Raw moments of 3-parameter Weibull distribution

#1
Hi,
I know that the formula for computing the raw moments of 2-parameter Weibull distribution is:
Mu'n=b^n*Gamma(1+n/c), where b and c are scale and shape parameters, respectively.

However, I couldn't find any exact formula for a 3-parameter Weibull distribution. Is there any simple formula for it???

Thank you in advance
 

BGM

TS Contributor
#2
First I need to make sure that the 3rd parameter you mentioned is the location parameter.

https://en.wikipedia.org/wiki/Weibull_distribution#Related_distributions

If that's the case, then we have the following:

Let \( X \sim \text{Weibull}(k, \lambda, \theta) \) and \( Y \sim \text{Weibull}(k, \lambda, 0) \) be two 3-parameters Weibull random variable.

Then
\( Y \) is a 2-parameter Weibull random variable; and

\( X \stackrel {d} {=} Y + \theta \)

As you know raw moments

\( m_n \triangleq E[Y^n] = \lambda^n \Gamma\left(1 + \frac {n} {k}\right) \)

Therefore,

\( E[X^n] = E[(Y + \theta)^n] \)

\( = E\left[\sum_{r=0}^n \binom {n} {r} Y^r \theta^{n-r}\right] \)

\( = \sum_{r=0}^n \binom {n} {r} m_r \theta^{n-r} \)

\( = \sum_{r=0}^n \binom {n} {r} \lambda^r
\Gamma\left(1 + \frac {r} {k}\right) \theta^{n-r} \)

This is not simple or very nice, but at least this is exact and can be computed.