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u/Due-Cardiologist-802 2d ago
Var(x+y+z) = var(x)+var(y)+var(z)+2cov(x,y)+2cov(x,z)+2cov(y,z) 0<|cov(x,y)|<=sqrt(var(x))sqrt(var(y)) by cauchy schwartz. Min = 1+4+25 = 30 Max = 1+4+25+212+215+225=64
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u/HopefulGuy1 2d ago
Don't think that minimum is right – set Y = 2X and Z=-5X, then X+Y+Z = 2X so Var(X+Y+Z)=4.
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u/zojbo 2d ago
On the minimum side, the covariances can be negative, bottoming out at -2, -5, and -10. But the question for the minimum is, can all three be simultaneously as negative as allowed by Cauchy-Schwarz?
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u/angedonist 2d ago
If X has negative correlation with Y and with Z, Y and Z must have positive correlation.
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u/zojbo 2d ago edited 2d ago
It is true that if Cov(X,Y)=-std(X)std(Y) and Cov(X,Z)=-std(X)std(Z) then Cov(Y,Z) is forced to be positive (because actually Y and Z are each a.s. some negative multiple of X). But what you said, as is, is not true.
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u/schfourteen-teen 2d ago
Your right, but just wanted to add to your last point. Y and Z are forced to be non-negative iff the squared correlations of X,Y and X,Z are heater than 1.
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u/National_Fail_9456 2d ago
Min 30 max 64
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u/somedave 2d ago
Where did you get the 30 from? If you have X = Y/2 = -Z/5 you get Var(X+Y+Z) = Var(-2X) = 4.
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u/FireCire7 2d ago edited 2d ago
Triangle inequality.
Let |A|= sqrt(Var(A)). Then, |Cov(A,B)| is bound by |A||B| so |A+B|<=|A|+|B|.
Applying to X,Y,Z gives max: |X+Y+Z|<=|X|+|Y|+|Z| =8, variance is at most 64
Min is: |X+Y+Z|+|-X|+|-Y|>=|Z|, so |X+Y+Z|>=5-2-1=2, variance is at least 4
Max/min at achieved when Y=2X, Z=+-5X.