Question

For the following discrete random variable X with probability distribution:

X 0 1 2 3

P(X) 0.2 0.4 0.15 0.25

(a) Find the probability distribution for Y = 3X2 − 2X + 1
(i.e, all values of Y and P(Y )).
(b) Find E[Y ] using part (a).
(c) Find E[X] and E[X2].
(d) Find Pr(Y ≤ 2).
(e) Using part (c), verify your result from part (b) using
E[X] and E[X2].

Answer 1

(a) The probability distribution is shown in the attachment.

(b) The value of E (Y) is 7.85.

(c) The value of E (X) and E (X²) are 1.45 and 3.25 respectively.

(d) The value of P (Y ≤ 2) is 0.60.

(e) Verified that the value of E (Y) is 7.85.

Explanation:

(a)

The random variable Y is defined as:
`Y=3X^(2)-2X+1`

For X = {0, 1, 2, 3} the value of Y are:

`X=0;\ Y=3*(0)^(2)-2*(0)+1 =1`

`X=1;\ Y=3*(1)^(2)-2*(1)+1 =2`

`X=2;\ Y=3*(2)^(2)-2*(2)+1 =9`

`X=3;\ Y=3*(3)^(2)-2*(3)+1 =22`

The probability of Y for different values are as follows:

P (Y = 1) = P (X = 0) = 0.20

P (Y = 2) = P (X = 1) = 0.40

P (Y = 9) = P (X = 2) = 0.15

P (Y = 22) = P (X = 3) = 0.25

The probability distribution of Y is shown below.

(b)

The expected value of a random variable using the probability distribution table is:

`E(U)=\sum[u* P(U=u)]`

Compute the expected value of Y as follows:

`E(Y)=\sum [y* P(Y=y)]\\=(1*0.20)+(2*0.40)+(9*0.15)+(22*0.25)\\=7.85`

Thus, the value of E (Y) is 7.85.

(c)

Compute the expected value of X as follows:

`E(X)=\sum [x* P(X=x)]\\=(0*0.20)+(1*0.40)+(2*0.15)+(3*0.25)\\=1.45`

Compute the expected value of X² as follows:

`E(X^(2))=\sum [x^(2)* P(X=x)]\\=(0^(2)*0.20)+(1^(2)*0.40)+(2^(2)*0.15)+(3^(2)*0.25)\\=3.25`

Thus, the value of E (X) and E (X²) are 1.45 and 3.25 respectively.

(d)

Compute the value of P (Y ≤ 2) as follows:

`P (Y\leq 2)=P(Y=1)+P(Y=2)=0.20+0.40=0.60`

Thus, the value of P (Y ≤ 2) is 0.60.

(e)

The value of E (Y) is 7.85.

`E(Y)=E(3X^(2)-2X+1)=3E(X^(2))-2E(X)+1`

Use the values of E (X) and E (X²) computed in part (c) to compute the value of E (Y).

`E(Y)=3E(X^(2))-2E(X)+1\\=(3* 3.25)-(2*1.45)+1\\=7.85`

Hence verified.