Solving for form of CDF that satisfies $G^t(x) = G(t^{-\theta} x)$ For $\theta >0$, I want to solve for the CDF, $G$, that satisfies:
$$
G^t(x) = G(t^{-\theta} x)
$$
The solution given in my notes and states that it's easy to check that $G(x) = \exp(-x^{-1/\theta})$. While it is easy to check, I'm just not sure how I would go about to find this if i wasn't given the hint?
 A: Given the argument $t^{-\theta}x$, one might try the substitution $x\mapsto y^{-\theta}$, giving
$$G^t(y^{-\theta}) = G((ty)^{-\theta}).$$
Suppose we define a new function
$$H(x) := G(x^{-\theta}).$$
Then our equation becomes simpler: $H^t(y) = H(ty)$.
Now, what function has the property whereby raising it to a power $t$ is equivalent to multiplying the argument by $t$? An exponent has this property because $(a^b)^c = a^{bc}$ for $a,b,c\in\mathbb{R},\;a\geq 0$. So $H(x) = \alpha^x$ for some $\alpha\gt 0$.
Working backwards now, we see $G(x) = H(x^{-1/\theta}) = \alpha^{x^{-1/\theta}}.$
The function $x^{-1/\theta}$ is a decreasing one so $\alpha^{x^{-1/\theta}}$ is increasing (which we require for a CDF) if $0\lt\alpha\lt 1$.
We could say instead that $1\lt \alpha$ and $G(x) = \alpha^{-x^{-1/\theta}}.$
When $x\rightarrow\infty$ we have $G(x)\rightarrow 1$ and when $x\rightarrow 0$ we have $G(x)\rightarrow 0$, which is compatible with its being a CDF.
The choice of $\alpha=e$ is fine, and best it could be said, but any $\alpha\gt 1$ is valid.
