Question

Researchers randomly select a married couple in which both spouses are employed. Let $X$be the income of the husband and $Y$be the income of the wife. Suppose that you know the means ${\mu }_X\mathrm{and}{\mu }_Y$and the variances ${{\mathrm{\sigma }}_{\mathrm{X}}}^2\mathrm{and}{{\mathrm{\sigma }}_{\mathrm{Y}}}^2$of both variables.

a. Can we calculate the mean of the total income $X+Y\mathrm{to}\mathrm{be}{\mu }_X+{\mu }_Y$? Explain your answer.

b. Can we calculate the variance of the total income to be${{\mathrm{\sigma }}_{\mathrm{X}}}^2+{{\mathrm{\sigma }}_{\mathrm{Y}}}^2$ ? Explain your answer.

Short answer

a. Yes, we can calculate the mean of the total income as${\mu }_X+{\mu }_Y$

b. The variance of total revenue cannot be calculated as${{\sigma }^2}_X+{{\sigma }^2}_Y$.

Step-by-step solution

Part(a) Step 1 : Given Information 

Given :

$X$: The income of the husband

$Y$: The income of the wife.

Part(a) Step 2 : Simplification   

The sum of the means of two random variables is the total of their means.

${\mu }_{X+Y}={\mu }_X+{\mu }_Y$

We may calculate the mean of total income $X+Y\mathrm{to}\mathrm{be}{\mu }_X+{\mu }_Y$since the property holds for any two random variables, and because the means for both $X\mathrm{and}Y$are known.

Part(b) Step 1 : Given Information 

Given :

$X$: The income of the husband

$Y$: The income of the wife.

Part(b) Step 2 : Simplification   

The variance of the total is equal to the sum of the variances of the two random variables.

However, this is only conceivable if two random variables are independent; otherwise, this is not true.

${{\sigma }^2}_{X+Y}={{\sigma }^2}_X+{{\sigma }^2}_Y$

People who are intelligent tend to grow even more intelligent.

Furthermore, clever people are more likely to make more money.

As a result, it is reasonable to conclude that the husband's income $X$is unrelated to the wife's $Y$income.

As a result, the variance of total revenue cannot be calculated as${{\sigma }^2}_X+{{\sigma }^2}_Y$