Can I do expectation like this

  • Thread starter Thread starter aegea
  • Start date Start date
  • Tags Tags
    Expectation
Click For Summary
SUMMARY

The discussion focuses on calculating the expected value of a transformed normal variable, specifically E(y) where y = 160*x^2 and x follows a normal distribution x ~ N(a, b). The correct approach is to utilize the linearity of expectation, leading to E(y) = 160*E(X^2). Additionally, the variance formula V(X) = E(X^2) - [E(X)]^2 is highlighted as a crucial concept in this calculation. The participants confirm that substituting the density function of x directly into the integral is valid but may complicate the process unnecessarily.

PREREQUISITES
  • Understanding of normal distribution, specifically x ~ N(a, b)
  • Knowledge of expected value and its properties
  • Familiarity with variance calculations, including V(X) = E(X^2) - [E(X)]^2
  • Basic integration techniques for probability density functions
NEXT STEPS
  • Study the properties of linear operators in expectation calculations
  • Learn about transformations of random variables and their expected values
  • Explore the derivation and application of the variance formula in statistics
  • Investigate advanced integration techniques for probability distributions
USEFUL FOR

Statisticians, data analysts, and students in probability theory who are working with normal distributions and expected value calculations.

aegea
Messages
2
Reaction score
0
Many thanks in advance

Suppose x is normal variable x~N(a,b)
and y=160*x^2
I need calculate E(y)=∫yf(y)d(y)
f(y) is the density function of y
how can I write it as an integral of x since we know x's distribution, I mean use the density function of x to substitute the original integral

Can I just use E(y)=E(160 x^2)= ∫ 160 x^2 f(x) d(x)
then I can use f(x), which is the density function of normal.

Seems right, but seems wrong, seems I use f(x) to substitute x directly!
 
Physics news on Phys.org
You can do it the way you suggested (it is logically correct), but I believe it will make more work for yourself (but I didn't look at it because there is a cleaner way).

Here are a few hints that should help:

Remember that the expected value function is a linear operator so

[tex]E(Y)=E(160X^2)=160E(X^2)[/tex]

and

[tex]V(X)=E(X^2)-[E(X)]^2[/tex]
 
Last edited:

Similar threads

Replies
5
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
26
Views
4K
Replies
8
Views
2K
  • · Replies 18 ·
Replies
18
Views
3K
Replies
4
Views
2K