Consider a random variable $X$ having pdf

$$f_X(x) = \begin{cases} {3\over{x^4}} & \text{$x \gt 1$} \\ {0} & \text{$x\leq 1$}\end{cases}$$

and consider $U\sim Unif (0,1)$. Give a function of $U$ which has the same distribution as $X$.

I guess I'm just confused on what the question is asking. I suppose it wants $f_X(x) = f_u(u)$?

A similar problem I found is this. From this, I am trying to make use of the following theorem: Let $X$ be a continuous random variable with monotonic increasing $cdf$ $F_X(x)$. Let $Y=F_X(x)$. Then $Y$ is uniformly distributed on $[0,1]$.

Since $f_X(x)$ $=$ $3\over{x^4}$ then $F_X(x)$ $=$ $-1\over{x^3}$ which monotonically increases when $x\gt 1$. Then inverting $F_X(x)$ I get that $x=\sqrt[3]{-1\over{y}}$. But the graph of this is not uniform on $(0,1)$.


1 Answer 1


You misunderstood the question. What is being asked is given a distribution $U$, try to find a function $g$ such that $g(U)$ follows the same distribution as $X$.

You are heading towards the right direction though with a careless mistake.

$$f_X(x) = \frac{3}{x^4}$$

if $x>1$, $$F_X(x) = \int_1^xf_X(t) \, dt=-\frac1{t^3}\vert_{t=1}^{t=x}=\color{blue}{1-}\frac1{x^3}$$

Try to invert the function to find $g$.

  • $\begingroup$ I got now that $x$ $=$ $\sqrt[3]{-1\over{y-1}}$ $\endgroup$
    – Remy
    Commented Oct 29, 2017 at 20:11
  • $\begingroup$ However, is this uniform on $(0,1)$? The graph doesn't look linear on this interval. Is that a requirement for the $cdf$? $\endgroup$
    – Remy
    Commented Oct 29, 2017 at 20:12
  • $\begingroup$ $x=\sqrt[3]{\frac{-1}{y-1}}= \sqrt[3]{\frac{1}{1-y}}$, try to verify the following, if $y \sim \operatorname{Uni}(0,1)$ then $x \sim X$. $\endgroup$ Commented Oct 29, 2017 at 20:16
  • $\begingroup$ Oh okay, I think I understand the question now. $\endgroup$
    – Remy
    Commented Oct 29, 2017 at 20:17

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