苏·Xu · 2021年01月20日
问题如下：
According to the above table, what is the correlation of X and Y, given the joint probability table above?
选项：
A. -0.98.
B. 0.16.
C. 0.98.
解释：
A is correct
,
Cov(X,Y)=-4.8, standard deviations of X and Y are 1.90 and 2.58, as calculated before,
thus correlation of X and Y is -0.98
这道题可以详细讲解一下每一步的解题过程吗？
星星_品职助教 · 2021年01月21日
同学你好，
本题要求correlation，根据公式，Corr(X,Y)=ρ(X,Y)=Cov(X,Y)/σxσy。需要知道①Cov(X,Y)；②σx；③σy
Cov(X,Y)=E[(（X-E(X）)（Y-E(Y)））]，所以问题先转化为求E(X)和E(Y)
根据表格，X一共有三个值。其中X=-2的概率就是当X=-2时的边际概率，0.2+0+0=0.2；同理，X=1的概率为0+0.6+0=0.6，X=4的概率为0+0+0.2=0.2.
求期望就是求（加权）平均的过程，所以E(X)=0.2×(-2)+0.6×1+0.2×4=1，
同理，E(Y)=0.2×5+0.6×2+0.2×(-3)=1.6.
所以Cov(X,Y)=0.2×（-2-1）×（5-1.6）+0.6×（1-1）×（2-1.6）+0.2×（4-1）×（-3-1.6）=-4.8 ①
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通过方差的公式得到Var(X)=E[（X-E(X）)^2]=0.2×（-2-1）^2+0.6×（1-1）^2+0.2×（4-1）^2=3.6，即σx=1.8974 ②
Var(Y)=0.2×（5-1.6）^2+0.6×（2-1.6）^2+0.2×（-3-1.6）^2=6.64，即σY=2.5768 ③
结合①，②，③，可得ρ(X,Y)=-4.8/（1.8974×2.5768）=-0.9818
NO.PZ2015120604000064问题如下 Accorng to the above table, whis the correlation of X anY, given the joint probability table above?A.-0.98.B.0.16.C.0.98.A is correctCorr(X,Y)=Cov(X,Y)σxσyCorr(X,Y)=\fr{ Cov(X,Y) }{ { \sigma }_{ x }{ \sigma }_{ y } } Corr(X,Y)=σxσyCov(X,Y),Cov(X,Y)=-4.8, stanrviations of X anY are 1.90 an2.58, calculatebefore,thus correlation of X anY is -0.98这道题不难理解 但是感觉写了半张纸 费了很多分钟。请问有没有简便方法 或者这种计算量是真题会有的吗。
NO.PZ2015120604000064 问题如下 Accorng to the above table, whis the correlation of X anY, given the joint probability table above? A.-0.98. B.0.16. C.0.98. A is correctCorr(X,Y)=Cov(X,Y)σxσyCorr(X,Y)=\fr{ Cov(X,Y) }{ { \sigma }_{ x }{ \sigma }_{ y } } Corr(X,Y)=σxσyCov(X,Y),Cov(X,Y)=-4.8, stanrviations of X anY are 1.90 an2.58, calculatebefore,thus correlation of X anY is -0.98 如题，看了之前的解析，还是不知道X和Y 的标准差怎么求出来的，能不能仔细讲解一下
NO.PZ2015120604000064 问题如下 Accorng to the above table, whis the correlation of X anY, given the joint probability table above? A.-0.98. B.0.16. C.0.98. A is correctCorr(X,Y)=Cov(X,Y)σxσyCorr(X,Y)=\fr{ Cov(X,Y) }{ { \sigma }_{ x }{ \sigma }_{ y } } Corr(X,Y)=σxσyCov(X,Y),Cov(X,Y)=-4.8, stanrviations of X anY are 1.90 an2.58, calculatebefore,thus correlation of X anY is -0.98 因为要求correlation，所以要分别求出公式里面的covariance和stanrviation分别求E(x)和E(y)，得出1和1.6求方差variance，然后开根号得出标准差stanrviation，-- 1.8974和2.5768求covariance: - 4.8把第二点的标准差和第三点的协方差带入correlation的公式求出结果
NO.PZ2015120604000064 问题如下 Accorng to the above table, whis the correlation of X anY, given the joint probability table above? A.-0.98. B.0.16. C.0.98. A is correctCorr(X,Y)=Cov(X,Y)σxσyCorr(X,Y)=\fr{ Cov(X,Y) }{ { \sigma }_{ x }{ \sigma }_{ y } } Corr(X,Y)=σxσyCov(X,Y),Cov(X,Y)=-4.8, stanrviations of X anY are 1.90 an2.58, calculatebefore,thus correlation of X anY is -0.98 可以告诉一下公式吗
NO.PZ2015120604000064问题如下Accorng to the above table, whis the correlation of X anY, given the joint probability table above?A.-0.98.B.0.16.C.0.98.A is correctCorr(X,Y)=Cov(X,Y)σxσyCorr(X,Y)=\fr{ Cov(X,Y) }{ { \sigma }_{ x }{ \sigma }_{ y } } Corr(X,Y)=σxσyCov(X,Y),Cov(X,Y)=-4.8, stanrviations of X anY are 1.90 an2.58, calculatebefore,thus correlation of X anY is -0.98为什么计算出VarX 和VarY后，最后一步代入公式 不用开根号？
