浅墨 · 2025年07月04日
NO.PZ2015120604000064
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
首先先计算期望值。
同理E（Y）=5*0.2+2*0.6+（-3）*0.2=1.6
第二步计算方差，
同理计算Y的方差=0.2*（5-1.6）^2+0.6*(2-1.6)^2+0.2*(-3-1.6)^2=2.312+0.096+4.232=6.64
第三步计算协方差
cov=(-2-1)*(5-1.6)*0.2+(1-1)*(2-1.6)*0.6+(4-1)*(-3-1.6)*0.2=-4.8
corr(x,y)=-4.8/(3.6^0.5*6.64^0.5）=-0.98
最后一行的公示计算相关系数的时候0.5是怎么得到的 是因为两个变量 所以每个各0.5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？
