Integration by substitution: Expectation and Variance of Weibull distribution

Question

I am calculating expectation and variance of Weibull distribution. PDF is $$f_Y(y) = \frac{\gamma}{\beta} e^{- y^\gamma / \beta} y^{ \gamma -1}$$

I firstly tried to calculate ${\rm E}Y^n$ using the integration by substitution: $z = y^\gamma / \beta$.

$$ {\rm E}Y^n = \frac{\gamma}{\beta} \int^{\infty}_{0} y^{\gamma + n -1 } e^{- y^\gamma /\beta} dy \ \ \ \ \ \ \ \ \ (1) $$

My book gives a hint: $$ {\rm E}Y^n = \beta^{n/\gamma} \int^{\infty}_{0} z^{n/\gamma} e^{-z} dz = \beta^{n/\gamma} \Gamma \left(\frac{n}{\gamma} + 1 \right) \ \ \ \ \ \ (2) $$

I understand the last equation comes from the definition of Gamma function, but how can I apply integration by substitution to (1) to get (the first equation in) (2)?

Answer 1

If $z=y^\gamma/\beta$, then $dz=\gamma y^{\gamma-1}dy/\beta$ and $y=(\beta z)^{1/\gamma}$, which implies that $y^{\gamma+n-1}dy$ in (1) can be written as $$\underbrace{y^{\gamma+n-1}y^{1-\gamma}}_{(\beta z)^{\frac{n}{\gamma}}}\underbrace{y^{\gamma-1}dy}_{(\beta/\gamma)dz}$$. So $$ {\rm E}Y^n = \frac{\gamma}{\beta} \int^{\infty}_{0} y^{\gamma + n -1 } e^{- y^\gamma /\beta} dy=\frac{\gamma}{\beta}\frac{\beta}{\gamma}\beta^{n/\gamma}\int_0^\infty dz\ e^{-z}z^{n/\gamma}\ . $$