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A straight forward question on variances

  1. Jun 9, 2010 #1
    1. The problem statement, all variables and given/known data

    suppose you have the following function:

    w=a+b(e+z)

    a, b and e are constants and z is a random variable distributed by some density function g(z).

    What is the variance of w?

    i.e. var(w)

    2. Relevant equations

    Suppose E(z) = 0 (expectation of z is 0) and var(z)=[tex]\sigma^{2}[/tex]

    3. The attempt at a solution

    The solution is var(w)= [tex]b^{2} \sigma^{2}[/tex], but I don't understand why. I appreciate your input.

    M
     
  2. jcsd
  3. Jun 10, 2010 #2
    Two facts. First, the variance of a constant plus a random variable is equal to the variance of the random variable. This makes sense if you think of variance as a measure of spread, since adding a constant to every observation doesn't change the spread. Second, the variance of a constant times a random variable is equal to the constant squared times the variance of the random variable. That is, if x is your random variable, var(ax) = a2var(x). Do you understand how the answer follows?
     
  4. Jun 10, 2010 #3

    Mark44

    Staff: Mentor

    Let's use caps for random variables to help keep them separate from constants.
    From the given information,
    E(Z) = 0 and Var(Z) = [itex]\sigma^2[/itex]

    Also, by definition, Var(Z) = E(Z2) - [itex]\mu^2[/itex].
    Since E(Z) = [itex]\mu[/itex] = 0, then Var(Z) = E(Z2).

    You're also given that W = a + b(Z + e). Using the properties of expectation, it can be seen that E(W) = a + bE(Z) + be = a + be, since E(Z) = 0.

    With all that out of the way, we can tackle Var(W).

    Var(W) = E(W2) - (E(W))2.

    If you replace W with a + b(Z + e) in the first term on the right, and work things through, you get the result you're supposed to get.
     
  5. Jun 10, 2010 #4

    statdad

    User Avatar
    Homework Helper

    If X is a random variable and [itex] Y = c + dX [/itex], then

    [tex]
    Vary(Y) = d^2 Var(X)
    [/tex]
     
  6. Jun 10, 2010 #5
    Thanks guys! I appreciate your help.
     
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