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YolandaQ · 2020年03月18日

问一道题:NO.PZ2020010304000051

问题如下:

You collect 50 years of annual data on equity and bond returns. The estimated mean equity return is 7.3% per year, and the sample mean bond return is 2.7% per year. The sample standard deviations are 18.4% and 5.3%, respectively. The correlation between the two-return series is -60%. Are the expected returns on these two assets the same? Does your answer change if the correlation is 0?

选项:

解释:

The null hypothesis is H0:μE=μBH_0: \mu_E = \mu_B.

The alternative is H1:μEμBH_1: \mu_E ≠ \mu_B.

The test statistic is based on the difference of the average returns, δ\delta = 7.3% - 2.7% = 4.6%.

The estimator of the variance of the difference is σE2+σB22σBE\sigma_E^2+\sigma_B^2-2\sigma_{BE}, which is 0.1842+0.05322(0.6)0.1840.053=0.0483.0.184^2+0.053^2-2*(-0.6)*0.184 * 0.053 = 0.0483.

The test statistic is δ0.04350=1.47\frac\delta{\sqrt{\displaystyle\frac{0.043}{50}}}=1.47

The critical value for a two-sides test is ±1.96\pm1.96 using a size of 5%. The null is not rejected.

If the correlation was 0, then the variance estimate would be 0.0366, and the test statistic is 1.69. The null would still not be rejected if the size was 5%, although if the test size was 10%, then the critical value would be ±1.645\pm1.645 and the correlation would matter.

test statistics 为什么是用4.6%去除以根号下0.043/50?为什么不是0.0483除以根号下50?

1 个答案

小刘_品职助教 · 2020年03月19日

同学你好

感谢你的指正,答案中出现了typo,我们很快修正:-)

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NO.PZ2020010304000051问题如下You colle50 years of annuta on equity anbonreturns. The estimatemeequity return is 7.3% per year, anthe sample mebonreturn is 2.7% per year. The sample stanrviations are 18.4% an5.3%, respectively. The correlation between the two-return series is -60%. Are the expectereturns on these two assets the same? es your answer change if the correlation is 0? The null hypothesis is H0:μE=μBH_0: \mu_E = \mu_BH0​:μE​=μB​.The alternative is H1:μE≠μBH_1: \mu_E ≠ \mu_BH1​:μE​​=μB​.The test statistic is baseon the fferenof the average returns, δ\lt= 7.3% - 2.7% = 4.6%.The estimator of the varianof the fferenis σE2+σB2−2σBE\sigma_E^2+\sigma_B^2-2\sigma_{BE}σE2​+σB2​−2σBE​, whiis 0.1842+0.0532−2∗(−0.6)∗0.184∗0.053=0.0483.0.184^2+0.053^2-2*(-0.6)*0.184 * 0.053 = 0.0483.0.1842+0.0532−2∗(−0.6)∗0.184∗0.053=0.0483.The test statistic is δ0.048350=1.47\frac\lta{\sqrt{\splaystyle\frac{0.0483}{50}}}=1.47500.0483​​δ​=1.47The criticvalue for a two-sis test is ±1.96\pm1.96±1.96 using a size of 5%. The null is not rejecteIf the correlation w0, then the varianestimate woul0.0366, anthe test statistic is 1.69. The null woulstill not rejecteif the size w5%, although if the test size w10%, then the criticvalue woul±1.645\pm1.645±1.645 anthe correlation woulmatter.老师,哪里能看出来crticvalue是1.96,和1.65呢

2024-05-13 16:53 1 · 回答

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2022-06-23 14:31 1 · 回答

NO.PZ2020010304000051 这个题目里没说confinlevel 5%啊, 这个是从哪里来的?

2021-08-09 02:07 2 · 回答

NO.PZ2020010304000051 老师好,请问一下estimator of the varianof the fference是拿什么公式算的呀?谢谢!

2021-04-07 13:33 1 · 回答