quietvivian · 2020年05月10日
问题如下：
How are the mean and variance of a mixture of normal random variables related to the mean and variance of the components of the mixture?
选项：
解释：
The mean is the weighted average of the means of the components. The variance is more complicated. The second non-central moment, of the mixture is the weighted average of the second non-central moments of the components. The variance is then , which depends the first and second moments of the mixture.
答案没有看懂，老师可以讲一下吗谢谢
NO.PZ2020010303000008 问题如下 How are the meanvarianof a mixture of normranm variables relateto the meanvarianof the components of the mixture? The meis the weighteaverage of the means of the components. The varianis more complicate The seconnon-centrmoment, E[X2]E[X^2]E[X2] of the mixture is the weighteaverage of the seconnon-centrmoments of the components. The varianis then E[X2]−E[X]2E[X^2] - E[X]^2E[X2]−E[X]2, whipen the first anseconmoments of the mixture. How are the meanvarianof a mixture of normranm variables relateto the meanvarianof the components of the mixture?1.请问这道题是这样理解吗混合正态分布的均值、方差与混合正态分布各组成部分分布的均值、方差的联系如何?2.课上有一点提到，即使混合分布的各组成分布无偏差、峰度为0，混合分布也可能有偏差、有峰度。也就是说各组成分布为正态分布，但混合分布不一定为正态分布。这样理解对吗？3.这里由各组成分布的均值、方差，推出混合分布的均值、方差为考点吗？需要记住什么公式和概念呢？
NO.PZ2020010303000008 问题如下 How are the meanvarianof a mixture of normranm variables relateto the meanvarianof the components of the mixture? The meis the weighteaverage of the means of the components. The varianis more complicate The seconnon-centrmoment, E[X2]E[X^2]E[X2] of the mixture is the weighteaverage of the seconnon-centrmoments of the components. The varianis then E[X2]−E[X]2E[X^2] - E[X]^2E[X2]−E[X]2, whipen the first anseconmoments of the mixture. 这道题视频里没有讲到，现在还会考察么？
NO.PZ2020010303000008老师 我先想确认一下mixture stribution的定义里是不是有要求weight必须是ranm variable而非常数？如果是的话，那题干中mixture of normstribution和linecombination of normstribution不是就不一样了吗，这个时候expectation还能直接根据weight这个ranm variable的每种取值对应的概率直接加权平均吗，weight已经不是常数了，那答案中说的期望是加权平均具体是怎么算的呢。或者说老师能否举一个有3个component stribution的 mixture的例子呢，这个时候W就不能是Bernoulli而是其他分布了，那求期望的时候不是要按定义计算，一定能够推导到加权平均的形式吗？我不要太理解为什么答案中关于期望和方差的部分是成立的，感觉题干问得是一种比较general的情况。谢谢老师！
老师，该题求Variance的没看懂能再说仔细点么？
