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Question: The covariance of two random variables \(X\) and \(Y\) is given by \(\operatorname{cov} (X,Y)\), pro...

The covariance of two random variables XX and YY is given by cov⁡(X,Y)\operatorname{cov} (X,Y), properties of cov⁡(X,Y)\operatorname{cov} (X,Y) are:-
(i)cov⁡(aX+b,cY+d)=ac×cov⁡(X,Y)where a,b,c,d are constant(i)\operatorname{cov} (aX + b,cY + d) = ac \times \operatorname{cov} (X,Y){\text{where a,b,c,d are constant}}
(ii)cov⁡(X1+X2,Y)=cov⁡(X1,Y)+cov⁡(X2,Y)(ii)\operatorname{cov} ({X_1} + {X_2},Y) = \operatorname{cov} ({X_1},Y) + \operatorname{cov} ({X_2},Y)
(iii)X and Y are independent⇒cov⁡(X,Y)=cov⁡(X1,Y)=0(iii){\text{X and Y are independent}} \Rightarrow \operatorname{cov} (X,Y) = \operatorname{cov} ({X_1},Y) = 0
(A) only (i) is correct{\text{(A) only (i) is correct}}
(B) All the three properties are correct{\text{(B) All the three properties are correct}}
(C) only (iii) is wrong{\text{(C) only (iii) is wrong}}
(D) only (i) and (ii) is correct{\text{(D) only (i) and (ii) is correct}}

Explanation

Solution

First we have to verify the given properties are satisfied covariance. Also, we use the covariance formula to satisfy the properties. Finally we conclude the required answer

Formula used: cov⁡(X,Y)=E([X−E(X)][Y−E(Y)])\operatorname{cov} (X,Y) = E([X - E(X)][Y - E(Y)])

Complete step by step solution:
It is given the question stated as the covariance of two random variables XX and YY is given by cov⁡(X,Y)\operatorname{cov} (X,Y)
Now we have to use the formula of covariance is: cov⁡(X,Y)=E([X−E(X)][Y−E(Y)])\operatorname{cov} (X,Y) = E([X - E(X)][Y - E(Y)])
Now consider the first property:
cov⁡(aX+b,cY+d)=ac⋅cov⁡(X,Y)where a,b,c,d are constant\operatorname{cov} (aX + b,cY + d) = ac \cdot \operatorname{cov} (X,Y){\text{where a,b,c,d are constant}}
Now we know a property of covariance which is:
if ab∈R then cov(a+bX,Y)=b⋅cov⁡(X,Y)→(1){\text{if ab}} \in {\text{R then cov}}(a + bX,Y) = b \cdot \operatorname{cov} (X,Y) \to (1)
Since the addition of constant value does not change the correlation because a constant value is independent of any random variable and from property (1)(1) we can conclude that the given property (i)(i) is true.
Now, consider the second property:
cov⁡(X1+X2,Y)=cov⁡(X1,Y)+cov⁡(X2,Y)\operatorname{cov} ({X_1} + {X_2},Y) = \operatorname{cov} ({X_1},Y) + \operatorname{cov} ({X_2},Y)
We know that there is a transitive property of covariance which implies that when X1,X2,Y{X_1},{X_2},Y is random variable such that cov⁡(X1+X2,Y)\operatorname{cov} ({X_1} + {X_2},Y) then by using the property of transitivity,
cov⁡(X1+X2,Y)=cov⁡(X1,Y)+cov⁡(X2,Y)\operatorname{cov} ({X_1} + {X_2},Y) = \operatorname{cov} ({X_1},Y) + \operatorname{cov} ({X_2},Y) Therefore, this property is true.
Now, the third property:
X and Y are independent⇒cov⁡(X,Y)=cov⁡(X1,Y)=0{\text{X and Y are independent}} \Rightarrow \operatorname{cov} (X,Y) = \operatorname{cov} ({X_1},Y) = 0
Since it is mentioned that the two variables XX and YY are independent then is no correlation between them because the randomness of XX does not affect YY and the randomness of YY does not affect XX. Therefore, the total correlation between both of them will be 00.
Therefore, the above given property is true.

Since all the three given properties are correct, all the given properties are correct, therefore the correct option is (B)(B).

Note: Covariance is an advancement of the concept of variance, which helps in finding the relation the variance of not singular but multiple random variables. Instead of measuring the fluctuation of a random variable which is singular, it finds the fluctuation of multiple random variables.
Covariance uses the means of the multiple distributions and their variances to find the covariance between them.
There are many applications of covariance in statistical data analysis.