# Raw moments of 3-parameter Weibull distribution

#### mehdi_m990

##### New Member
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???

#### BGM

##### TS Contributor
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.