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NO.PZ2023091601000073

问题如下：

Suppose that the current daily volatilities of asset X and asset Y are 1.0% and 1.2%, respectively. The prices of the assets at close of trading yesterday were $30 and $50 and the estimate of the coefficient of correlation between the returns on the two assets made at this time was 0.50. Correlations and volatilities are updated using a GARCH (1, 1) model. The estimates of the model's parameters are α = 0.04 and β = 0.94. For the correlation ω = 0.000001, and for the volatilities ω = 0.000003. If the prices of the two assets at close of trading today are $31 and $51, how is the correlation estimate updated?

选项：

0.539

0.549

0.559

0.569

解释：

The estimated covariance of n-1 is:

The covariance of n i:

The asset X’s variance of n is:

因此，

The asset Y’s variance of n is:

So,

The correlation is: 0.0000841 / (0.01189 × 0.01242) = 0.569

老师你好，还是不明白，可以再解释吗

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1 个答案

李坏_品职助教 · 2024年11月11日

嗨，从没放弃的小努力你好：

题目让你用GARCH(1,1)去推算资产X的return（X的return是u）和资产Y的return（Y的return是v）的相关系数。就是求u和v的在t时刻的相关系数。

思路是：用t-1时刻的协方差cov t-1，代入GARCH模型求出t时刻的cov，然后用t时刻的cov除以t时刻u和v的标准差，得出t时刻的相关系数。

根据条件，ut-1 = (31-30)/30 = 1/30 = 0.03333, vt-1 = (51-50)/50 = 1/50= 0.02.
而t-1时刻的u和v的协方差是cov t-1, cov t-1 = t-1时刻的相关系数0.5 * t-1时刻u的标准差1% * t-1时刻v的标准差1.2% = 0.00006.
为了求出相关系数，还要求出t时刻的u和v的协方差也就是cov t。按照GARCH(1,1)模型，cov t = w for correlation + α* ut-1 * vt-1 + β * cov t-1 = 0.000001 + 0.04 * 0.03333 * 0.02 + 0.94 * 0.00006 = 0.0000841.
由于根据GARCH(1,1)模型，t时刻的u的方差 = w for volatilities + α * u^2 t-1 + β * t-1时刻u的方差 = 0.000003+0.04*0.03333^2 + 0.94*0.01^2 = 0.0001414, 所以t时刻u的标准差 = 根号0.0001414 = 0.01189. 同理可以求出t时刻v的标准差是0.01242.
所以t时刻u和v的相关系数 = t时刻的协方差 0.0000841 / (0.01189 × 0.01242) = 0.569

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虽然现在很辛苦，但努力过的感觉真的很好，加油！

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NO.PZ2023091601000073 问题如下 Suppose ththe current ilyvolatilities of asset X anasset Y are 1.0% an1.2%, respectively. The pricesof the assets close of trang yestery were $30 an$50 anthe estimateof the coefficient of correlation between the returns on the two assets ma atthis time w0.50. Correlations anvolatilities are upteusing a GAR(1,1) mol. The estimates of the mol's parameters are α = 0.04 anβ = 0.94.For the correlation ω = 0.000001, anfor the volatilities ω = 0.000003. If theprices of the two assets close of trang toy are $31 an$51, how is thecorrelation estimate upte A.0.539 B.0.549 C.0.559 0.569 The estimateovarianof n-1 is: The covarianof n i: The asset X’s varianof n is: 因此，The asset Y’s varianof n is: So, The correlation is: 0.0000841 / (0.01189 × 0.01242)= 0.569

2024-05-05 10:50 1 · 回答