTom AI · Accepted Answer

To tackle this prediction problem, I’ll walk through how the next period’s conditional variance h_{T+1} is formed in a GARCH(1,1) setting as described by the given parameter estimates and the observed data.

First, identify the components you need for the forecast. In a typical GARCH(1,1) framework adapted to the notation in the prompt, the next-period conditional variance is computed as:
h_{T+1} = μ + δ h_T + φ ε_T^2,
where:
- μ is the intercept term,
- δ is the coefficient on the lagged variance h_T,
- φ is the coefficient on the squared shock ε_T^2, and
- ε_T is the innovation (shock) at time T, which can be obtained from the return equation once you know the parameters and the last two returns.

Step by step, evaluate each part using the provided values:
- Given parameters: μ = 0.0011, δ = 0.9321, φ = 0.0511, and h_T = 0.75.
- The return equation provided is r_t = α + β r_{t-1} + ε_t. With the last two observed returns r_T = 0.27 and r_{T-1} = 0.02, compute ε_T:
  ε_T = r_T − α − β r_{T-1} = 0.27 − 0.111 − 0.8122 × 0.02.
  Calculate β r_{T-1}: 0.8122 × 0.02 = 0.016244. So ε_T = 0.27 − 0.111 − 0.016244 ≈ 0.142756.
- Next, ε_T^2 ≈ (0.142756)^2 ≈ 0.0204 (approximately).

Now plug these into the forecast formula:
h_{T+1} = μ + δ h_T + φ ε_T^2 ≈ 0.0011 + 0.9321 × 0.75 + 0.0511 × 0.0204.
Compute each term:
- δ h_T = 0.9321 × 0.75 ≈ 0.699075,
- μ = 0.0011,
- φ ε_T^2 ≈ 0.0511 × 0.0204 ≈ 0.001043.
Sum them: h_{T+1} ≈ 0.0011 + 0.699075 + 0.001043 ≈ 0.701218.

Now evaluate the answer choices in light of this calculation:
- Option 1: 0.0729. This is far too small given h_T ≈ 0.75 and the positive contributions from δ h_T and μ. It would correspond to mistakenly omitting the h_T term or miscomputing the components.
- Option 2: 0.701216. This is extremely close to the computed value 0.701218, differing only by rounding. It matches the structure h_{T+1} = μ + δ h_T + φ ε_T^2 with the provided numbers.
- Option 3: There is not enough data to compute h_{T+1}. In fact, all required pieces are available: μ, δ, h_T, φ, and ε_T^2 derived from r_T and r_{T-1} along with α and β.
- Option 4: 0.519615. This would imply a smaller impact from the lagged variance than δ h_T provides, or an incorrect ε_T^2 term; it does not align with the explicit computation above.
- Option 5: 0.866025. This would arise from an overestimated contribution, perhaps treating ε_T^2 as 1 or misapplying coefficients; it deviates significantly from the calculated forecast.
- Option 6: 0.75. This simply equals h_T and ignores the μ and φ ε_T^2 contributions, which is inconsistent with the dynamic updating formula.

In summary, the only option that coherently implements the given GARCH(1,1) update with the provided parameters and observed data is the one yielding approximately 0.7012, i.e., the value near 0.701216, which corresponds to the detailed calculation above.

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