
The covariance of two random variables $X$ and $Y$ is given by $\operatorname{cov} (X,Y)$, properties of $\operatorname{cov} (X,Y)$ are:-
$(i)\operatorname{cov} (aX + b,cY + d) = ac \times \operatorname{cov} (X,Y){\text{where a,b,c,d are constant}}$
$(ii)\operatorname{cov} ({X_1} + {X_2},Y) = \operatorname{cov} ({X_1},Y) + \operatorname{cov} ({X_2},Y)$
$(iii){\text{X and Y are independent}} \Rightarrow \operatorname{cov} (X,Y) = \operatorname{cov} ({X_1},Y) = 0$
${\text{(A) only (i) is correct}}$
${\text{(B) All the three properties are correct}}$
${\text{(C) only (iii) is wrong}}$
${\text{(D) only (i) and (ii) is correct}}$
Answer
554.4k+ views
Hint: 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: $\operatorname{cov} (X,Y) = E([X - E(X)][Y - E(Y)])$
Complete step by step solution:
It is given the question stated as the covariance of two random variables $X$ and $Y$ is given by $\operatorname{cov} (X,Y)$
Now we have to use the formula of covariance is: $\operatorname{cov} (X,Y) = E([X - E(X)][Y - E(Y)])$
Now consider the first property:
$\operatorname{cov} (aX + b,cY + d) = ac \cdot \operatorname{cov} (X,Y){\text{where a,b,c,d are constant}}$
Now we know a property of covariance which is:
${\text{if ab}} \in {\text{R then cov}}(a + bX,Y) = b \cdot \operatorname{cov} (X,Y) \to (1)$
Since the addition of constant value does not change the correlation because a constant value is independent of any random variable and from property $(1)$ we can conclude that the given property $(i)$ is true.
Now, consider the second property:
$\operatorname{cov} ({X_1} + {X_2},Y) = \operatorname{cov} ({X_1},Y) + \operatorname{cov} ({X_2},Y)$
We know that there is a transitive property of covariance which implies that when ${X_1},{X_2},Y$ is random variable such that $\operatorname{cov} ({X_1} + {X_2},Y)$ then by using the property of transitivity,
$\operatorname{cov} ({X_1} + {X_2},Y) = \operatorname{cov} ({X_1},Y) + \operatorname{cov} ({X_2},Y)$ Therefore, this property is true.
Now, the third property:
${\text{X and Y are independent}} \Rightarrow \operatorname{cov} (X,Y) = \operatorname{cov} ({X_1},Y) = 0$
Since it is mentioned that the two variables $X$ and $Y$ are independent then is no correlation between them because the randomness of $X$ does not affect $Y$ and the randomness of $Y$ does not affect $X$. Therefore, the total correlation between both of them will be $0$.
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 $(B)$.
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.
Formula used: $\operatorname{cov} (X,Y) = E([X - E(X)][Y - E(Y)])$
Complete step by step solution:
It is given the question stated as the covariance of two random variables $X$ and $Y$ is given by $\operatorname{cov} (X,Y)$
Now we have to use the formula of covariance is: $\operatorname{cov} (X,Y) = E([X - E(X)][Y - E(Y)])$
Now consider the first property:
$\operatorname{cov} (aX + b,cY + d) = ac \cdot \operatorname{cov} (X,Y){\text{where a,b,c,d are constant}}$
Now we know a property of covariance which is:
${\text{if ab}} \in {\text{R then cov}}(a + bX,Y) = b \cdot \operatorname{cov} (X,Y) \to (1)$
Since the addition of constant value does not change the correlation because a constant value is independent of any random variable and from property $(1)$ we can conclude that the given property $(i)$ is true.
Now, consider the second property:
$\operatorname{cov} ({X_1} + {X_2},Y) = \operatorname{cov} ({X_1},Y) + \operatorname{cov} ({X_2},Y)$
We know that there is a transitive property of covariance which implies that when ${X_1},{X_2},Y$ is random variable such that $\operatorname{cov} ({X_1} + {X_2},Y)$ then by using the property of transitivity,
$\operatorname{cov} ({X_1} + {X_2},Y) = \operatorname{cov} ({X_1},Y) + \operatorname{cov} ({X_2},Y)$ Therefore, this property is true.
Now, the third property:
${\text{X and Y are independent}} \Rightarrow \operatorname{cov} (X,Y) = \operatorname{cov} ({X_1},Y) = 0$
Since it is mentioned that the two variables $X$ and $Y$ are independent then is no correlation between them because the randomness of $X$ does not affect $Y$ and the randomness of $Y$ does not affect $X$. Therefore, the total correlation between both of them will be $0$.
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 $(B)$.
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.
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