问题如下图：
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老师好，为什么第n天的协方差可以用画绿色的这个公式来计算？
NO.PZ2019040801000027问题如下The pripercent changes of stoX anY were 5.0% an1.0%, respectively. The correlation estimate baseon the historicta of the two on y n-1 is 0.6, the estimatestanrviations of priof X anY on y n-1 were 2.3% an1.7%, respectively. Suppose the analyst uses the EWMA mol with λ = 0.97 to upte the correlation ancovariance. Whis the new estimate of the correlation between X anY on y n?A.0.34.B.0.42.C.0.60.0.68.C is correct.考点EWMA模型解析先计算y n-1时候的协方差cov(X, Y) = ΡX,Y x σXσY = 0.6 * 2.3% * 1.7% = 0.000235然后通过EWMA模型估计y n的协方差covn = 0.97 * 0.000235 + 0.03 * 5% * 1% = 0.00024295估计X的方差σ2X,n = 0.97 x 0.023^2 + 0.03 x 0.05^2 = 0.00058813X的标准差就是0.00058813^0.5=0.02425估计Y的方差σ2Y,n = 0.97 x 0.0172 + 0.03 x0.012 = 0.0003078 + 0.000003 = 0.0002833Y的标准差就是0.0002833^0.5=0.01683最后新的相关系数就是协方差除以X标准差再除以Y的标准差0.00024295/（0.02425*0.01683）=0.5952老师好，请问EWMA和GARCH的covariance是在课程哪里讲的呢？看了下笔记和课程没找到。
NO.PZ2019040801000027问题如下The pripercent changes of stoX anY were 5.0% an1.0%, respectively. The correlation estimate baseon the historicta of the two on y n-1 is 0.6, the estimatestanrviations of priof X anY on y n-1 were 2.3% an1.7%, respectively. Suppose the analyst uses the EWMA mol with λ = 0.97 to upte the correlation ancovariance. Whis the new estimate of the correlation between X anY on y n?A.0.34.B.0.42.C.0.60.0.68.C is correct.考点EWMA模型解析先计算y n-1时候的协方差cov(X, Y) = ΡX,Y x σXσY = 0.6 * 2.3% * 1.7% = 0.000235然后通过EWMA模型估计y n的协方差covn = 0.97 * 0.000235 + 0.03 * 5% * 1% = 0.00024295估计X的方差σ2X,n = 0.97 x 0.023^2 + 0.03 x 0.05^2 = 0.00058813X的标准差就是0.00058813^0.5=0.02425估计Y的方差σ2Y,n = 0.97 x 0.0172 + 0.03 x0.012 = 0.0003078 + 0.000003 = 0.0002833Y的标准差就是0.0002833^0.5=0.01683最后新的相关系数就是协方差除以X标准差再除以Y的标准差0.00024295/（0.02425*0.01683）=0.5952有快速计算方法吗，或者计算机简便方式
NO.PZ2019040801000027 问题如下 The pripercent changes of stoX anY were 5.0% an1.0%, respectively. The correlation estimate baseon the historicta of the two on y n-1 is 0.6, the estimatestanrviations of priof X anY on y n-1 were 2.3% an1.7%, respectively. Suppose the analyst uses the EWMA mol with λ = 0.97 to upte the correlation ancovariance. Whis the new estimate of the correlation between X anY on y n? A.0.34. B.0.42. C.0.60. 0.68. C is correct.考点EWMA模型解析先计算y n-1时候的协方差cov(X, Y) = ΡX,Y x σXσY = 0.6 * 2.3% * 1.7% = 0.000235然后通过EWMA模型估计y n的协方差covn = 0.97 * 0.000235 + 0.03 * 5% * 1% = 0.00024295估计X的方差σ2X,n = 0.97 x 0.023^2 + 0.03 x 0.05^2 = 0.00058813X的标准差就是0.00058813^0.5=0.02425估计Y的方差σ2Y,n = 0.97 x 0.0172 + 0.03 x0.012 = 0.0003078 + 0.000003 = 0.0002833Y的标准差就是0.0002833^0.5=0.01683最后新的相关系数就是协方差除以X标准差再除以Y的标准差0.00024295/（0.02425*0.01683）=0.5952 估计y n的协方差covn = 0.97 * 0.000235 + 0.03 * 5% * 1% = 0.00024295这步是什么意思？具体使用的公式是哪个公式？XY的协方差不是已经在上一步计算出来是0.000235吗？此处的协方差是什么的协方差？
NO.PZ2019040801000027 问题如下 The pripercent changes of stoX anY were 5.0% an1.0%, respectively. The correlation estimate baseon the historicta of the two on y n-1 is 0.6, the estimatestanrviations of priof X anY on y n-1 were 2.3% an1.7%, respectively. Suppose the analyst uses the EWMA mol with λ = 0.97 to upte the correlation ancovariance. Whis the new estimate of the correlation between X anY on y n? A.0.34. B.0.42. C.0.60. 0.68. C is correct.考点EWMA模型解析先计算y n-1时候的协方差cov(X, Y) = ΡX,Y x σXσY = 0.6 * 2.3% * 1.7% = 0.000235然后通过EWMA模型估计y n的协方差covn = 0.97 * 0.000235 + 0.03 * 5% * 1% = 0.00024295估计X的方差σ2X,n = 0.97 x 0.023^2 + 0.03 x 0.05^2 = 0.00058813X的标准差就是0.00058813^0.5=0.02425估计Y的方差σ2Y,n = 0.97 x 0.0172 + 0.03 x0.012 = 0.0003078 + 0.000003 = 0.0002833Y的标准差就是0.0002833^0.5=0.01683最后新的相关系数就是协方差除以X标准差再除以Y的标准差0.00024295/（0.02425*0.01683）=0.5952 助教你好一基础班讲的ARCH, EWMGARmol全是关于volatility，所有例题也是关于volatility，而这题涉及到的东西为什么没在基础班里面讲过？二这题所说的东西是指下方框架图这个知识吗？框架图上这块内容根本没出现在基础班讲义里面。基础班是漏了东西？还是这框架图和题目是多余的东西？
NO.PZ2019040801000027问题如下The pripercent changes of stoX anY were 5.0% an1.0%, respectively. The correlation estimate baseon the historicta of the two on y n-1 is 0.6, the estimatestanrviations of priof X anY on y n-1 were 2.3% an1.7%, respectively. Suppose the analyst uses the EWMA mol with λ = 0.97 to upte the correlation ancovariance. Whis the new estimate of the correlation between X anY on y n?A.0.34.B.0.42.C.0.60.0.68.C is correct.考点EWMA模型解析先计算y n-1时候的协方差cov(X, Y) = ΡX,Y x σXσY = 0.6 * 2.3% * 1.7% = 0.000235然后通过EWMA模型估计y n的协方差covn = 0.97 * 0.000235 + 0.03 * 5% * 1% = 0.00024295估计X的方差σ2X,n = 0.97 x 0.023^2 + 0.03 x 0.05^2 = 0.00058813X的标准差就是0.00058813^0.5=0.02425估计Y的方差σ2Y,n = 0.97 x 0.0172 + 0.03 x0.012 = 0.0003078 + 0.000003 = 0.0002833Y的标准差就是0.0002833^0.5=0.01683最后新的相关系数就是协方差除以X标准差再除以Y的标准差0.00024295/（0.02425*0.01683）=0.5952估计协方差在何处需确定公示
