Answer to Question #214793 in Statistics and Probability for Ashish Kumar Swain

Question

Let X and Y be independent random variable with means 3 ,5 and variances 16,25 respectively. Find the mean and standard deviations of 5X-3Y?

Answer 1

Solution:

$\mu_X=3,\mu_Y=5 \\\sigma_X^2=16, \sigma_Y^2=25$

$\Rightarrow \sigma_X=4, \sigma_Y=5$

For $5X-3Y$:

Mean$=E[5X-3Y]=5E[X]-3E[Y]=5(3)-3(5)=0$

Variance$=Var[5X-3Y]=25Var[X]+9Var[Y]-2(5)(3)Cov(X,Y)$

$Cov(X,Y)=0$ for independent random variable.

Thus, $Var[5X-3Y]=25(4)+9(5)-0=100-45=55$

Then, standard deviation$=\sqrt{55}$