A straight forward question on variances

  • Thread starter Thread starter michonamona
  • Start date Start date
Click For Summary

Homework Help Overview

The discussion revolves around the variance of a function defined as w = a + b(e + z), where a, b, and e are constants, and z is a random variable with a specified density function. Participants are exploring the implications of variance in the context of this function, particularly under the assumption that the expectation of z is zero and its variance is σ².

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the properties of variance concerning constants and random variables, questioning how these properties lead to the conclusion that var(w) = b²σ². There is an exploration of the definitions and calculations related to variance and expectation.

Discussion Status

Some participants have provided insights into the properties of variance, suggesting that the variance of a constant plus a random variable remains unchanged, and that the variance of a constant multiplied by a random variable involves squaring the constant. The original poster expresses confusion about the reasoning behind the solution, indicating an ongoing exploration of the topic.

Contextual Notes

There is a repeated emphasis on the assumptions regarding the expectation and variance of the random variable z, which are central to the discussion. The original poster's request for clarification suggests that they are navigating through the mathematical properties rather than seeking a direct answer.

michonamona
Messages
120
Reaction score
0

Homework Statement



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)

Homework Equations



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

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
 
Physics news on Phys.org
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?
 
michonamona said:

Homework Statement



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)

Homework Equations



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

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

[tex] Vary(Y) = d^2 Var(X)[/tex]
 
Thanks guys! I appreciate your help.
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
3K
Replies
0
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
6K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 42 ·
2
Replies
42
Views
6K
Replies
9
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K