叶赫那拉坤坤 · 2020年02月08日
问题如下：
According to the above table, what is the variance of X?
选项：
A.3.6.
B.1.90.
C.6.64.
解释：
A is correct
Firstly ,We calulate expected return of X, E(X) = (0.2)(-2) + (0.6)(1) + (0.2)(4) = 1,
Then,Var(X) = 0.2(-2-1)2 + 0.6(1-1)2 + 0.2(4-1)2 = 3.6
这个题与y有什么关系呀？
第一步的期望不知道怎么算
NO.PZ2015120604000062 问题如下 Accorng to the above table, whis the varianof X? A.3.6. B.1.90. C.6.64. A is correctFirstly ,We calulate expectereturn of X, E(X) = (0.2)(-2) + (0.6)(1) + (0.2)(4) = 1, Then,Var(X) = 0.2(-2-1)2 + 0.6(1-1)2 + 0.2(4-1)2 = 3.6 如上
NO.PZ2015120604000062 问题如下 Accorng to the above table, whis the varianof X? A.3.6. B.1.90. C.6.64. A is correctFirstly ,We calulate expectereturn of X, E(X) = (0.2)(-2) + (0.6)(1) + (0.2)(4) = 1, Then,Var(X) = 0.2(-2-1)2 + 0.6(1-1)2 + 0.2(4-1)2 = 3.6 答案中第一步算均值理解， 第二步计算VarX不太明白，概率是Joint 也不是X单独的，为什么可以用来算X维度的Var呢
NO.PZ2015120604000062问题如下Accorng to the above table, whis the varianof X?A.3.6.B.1.90.C.6.64.A is correctFirstly ,We calulate expectereturn of X, E(X) = (0.2)(-2) + (0.6)(1) + (0.2)(4) = 1, Then,Var(X) = 0.2(-2-1)2 + 0.6(1-1)2 + 0.2(4-1)2 = 3.6如题，看答案解析以后，是要理解成这道题目中，P（XY）=P（X）=P（Y）吗？主要不理解，为什么可以用Joint Probability的概率，直接用来当做X发生的概率。
NO.PZ2015120604000062 问题如下 Accorng to the above table, whis the varianof X? A.3.6. B.1.90. C.6.64. A is correctFirstly ,We calulate expectereturn of X, E(X) = (0.2)(-2) + (0.6)(1) + (0.2)(4) = 1, Then,Var(X) = 0.2(-2-1)2 + 0.6(1-1)2 + 0.2(4-1)2 = 3.6 記得之前有到道題要求x的，最後要開根號.爲何這裡算出y以後，不用開根號？謝謝
NO.PZ2015120604000062问题如下Accorng to the above table, whis the varianof X?A.3.6.B.1.90.C.6.64.A is correctFirstly ,We calulate expectereturn of X, E(X) = (0.2)(-2) + (0.6)(1) + (0.2)(4) = 1, Then,Var(X) = 0.2(-2-1)2 + 0.6(1-1)2 + 0.2(4-1)2 = 3.6根据讲义的公式算出来是需要除以n，这里算出3.6后为啥不需要除3？
