Distribution of function of random variable $y = (x_{0}-x)^n$ where $x$ is $N(0,1)$ Need to compute distribution of function of random variable $y = (x_{0}-x)^n$ where $x$ and $x_0$ are both less than -3, $x<x_0$ and $x$ is distributed as $N(0,1)$
I have pages of ink on paper attempts but am overwhelmed (and frankly, a bit out of practice) ...
My first round through gives me that the probability density function for $y$ is
$$f(y)=\frac{1}{n\sqrt{2\pi}}y^{\frac{1}{n}-1}e^{-\frac{1}{2}\left(x_0^2+y^2+2x_0y\right)^\frac{1}{n}}$$
How would I apply the fact that we're out in the left tail
 A: I am not very sure if I am doing right, but let me give a try by using the defintions: first I would like to write $X$ and $Y$ for the $x$ given in the OP, since for me it is more comfortable to use $X$ and $Y$ for a random variables... Now you want to derive the distribution of $Y$. Then it suffices to derive the distribution function of $Y$, namely, you want to derive the explicit form of $P(Y\leq t)$ for $t\in\mathbb{R}$. Due to the defintion given by the OP, that is to calculate $P((x_0-X)^n\leq t)$. Due to the definition, we should notice that $P(X\geq x_0)=0$, therefore $P((x_0-X)^n\leq t)=P((x_0-X)^n\leq t,X<x_0)$. For $t<0$, this is obviously an empty set; for $t>0$, we have $P(Y\leq t)=P(x_0-t^{\frac{1}{n}}\leq X<x_0)=\int_{x_0-t^{\frac{1}{n}}}^{x_0}f(\sigma)d\sigma=\Phi(x_0)-\Phi(x_0-t^{\frac{1}{n}})$, where $f$ is the density of $X$ and $\Phi$ is the distribution function of $X$. Now calculating the derivative of the distribution function you can obtain the density, and in our case, we can apply the chain rule: $\frac{d}{dt}P(Y\leq t)=f(x_0-t^{\frac{1}{n}})t^{\frac{1-n}{n}}$. It seems that this is in fact a very easy derivation, so I am not sure if I am mistaken or misunderstood the meaning of the problem. Please tell me if I made a mistake.
