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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

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  • $\begingroup$ 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 $U$. $\endgroup$ – Dilip Sarwate May 12 '13 at 14:51
  • $\begingroup$ @DilipSarwate Thank you very much $\endgroup$ – time May 13 '13 at 13:21
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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 $$ {\rm E}[X^n]={\rm E}[Y^n],\quad \text{for all }n\geq 1. $$

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  • $\begingroup$ " 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? $\endgroup$ – time May 13 '13 at 13:28
  • $\begingroup$ 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. $\endgroup$ – Stefan Hansen May 13 '13 at 13:38
  • $\begingroup$ thanks a lot. Now i have understood your note. thank you again. $\endgroup$ – time May 13 '13 at 13:44

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