弱问：线性模型误差项不服从正态分布

是 否 +2 论坛币

k人 参与回答

经管之家送您一份

应届毕业生专属福利!

求职就业群

赵安豆老师微信：zhaoandou666

经管之家联合CDA

送您一个全额奖学金名额~ !

立即领取

感谢您参与论坛问题回答

经管之家送您两个论坛币！

+2 论坛币

的实践后果是什么呀 好像没有多大关系吧

扫码加我 拉你入群

请注明：姓名-公司-职位

以便审核进群资格，未注明则拒绝

分享0 收藏0 回帖

关键词：正态分布 线性模型 不服从 误差项 模型 正态分布 线性 误差

as long as the first two moment conditions on the error term hold, most of the properties of OLS technique. However, t and F test are no longer valid, or at least no exact distribution. but you can utilize CLT to have asymptotic results.

read the text!

as long as the first two moment conditions on the error term hold, most of the properties of OLS technique. However, t and F test are no longer valid, or at least no exact distribution. but you can utilize CLT to have asymptotic results.

read the text!

Thanks! But waht I'm asking is, if there is any practical consequense.

What I know is, when we appeal to large (asymptotic) sampal properties, we should use Z-test instead of t and Chi-square instead of F.

However, it seems to me that, practically few people use Z and Chi; and for most of the time, the test results of t and F are pretty close to Z and Chi. If this is the case, does that mean, the normality of error term is only a theoretial issue and we can actually ignore it in practice?

在样本量较小的情况下，如果扰动项不服从正太分布，那么此时被解释变量也为非正太分布，而参数估计建立在t分布的基础上构造的会存在分布的设定偏误，但当样本量较大时，参数分布会趋于正太分布，相对影响不大。

看一下大样本性质就知道了，大样本不需要误差项正态性假定