E(X^r) from Weibull Distribution is equal to Gamma Fn

  • Thread starter Thread starter Nick Jarvis
  • Start date Start date
  • Tags Tags
    Distribution Gamma
Click For Summary
SUMMARY

The forum discussion centers on proving that the expected value E(X^r) from the Weibull distribution is equal to Γ(r/B + 1). The user begins by substituting u = x^B and seeks confirmation on whether this approach is valid. The discussion clarifies that if u = x^B, then X^r can be expressed as u^(B/r). Additionally, it emphasizes the importance of correctly distinguishing between u = x^B and u = xB in the calculations. The correct formula to prove is E(X^r) = Γ(r/B + 1), not E(X^r) = Γ(r/(B + 1)).

PREREQUISITES
  • Understanding of the Weibull distribution and its probability density function.
  • Familiarity with the concept of expected value in probability theory.
  • Knowledge of integration techniques, particularly substitution methods.
  • Basic proficiency in LaTeX for formatting mathematical equations.
NEXT STEPS
  • Study the properties of the Weibull distribution and its applications in reliability engineering.
  • Learn about the Gamma function and its significance in probability distributions.
  • Practice integration techniques, focusing on substitution and integration by parts.
  • Explore LaTeX tutorials to improve mathematical expression formatting in discussions.
USEFUL FOR

Students studying probability and statistics, mathematicians working with distributions, and anyone involved in data analysis requiring knowledge of the Weibull distribution and expected values.

Nick Jarvis
Messages
29
Reaction score
2

Homework Statement



I have f(x) = BxB-1e-x^B

I need to show that E(Xr) = Ƭ(Gamma)(R/B + 1)

Homework Equations


I know that E(Xr) = f(x)


The Attempt at a Solution



Attempt at part of solution

I started by saying let u = xB so du = BxB-1dx

Can I say then that if u = xB, then Xr = ur/B

That’s my first question. If I am on the right lines I will integrate. I think I have asked this before, but it got removed as I hadn’t followed the guidelines. Hoping that is set out properly. I am integrating by substitution. Assuming this is possible and I don’t HAVE to integrate by parts.Many thanks
[/B]
 
Physics news on Phys.org
Nick Jarvis said:

Homework Statement



I have f(x) = BxB-1e-x^B

I need to show that E(Xr) = Ƭ(Gamma)(R/B + 1)

Homework Equations


I know that E(Xr) = f(x)


The Attempt at a Solution



Attempt at part of solution

I started by saying let u = xB so du = BxB-1dx

Can I say then that if u = xB, then Xr = ur/B

That’s my first question. If I am on the right lines I will integrate. I think I have asked this before, but it got removed as I hadn’t followed the guidelines. Hoping that is set out properly. I am integrating by substitution. Assuming this is possible and I don’t HAVE to integrate by parts.Many thanks[/B]
(1) Please refrain from putting all of your message in a bold font; it looks like you are yelling at us.
(2) When writing you must distinguish clearly between ##u = xB## and ##u = x^B##.
(3) And, yes, of course, if ##u = x^B## then ##x^r = u^{B/r}## for any ##x > 0##. How could it be otherwise?
(4) I hope you were are being asked to prove that
$$ E X^r = \Gamma \left( \frac{r}{B}+1 \right) \; ?$$
rather than
$$ E X^r = \Gamma \left( \frac{r}{B+1} \right) \: ?$$
because the second one of these is false.
 
Last edited:
  • Like
Likes Nick Jarvis
Thanks. I copied and pasted from Word and assume the bold was inherited from the the 3 titles in the template. Apologies for that.

I cannot work out to insert nice equations like the two that you have inserted above. And yes, I need to prove that:

E(Xr)=Γ((r/B)+1) - your first equation

When I ask 'Can I say then that if u = xB, then Xr = ur/B' I meant is this the correct way of starting to solve it? Or am I on a hiding to nothing?

Many thanks
 
Nick Jarvis said:
Thanks. I copied and pasted from Word and assume the bold was inherited from the the 3 titles in the template. Apologies for that.

I cannot work out to insert nice equations like the two that you have inserted above. And yes, I need to prove that:

E(Xr)=Γ((r/B)+1) - your first equation

When I ask 'Can I say then that if u = xB, then Xr = ur/B' I meant is this the correct way of starting to solve it? Or am I on a hiding to nothing?

Many thanks

You can get rid of the bold font, just by making sure your input occurs after the "[/B]" delimiter. In this forum, "[B ]" turns on bold and "[/B ]" turns it off. (Note: I inserted extra space after the "B" and before the "]" to prevent the processor from actually switching to bold, but there should be no space between them.)

To insert "nice" equations, just use LaTeX; a stripped-down version of it comes loaded into this Forum. For an in-line equation, use # # d = a + b c^2 \int_0^1 x^3 dx # # (with no space between the two #'s at the start and at the end); that produces ##d = a + b c^2 \int_0^1 x^3 \, dx##. For a "dsplayed" equation (on its own, separate line) use two $ signs (with no space between them) at the start and at the end. That gives
$$d = a + b c^2 \int_0^1 x^3 \, dx$$
If you search in this Forum for a "LaTeX tutorial", I am sure you will find one. To see the actual typed commands for a LaTeX expression, just right-click on the expression or equation and ask for a display math as tex.

As to your question: the fall-back position is to always try it yourself, to see what you get. If it works, you are done; if it fails, you need to try something else. But, try it first.
 
Last edited:
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

Replies
3
Views
1K
  • · Replies 4 ·
Replies
4
Views
13K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
4
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
6K
  • · Replies 11 ·
Replies
11
Views
10K