Question

X is a discrete random variable with mean 1 and variance 1 . Which of the following generates Y with mean 4 and variance 4 .

(a) Y=-2 X+6
(b) Y=2 X+4
(c) Y= -4X+8
(d) Y= 4X+4

Answer 1

To find a random variable Y with mean 4 and variance 4 from a random variable X with mean 1 and variance 1, one must apply the right linear transformation. The transformation that meets the criteria is Y = -2X + 6, which gives us the correct new mean and variance.

Step-by-step explanation:

To find the new random variable Y with mean 4 and variance 4 from the given variable X (which has mean 1 and variance 1), we need to apply a linear transformation of the form Y = aX + b.

Properties of Linear Transformation

For a linear transformation:

• The new mean is given by μY = aμX + b.
• The new variance is σY2 = a2σX2.

Given that X has mean (= μX) 1 and variance (= σX2) 1, Y must have mean 4 (μY = 4) and variance 4 (σY2 = 4).

Applying these properties to the options:

For option (a), Y = -2X + 6, the mean would be -2 * 1 + 6 = 4 and the variance would be (-2)2 * 1 = 4. So, this transformation gives the correct mean and variance.

1. For option (b), Y = 2X + 4, the mean would be 2 * 1 + 4 = 6, which does not match the required mean of 4.
2. For option (c), Y = -4X + 8, the mean would be -4 * 1 + 8 = 4, but the variance would be (-4)2 * 1 = 16, which is not the required variance of 4.
3. For option (d), Y = 4X + 4, the mean would be 4 * 1 + 4 = 8, which also does not match the required mean.

Thus, the correct option that transforms X into Y with the desired mean and variance is option (a).