Problem with unknown functions in integrals

In summary: That gives\int (x- \mu)^2f(x)dx= \int x^2f(x)dx- 2\mu^2+ \mu^2= \int x^2f(x)dx- \mu^2.In summary, the conversation discusses the problem of dealing with unknown functions in derivations, specifically in the context of finding the variance of probability density functions (pdfs). It involves an integral that does not hold true and the suggestion of using specific values for μ to make it true. The conversation concludes by stating the general result for finding the variance of a random variable, which involves using the mean and squared mean of the variable.
  • #1
pat804
2
0

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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  • #2
Try f(x) = 1 and the equation does not hold.
 
  • #3
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##.
 
  • #4
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.
 
  • #5
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:
  • #6
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].
 

FAQ: Problem with unknown functions in integrals

What is a problem with unknown functions in integrals?

A problem with unknown functions in integrals occurs when the function being integrated cannot be easily identified or expressed in terms of known mathematical functions.

How do you solve integrals with unknown functions?

Solving integrals with unknown functions often requires advanced mathematical techniques such as substitution, integration by parts, or numerical methods.

Can all integrals with unknown functions be solved?

No, not all integrals with unknown functions can be solved analytically. Some integrals may have no closed-form solution and can only be approximated.

What are some real-world applications of integrals with unknown functions?

Integrals with unknown functions are commonly used in physics, engineering, and economics to model various real-world phenomena such as fluid flow, electrical circuits, and optimization problems.

How do integrals with unknown functions relate to the concept of antiderivatives?

Integrals with unknown functions are essentially the inverse of derivatives, which are used to find the rate of change of a function. The solution to an integral with unknown functions is the antiderivative of the function being integrated.

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