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

You are being asked to prove something, so you need to start with what you are given, do some manipulations in an orderly fashion, and derive the result you are asked to prove. The fact that what you are doing is called "substitution" should not be the first thing you thing of; instead, you should have a method, a plan, an approach, to attack the problem. (Otherwise, you are just floundering, and that is not productive.) I think you should probably start with the definition of E(X^r) (the expected value of X^r, not the expected value of X-to-the-power-of-r).
  • #1
Nick Jarvis
29
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]
 
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  • #2
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.
 
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  • Like
Likes Nick Jarvis
  • #3
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
 
  • #4
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.
 
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FAQ: E(X^r) from Weibull Distribution is equal to Gamma Fn

1. What is the Weibull distribution?

The Weibull distribution is a probability distribution commonly used to model the time it takes for a certain event to occur. It is commonly used in reliability and survival analysis, as well as in engineering, finance, and other fields.

2. What is E(X^r) from Weibull Distribution?

E(X^r) from Weibull Distribution refers to the expected value of the random variable X raised to the power of r, which follows a Weibull distribution. It is a measure of central tendency for the Weibull distribution and can be used to calculate various statistical properties.

3. What is the Gamma Function?

The Gamma Function is a mathematical function used to extend the factorial function to non-integer values. It is denoted by Γ(x) and is defined for all positive real numbers. It is commonly used in probability and statistics to calculate integrals and as a part of various probability distributions.

4. How is E(X^r) from Weibull Distribution related to the Gamma Function?

The relationship between E(X^r) from Weibull Distribution and the Gamma Function is that E(X^r) can be expressed in terms of the Gamma Function. Specifically, the expected value of X^r from a Weibull distribution is equal to the Gamma Function of (r+1) divided by the Gamma Function of (r).

5. Why is E(X^r) from Weibull Distribution equal to the Gamma Function?

This is because the Weibull distribution is a special case of the Gamma distribution, and the expected value of X^r from a Gamma distribution is equal to the Gamma Function of (r+1) divided by the Gamma Function of (r). Therefore, for a Weibull distribution, which is a special case of the Gamma distribution, the expected value is also equal to the Gamma Function.

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