问题如下：
In the MA(2), , where , what is the ACF?
选项：
解释：
and . The autocorrelations are then \rho_0 = 1, \rho_1 = 0.557\rho_2 = 0.096$$.
r2是6.25倍的方差吧？
NO.PZ2020011101000014 问题如下 In the MA(2), Yt=4.1+5ϵt−1+6.25ϵt−2+ϵtY_t = 4.1 + 5\epsilon_{t - 1} + 6.25\epsilon_{t - 2} + \epsilon_tYt=4.1+5ϵt−1+6.25ϵt−2+ϵt, where ϵt∼WN(0,σ2)\epsilon_t ∼ WN(0, \sigma^2)ϵt∼WN(0,σ2), whis the ACF? γ0=(1+52+6.252)σ2,γ1=(5+5∗6.25)σ2\gamma_0 = (1 + 5^2 + 6.25^2)\sigma^2, \gamma_1 = (5 + 5 * 6.25)\sigma^2γ0=(1+52+6.252)σ2,γ1=(5+5∗6.25)σ2 anγ2=6.25σ2\gamma_2 = 6.25\sigma^2γ2=6.25σ2. The autocorrelations are then ρ0=1,ρ1=0.557\rho_0 = 1, \rho_1 = 0.557ρ0=1,ρ1=0.557 anρ2=0.096\rho_2 = 0.096ρ2=0.096. 老师好，请问这个公式在讲义哪里有呀。解题思路是什么呢？
NO.PZ2020011101000014 问题如下 In the MA(2), Yt=4.1+5ϵt−1+6.25ϵt−2+ϵtY_t = 4.1 + 5\epsilon_{t - 1} + 6.25\epsilon_{t - 2} + \epsilon_tYt=4.1+5ϵt−1+6.25ϵt−2+ϵt, where ϵt∼WN(0,σ2)\epsilon_t ∼ WN(0, \sigma^2)ϵt∼WN(0,σ2), whis the ACF? γ0=(1+52+6.252)σ2,γ1=(5+5∗6.25)σ2\gamma_0 = (1 + 5^2 + 6.25^2)\sigma^2, \gamma_1 = (5 + 5 * 6.25)\sigma^2γ0=(1+52+6.252)σ2,γ1=(5+5∗6.25)σ2 anγ2=6.25σ2\gamma_2 = 6.25\sigma^2γ2=6.25σ2. The autocorrelations are then ρ0=1,ρ1=0.557\rho_0 = 1, \rho_1 = 0.557ρ0=1,ρ1=0.557 anρ2=0.096\rho_2 = 0.096ρ2=0.096. 公式中包含σ²，白噪声中的σ²怎么求？
NO.PZ2020011101000014 老师，请帮助理解一下问号部分。感恩
NO.PZ2020011101000014 公式我知道，但是看不懂怎么带的，按照公式还有一个 θ0 ，但是公式里面哪里有这个值？不懂
h>=2,ACF不是等于0吗？
