Suppose that X has mean 43 and variance 16. Set Y = 3X + 17, where a > 0. Find E

Question

Suppose that X has mean 43 and variance 16. Set Y = 3X + 17, where a > 0. Find E Y and
Var Y .

Problem:

Please provide all work so I understand and also please complete entire problem.

Note: One way to derive an estimator for a parameter 0 is the method of moments. It has the following steps 1. Making use of the population distribution, find the formula for the population mean (1) EX This formula will generally involve 0. 2. Make two substitutions in (1): substitute X n for EX and substitute for 0 3. Solve for 8mom. This is the method of moments estimator for 0 Example: suppose that the sample X1, A2, Xn is from the uniform distribution on 0,0 1. If X follows that distribution, then (2) EX 2. Making the substitutions, we get: 0mom (3) 3. Solving for mom in (3), we get (4) 0mom which is the method of moments estimator for 0. This example was done slightly differently in class on Thursday. Other example(s) of the method of moments will be discussed in class next Tuesday.

Explanation / Answer

As we are given the mean and the variance of X, we get:

E(X) = 43, and Var(X) = 16

Now we are given that: Y = 3X + 17

From this we get:

E(Y) = E( 3X + 17) = 3E(X) + 17 = 3*43 + 17 = 129 + 17 = 146

Therefore the expected value of Y is 146

Note that the linear terms inside the expectation operator could be separated into separate expectations.

Now for variance note that when we take out the constant out of the variance operator, it is squared. Also the variance of a constant is 0. Therefore we get:

Var(Y) = Var( 3X + 17) = 32Var(X) + Var(17) = 9Var(X) = 9*16 = 144

Therefore the variance of the random variable Y is 144

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.