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Question: . If \[x,y\] are independent variable , then A.\[Cov\left( {x,y} \right) = 1\] B.\[{r_{xy}} = 0...

. If x,yx,y are independent variable , then
A.Cov(x,y)=1Cov\left( {x,y} \right) = 1
B.rxy=0{r_{xy}} = 0
C.rxy=1{r_{xy}} = 1
D.Cov(x,y)=0Cov\left( {x,y} \right) = 0

Explanation

Solution

Hint : In the given question we have given with Covariance Cov(X,Y)Cov\left( {X,Y} \right) and Coefficient of correlation rXY{r_{XY}} , so we have to solve for both the terms . We will use theorems like E(XY)=E(X)E(Y)E\left( {XY} \right) = E\left( X \right)E\left( Y \right) , where EE is the expected value or mean value and X&YX\& Y are independent variables. Another theorem will be Cov(X,Y)=E(XY)−E(X)E(Y)Cov\left( {X,Y} \right) = E\left( {XY} \right) - E\left( X \right)E\left( Y \right) , where CovCov is the covariance and X&YX\& Y are independent variables

** Complete step-by-step answer** :
Mean - The expected value of XX, where XX is a discrete random variable and is a weighted average of all the possible values that XX can take , each value being calculated according to the probability of that event occurring. The expected value of XX is written as E(X)E\left( X \right) .
Covariance is a measure of the association or dependency between two random variables X&YX\& Y . Covariance can be positive or negative both.
Now , using the above theorems we get ,
Cov(X,Y)=E(XY)−E(X)E(Y)................(i)Cov\left( {X,Y} \right) = E\left( {XY} \right) - E\left( X \right)E\left( Y \right)................\left( i \right)
Now putting the value of E(XY)E\left( {XY} \right) from first theorem we get in eqn (i) we get ,
Cov(X,Y)=E(X)E(Y)−E(X)E(Y)Cov\left( {X,Y} \right) = E\left( X \right)E\left( Y \right) - E\left( X \right)E\left( Y \right)
On solving we get ,
Cov(X,Y)=0............(ii)Cov\left( {X,Y} \right) = 0............\left( {ii} \right) .
Now for Coefficient of correlation rXY{r_{XY}} , we have a formula
rXY=Cov(X,Y)V(X)V(Y){r_{XY}} = \dfrac{{Cov\left( {X,Y} \right)}}{{\sqrt {V\left( X \right)} \sqrt {V\left( Y \right)} }}, where V(X)V\left( X \right) and V(Y)V\left( Y \right) is the variance of X&YX\& Y respectively .
Now from equation (ii) we have , Cov(X,Y)=0Cov\left( {X,Y} \right) = 0 , therefore putting this value in above equation we get ,
rXY=0V(X)V(Y){r_{XY}} = \dfrac{0}{{\sqrt {V\left( X \right)} \sqrt {V\left( Y \right)} }}
On solving we get ,
rXY=0{r_{XY}} = 0
So, the correct answer is “Option B and D”.

Note : Always check whether the given variables are independent or not. The term covariance is different with variance as the covariance can either be positive or negative but the variance will always be positive . You must remember the results of the theorem only instead of their proofs.