Densitiy function of $X^n$ for even add odd $n$ Please help, Let be $X$ a random variable with density function $f$. In what connection is $f$ with the density function of $X^n$?
The random variable $X$ is, we are told, non-negative.
Let $f$ be the density function of $X$, and let $F$ be the cumulative distribution function of $X$. Let $F_Y$ be the cdf of $Y := X^n$. For $y>0$, we have
$$ F_Y(y)=\def\Pr{\mathrm{Pr}}\Pr(Y≤y)=\Pr(X^n≤y)=\Pr(X≤y^{1/n})=F(y^{1/n}). $$
For the density function $f_Y$ of $y$, differentiate. For $y>0$, by the Chain Rule, we get
$$ f_Y(y)= \frac 1n y^{1/n−1}f(y^{1/n}). $$
But what about odd $n$?
 A: From the comments I assume that the question is about what changes if we drop the "non-negativity" assumption.
For odd $n$, almost nothing: For any $y \in \mathbb R$ we have now (note that $x \mapsto x^n$ is monotone on $\mathbb R$ if $n$ is odd)
$$ F_Y(y) = \def\P{\mathbb P}\P(Y \le y) = \P(X^n \le y) = \P(X \le y^{1/n}) = F(y^{1/n})$$
So 
$$ f_Y(y) = F_Y'(y) = \frac 1n y^{1/n - 1} f(y^{1/n}) $$
If $n$ is even, things change: Although $X$ can be negative, $Y = X^n$ is non-negative. Hence $F_Y(y) = 0$ for $y < 0$. For $y \ge 0$ we have 
\begin{align*}
   F_Y(y) &= \P(X^n \le y)\\
          &= \P(-y^{1/n} \le X \le y^{1/n})\\
          &= \P(X \le y^{1/n}) - \P(X < -y^{1/n}) 
\end{align*}
Now, as $X$ has a density function $\P(X = -y^{1/n})$ is zero, hence we can continue
\begin{align*}
  F_Y(y) &= \P(X \le y^{1/n}) - \P(X \le -y^{1/n})\\
         &= F(y^{1/n}) - F(-y^{1/n})
\end{align*}
Differentiating gives that for $y \ge 0$
$$ f_Y(y) = \frac 1n y^{1/n - 1} \bigl(f(y^{1/n}) + f(-y^{1/n})\bigr). $$
