Question
Two laboratory readings \(X\) and \(Y\), measured in their original units, satisfy \(\operatorname{Var}(X)=9\), \(\operatorname{Var}(Y)=16\) and \(\operatorname{Cov}(X,Y)=-6\). Neither variance is zero. The lab now publishes \(U=2X+10\) and \(V=100-3Y\). The subtraction in \(V\) is intentional: it records how far below a fixed reference level the Y reading lies.
What are \(\operatorname{Cov}(U,V)\) and \(\operatorname{Corr}(U,V)\)? Explain which parts of these transformations matter to each result.
Get hint
Hint 1
Subtracting fixed means removes the constants +10 and +100 from covariance.
Check your answer
Solution
The centered transforms are \(U-E[U]=2(X-E[X])\) and \(V-E[V]=-3(Y-E[Y])\). Therefore
\[
\operatorname{Cov}(U,V)=(2)(-3)\operatorname{Cov}(X,Y)=(-6)(-6)=36.
\]
The transformed variances are \(\operatorname{Var}(U)=2^2(9)=36\) and \(\operatorname{Var}(V)=(-3)^2(16)=144\). Thus
\[
\operatorname{Corr}(U,V)=\frac{36}{\sqrt{36}\sqrt{144}}=\frac{36}{6\cdot12}=\frac12.
\]
Originally \(\operatorname{Corr}(X,Y)=-6/(3\cdot4)=-1/2\), so reversing the direction of exactly one scale flips the sign without changing the absolute correlation. Constants do not affect centered moments; multiplying readings changes the covariance's units and magnitude, whereas standardization cancels the absolute scale factors in correlation.
For an explicit distribution check, take independent symmetric \(Z,W\in\{-1,1\}\) and set \(X=3Z\), \(Y=-2Z+\sqrt{12}\,W\). Then \(\operatorname{Var}(X)=9\), \(\operatorname{Var}(Y)=16\), \(\operatorname{Cov}(X,Y)=-6\), giving the transformed results directly.
Assumptions
All moments are finite and the specified covariance matrix is valid.
The offsets 10 and 100 are deterministic constants.
Final Answer
\(\operatorname{Cov}(U,V)=36\), \(\operatorname{Corr}(U,V)=+1/2\).