Problem with unknown functions in integrals

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Homework Help Overview

The discussion revolves around the integration of an expression involving an unknown probability density function (pdf) f(x) and its relationship to variance. The original poster is attempting to show the equality involving the integral of (x - μ)² f(x) and its components, but is struggling with the derivation and application of integration techniques.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the original poster's attempts to manipulate the integral using integration by parts and question the validity of the expression. Some suggest testing specific functions for f(x) to explore the equality further.

Discussion Status

There is ongoing exploration of the assumptions regarding the pdf and the limits of integration. Some participants have provided clarifications on the mean μ and the conditions under which the original statement may hold true, indicating a productive direction in the discussion.

Contextual Notes

Participants note the necessity of specifying limits for the integrals involved, as the behavior of the pdf can vary significantly based on its domain. The original poster's assumptions about the mean and the nature of the pdf are also under scrutiny.

pat804
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Homework Statement


I often have a problem dealing with unknown functions in derivations. Recently I was looking at variance of pdf's and tried to do the integral below with no success. Could someone suggest a method, or point out where I am going wrong.


Homework Equations


Show ∫(x - μ)2 f(x) dx = ∫x2 f(x) dx - μ2

The Attempt at a Solution


∫x2 f(x) dx -2μ∫x f(x) dx + μ2 ∫f(x) dx

then i try to solve by parts,
x2F(x) - 2∫xF(x)dx - 2μ(xF(x) - ∫F(x)dx) + μ2F(x)
where F(x) = ∫f(x)dx

this doesn't really get me anywhere, does anyone have any suggestions??
 
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Try f(x) = 1 and the equation does not hold.
 
pat804 said:

Homework Statement


I often have a problem dealing with unknown functions in derivations. Recently I was looking at variance of pdf's and tried to do the integral below with no success. Could someone suggest a method, or point out where I am going wrong.


Homework Equations


Show ∫(x - μ)2 f(x) dx = ∫x2 f(x) dx - μ2

The Attempt at a Solution


∫x2 f(x) dx -2μ∫x f(x) dx + μ2 ∫f(x) dx

then i try to solve by parts,
x2F(x) - 2∫xF(x)dx - 2μ(xF(x) - ∫F(x)dx) + μ2F(x)
where F(x) = ∫f(x)dx

this doesn't really get me anywhere, does anyone have any suggestions??

The result is fasle as written, but it can be made true if written properly (and is then an elementary property developed in Probability 101). I assume that μ is the mean of the distribution f(x)---is that right?

You need to specify limits on all the integrations. If f(x) is the pdf of a random variable on the whole line, the limits are -∞ to +∞. If the random variable is non-negative the limits are 0 and +∞, etc. In any case, for a random variable on ##(a,b)## just substitute in the known values of ##\int_a^b f(x) \, dx## and ##\int_a^b x f(x) \, dx##.
 
Ray Vickson said:
The result is fasle as written, but it can be made true if written properly (and is then an elementary property developed in Probability 101). I assume that μ is the mean of the distribution f(x)---is that right?

You need to specify limits on all the integrations. If f(x) is the pdf of a random variable on the whole line, the limits are -∞ to +∞. If the random variable is non-negative the limits are 0 and +∞, etc. In any case, for a random variable on ##(a,b)## just substitute in the known values of ##\int_a^b f(x) \, dx## and ##\int_a^b x f(x) \, dx##.

thanks, yes you're correct μ is the mean. So for definite integrals you just sub in μ=∫xf(x)dx and ∫f(x)=1.
 
pat804 said:
thanks, yes you're correct μ is the mean. So for definite integrals you just sub in μ=∫xf(x)dx and ∫f(x)=1.

Yes, exactly!

For the record: the result is true in general for any random variable X having finite variance---whether X is continuous, discrete or mixed. The discrete case involves sums instead of integrals, and the mixed case involves both (or Stieltjes' integrals instead of Riemann integrals). The general result is that
[tex]\text{Var} X \equiv E(X - EX)^2 = E(X^2) - (EX)^2[/tex]
 
Last edited:
Note that
[tex]\int (x- \mu)^2f(x)dx= \int (x^2- 2\mu x+ \mu^2)f(x) dx= \int x^2f(x)dx- 2\mu\int xf(x)dx+ \mu^2\int f(x)dx[/tex]

Now use the fact that, as you say, [itex]\int f(x)dx= 1[/itex] and [itex]\int xf(x)dx= \mu[/itex].
 

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