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Suppose $X$ is a random variable with Pareto distribution. There pdf and cdf are:

$$f_X(x) = \begin{cases} {\alpha x_m^\alpha \over x^{\alpha +1}}, & \text{if $x\ge x_m$ } \\ 0, & \text{if $x\lt x_m$} \\ \end{cases}$$

$$F_X(x) = \begin{cases} 1-\left({x_m \over x}\right)^\alpha , & \text{if $x\ge x_m$ } \\ 0, & \text{if $x\lt x_m$} \\ \end{cases}$$

with $x_m\gt 0$ and $\alpha \gt 0$

Suppose $y=kX$, where $k$ is a non-negative constant. What are the pdf and cdf of $y$? Are there any restrictions on the value of $k$?

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If $k=0$ then $Y$ is identically $0$. If $k\gt 0$, then for suitable $y$ we have $$F_Y(y)=\Pr(Y\le y)=\Pr(kX\le y)=\Pr(X\le y/k)=1-\left(\frac{kx_m}{y}\right)^{\alpha}.$$ The suitable $y$ are where $\frac{y}{k}\ge x_m$. Elsewhere we have $F_Y(y)=0$.

The density function $f_Y(y)$ can now be obtained by differentiating.

Alternately, look up the formula for $f_Y(y)$ given by the method of transformations, and integrate to find $F_Y(y)$.

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  • $\begingroup$ I want to express $F_Y(y)$ in terms of $k$ and $x$. I get back the cdf F_X(x) when I do that. Why? @AndréNicolas $\endgroup$
    – user167965
    Aug 9, 2014 at 15:32
  • $\begingroup$ If by $x$ you mean $y/k$, then the reason is written in the displayed line of the answer. For the computation we used the fact that $F_Y(y)=F_X(y/k)$. (The displayed line takes a couple of steps to show this.) But it is $F_Y(y)$ that we want, or $F_Y(t)$. To use the variable $x$ risks confusion. $\endgroup$ Aug 9, 2014 at 15:43
  • $\begingroup$ thanks a lot @AndréNicolas $\endgroup$
    – user167965
    Aug 9, 2014 at 16:39
  • $\begingroup$ You are welcome. This is not the only time you will need to find the distribution of a function $g(X)$ of a random variable. In most such questions, the algebra will be more complicated. $\endgroup$ Aug 9, 2014 at 16:44

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