what will be the procedure to prove the following relationship?

Question

Let $U$ follows standard uniform distribution , that is, $U\sim U(0,1)$ and $X$ follows Pareto distribution, that is, $X\sim Pa{(\alpha,a,h)}$

where , $a=$location parameter ; $-∞<a<∞$

$h=$scale parameter ; $h>0$

$\alpha=$shape parameter ; $\alpha>0$

then How can i prove the relationship that $X$ and $a+hU^{-\frac{1}{\alpha}}$ have same distribution

My procedure was :

i derived the pdf of $X$ when $X=a+hU^{-\frac{1}{\alpha}}$ then i found that the derived distribution of $X$ is nothing but a Pareto distribution $Pa(\alpha,a,h)$

so, i concluded that if $X$ follows Pareto distribution $Pa(\alpha,a,h)$ & $U$ follows standard Uniform distribution $U(0,1)$ then

$$ "X\quad and\quad a+hU^{-\frac{1}{\alpha}}\quad have\quad same\quad distribution\quad". $$

is that my procedure correct?

i am confused because i have been asked for prove the relationship that $$ "X\quad and\quad a+hU^{-\frac{1}{\alpha}}\quad have\quad same\quad distribution\quad". $$ not to derive the pdf of $X$ when $X=a+hU^{-\frac{1}{\alpha}}$

please tell me the procedure to prove the relationship that $$ "X\quad and\quad a+hU^{-\frac{1}{\alpha}}\quad have\quad same\quad distribution\quad". $$

i have thought of another process using moment generating function technique(mgf) but i couldn't compute the mgf of pareto distribution

if i generalized my problem "i actually want to know that"

How can i prove a relationship between two different distributions that they follow the same distribution after some transformation

Answer 1

If you want to show that two random variables follow the same distribution, there's not just one procedure to do that. This is because there are several quantities that uniquely determine the distribution of a random variable.

For instance, if $X$ is a random variable, then the cumulative distribution function $$F(x)=P(X\leq x),\quad x\in\mathbb{R},$$ determines the distribution of $X$ in the sense that if $Y$ is another random variable with cumulative distribution function given by $F$, then $X\sim Y$. Other quantities that determines the distribution uniquely are the moment-generating function, the characteristic function, the Laplace transform, the density function (if $X$ is absolutely continuous) and the probability mass function (if $X$ is discrete).

So in general if you want to show that $X\sim Y$ for two random variables $X$ and $Y$ you may freely choose any of the above quantities. For example we can choose to show that $\varphi_X(t)=\varphi_Y(t)$ for all $t\in\mathbb{R}$, where $\varphi$ is the characteristic function. Then we may conclude that $X\sim Y$.

Note that not all such quantities determines the distribution uniquely. For example, the sequence of moments $\{{\rm E}[X^n]\mid n\geq 1\}$ does not completely determine the distribution in the sense that one can find two random variables $X$ and $Y$ such that $X\not\sim Y$ but $$ {\rm E}[X^n]={\rm E}[Y^n],\quad \text{for all }n\geq 1. $$