Question

Suppose that y is a discrete random variable with mean μ and variance σ^2, and let x = y^2.

Options:
A) μ and σ^2
B) μ^2 and σ^4
C) μ and σ
D) μ^2 and σ^2

Answer 1

The random variable x = y^2 is a continuous random variable with mean μ^2 and variance σ^2. The correct answer is: (B) 2μ 2and 4σ 4

Step-by-step explanation:

To find the mean and variance of the random variable

x=y 2 , we need to consider the transformations of the mean and variance.

Mean of

E(x)=E(y 2 )=Var(y)+[E(y)] 2

Since y has a mean

E(y)=μ, and the mean of

Variance of x:

Var=E4)−[E(2)]2

Var(x)=E(y 4 )−[E(y 2 )] 2

Since Var(y)=σ=Var E(y 2 )=Var(y)+[E(y)]

2 =σ 2 +μ 2 .

Therefore,

[E(2)]2E(y 4 )=Var(y 2)+[E(y 2 )] 2

The term Var(2)Var(y 2 ) is the variance of 2y 2.

Since y is a discrete random variable, the variance of

Substituting these values into the formula:

Var(x)=Var(y) 2+(σ 2+μ 2 ) 2−(σ 2 +μ 2) 2

Simplifying, the variance of

x is 4σ 4 .

Therefore, the correct answer is:

B) 2μ 2and 4σ 4