# Can you do change of variables with quantiles? i.e. from quantile for the CDF $F(x)$ , get quantile for the CDF $F(x)^{\kappa}$?

I have a random variable $Z$ with some CDF $F(z)$ and quantile $Q(p) \equiv F^{-1}(p)$. I have $Q(p)$ in closed-form but not $F(z)$.

Create a new random variable $\hat{Z}$ defined so that the CDF $\hat{F}(z) \equiv F(z)^{\kappa}$ for some $\kappa > 0$. (it sounds like this is not necessarily the CDF of $Z^{\kappa}$, but that doesn't bother me.

Question: Can I get a quantile function for the random variable with CDF $\hat{Z}$, call it $\hat{Q}(p)$, from some transformation of the original quantile function $Q(p)$ I have in closed form?

• Does the closely related thread at CDF raised to a power help? And what specifically is your question? The function $F(z)^\kappa$ is not, in general, the CDF of a power of $Z$, so which construction is the one that really interests you?
– whuber
Jun 20, 2014 at 22:03
• Thank you, that helps me refine my question. I am defining the random variable by the cdf taken the power, and will edit the question. My question is: given a quantile for $F$, can I get a quantile for $\hat{F} = F^{\kappa}$ Jun 20, 2014 at 22:07
• The related thread is very helpful, but doesn't tackle the quantile question. Hopefully my edit clarified things. Jun 20, 2014 at 22:10

The quantile function $Q$ by definition solves the equation $F(x)=q$ for $x$. The quantile function for the new variable therefore solves $F^\kappa(x)=q$. Because this is equivalent to $F(x)=q^{1/\kappa}$, $$\hat{Q}(q)=Q(q^{1/\kappa}).$$