Drink H · 2020年03月15日
问题如下：
Suppose that the current volatility estimate is 3% per day and the long-run average volatility estimate is 2% per day. What are the volatility estimates in ten days and 100 days in a GARCH (1,1) model where ω= 0.000002, α= 0.04, and β= 0.94?
解释：
The expected variance rate in ten days is 0.02^2 + (0.04 + 0.94)^10 (0.03^2– 0.02^2) = 0.000809,
NO.PZ2020011303000086问题如下Suppose ththe current volatility estimate is 3% per y anthe long-run average volatility estimate is 2% per y. Whare the volatility estimates in ten ys an100 ys in a GAR(1,1) mol where ω= 0.000002, α= 0.04, anβ= 0.94? The expectevarianrate in ten ys is 0.02^2 + (0.04 + 0.94)^10 (0.03^2– 0.02^2) = 0.000809, whicorrespon to a volatility of 2.84%. The expectevarianrate in 100 ys is 0.02^2 + (0.04 + 0.94)^100 (0.03^2 – 0.02^2) = 0.000466, whicorrespon to a volatility of 2.16%.题目问现在是volatility=3%，long-run average volatility=2%，利用GARCH(1，1)来计算10天和100天的volatility，ω=0.000002, α= 0.04, anβ= 0.94。10天volatility=[0.02^2 + (0.04 + 0.94)^10 (0.03^2– 0.02^2)]^0.5 = 2.84%.100天volatility=[0.02^2 + (0.04 + 0.94)^100 (0.03^2 – 0.02^2)]^0.5 =2.16%. 老师好，看不懂答案，1、这道题哪里的表述能看出u啊？也就是收益率？2、还有我写出的表达式除了不知道u（n-1），有其他哪里不对的地方吗？
NO.PZ2020011303000086问题如下 Suppose ththe current volatility estimate is 3% per y anthe long-run average volatility estimate is 2% per y. Whare the volatility estimates in ten ys an100 ys in a GAR(1,1) mol where ω= 0.000002, α= 0.04, anβ= 0.94? The expectevarianrate in ten ys is 0.02^2 + (0.04 + 0.94)^10 (0.03^2– 0.02^2) = 0.000809, whicorrespon to a volatility of 2.84%. The expectevarianrate in 100 ys is 0.02^2 + (0.04 + 0.94)^100 (0.03^2 – 0.02^2) = 0.000466, whicorrespon to a volatility of 2.16%.题目问现在是volatility=3%，long-run average volatility=2%，利用GARCH(1，1)来计算10天和100天的volatility，ω=0.000002, α= 0.04, anβ= 0.94。10天volatility=[0.02^2 + (0.04 + 0.94)^10 (0.03^2– 0.02^2)]^0.5 = 2.84%.100天volatility=[0.02^2 + (0.04 + 0.94)^100 (0.03^2 – 0.02^2)]^0.5 =2.16%. 为什么和上一题不一样
