Question

let x and y be random variables find the mean and variance of z = 3x - 2y​

Answer 1

Tis is your answer

..........

Step-by-step explanation:

1

Answer 2

Complete Question: Let x and y be random variables with E[x] = 1, E[y] = 4, Var(x) = 4, Var(y) = 6 and cov(x, y) = 1/2. Find the mean and variance of $z = 3x - 2y$​.

Answer:

The mean is -5.

The variance is 54.

Step-by-step explanation:

Concept:-

• The expected value of a discreate random variable is denoted by E[x].
• E[x] is also called the mean of the probability distribution.

Given:-

Mean of x, E[x] = 1,

Mean of y, E[y] = 4,

Var(x) = 4,

Var(y) = 6 and cov(x, y) = 1/2.

Step 1 of 2

Find the mean of $z = 3x - 2y$.

Consider the given function as follows:

z = 3x - 2y

Then,

E[z] = E[3x - 2y]

E[z] = E[3x] - E[2y]

E[z] = 3E[x] - 2E[y]

Now, substitute the values of E[x] and E[y], we get

E[z] = $3 \times 1 - 2 \times 4$

      = $3 - 8$

      = - 5

Therefore, the mean is -5.

Step 2 of 2

Find the variance of $z = 3x - 2y$.

Consider the given function as follows:

z = 3x - 2y

Then,

Var(z) = Var(3x - 2y)

Var(z) = $3^2$Var(x) + $(-2)^2$Var(y) + 2×(coeff. of x)×(coeff. of y)×cov(x, y)

Now, substitute the given values, we get

Var(z) = $3^2$ × 4 + $(-2)^2$ × 6 + 2(3)(-2)$\frac{1}{2}$

          = $36 + 24 - 6$

          = 54

Therefore, the variance is 54.

#SPJ2