# Given pdf of $X$, find a function $U$ that has the same distribution as $X$ where $U\sim Unif (0,1)$

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 $$(0,1)$$.

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 $g$.

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