# what will be the procedure to prove the following relationship?

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

• Yes, your procedure is correct and you have nothing else to prove. You are confusing yourself needlessly. Why not write $Y = a+HU^{-\frac{1}{n}}$ and show that $Y$, not $X$, has the desired Pareto distribution? Then you will have shown that if $X$ has the specified Pareto distribution, then $Y = a+HU^{-\frac{1}{n}}$ also has the same specified Pareto distribution. What the question is setting up for further development is a method for simulating a Pareto random variable using a uniform random number generator which typically returns samples of $U$. – Dilip Sarwate May 12 '13 at 14:51
• @DilipSarwate Thank you very much – time May 13 '13 at 13:21

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 $${\rm E}[X^n]={\rm E}[Y^n],\quad \text{for all }n\geq 1.$$
• " 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 $${\rm E}[X^n]={\rm E}[Y^n],\quad \text{for all }n\geq 1.$$ " I have not understood this example.Would you explain me? – time May 13 '13 at 13:28
• It was just a note saying that not everything determines the distribution of a random variable. What the above expresses is that by knowing all moments ${\rm E}[X],{\rm E}[X^2],\ldots$ we do not know the distribution (not in general at least). In contrast, if we know the CDF (or MGF or...) then we do know the distribution of the variable. – Stefan Hansen May 13 '13 at 13:38