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If $X$ is uniform on $(0,1)$ how can I prove $2X$ is uniform on $(0,2)$?

I am struggling with this question and similar but harder variants.

It makes a lot of sense to me intuitively but I am unfamiliar with rigorous arguments that are needed to prove it.

I honestly don't know where to begin; do I need to look at the density function of $2X$ and show it is $1/2$ on $(0,2)$ and $0$ elsewhere or perhaps I need to look at the distribution formula maybe?

Could anyone help me with this example but also try to motivate the steps so I can appreciate the technique and hopefully be able to see how I can adapt to harder but similar problems.

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2 Answers 2

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What you say is a way, and here I give another point, but both are similar.

The distribution function $F_{2X}(x)=P(2X\le x)=P(X\le \frac{1}{2}x)=F_X(\frac{1}{2}x)$ and we can start from here.

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  • $\begingroup$ So does it just suffice to say that $F_{2X}(x)=F_X(x/2)$ so then just plug in different values of $x$ to show the distribution is as it should be if it was uniform $(0,2)$ for instance $F_{2X}(t)=F_X(t/2)=t/2$ for $t \in [0,2]$. I think I understand now but could you explicitly show me what I need to write down to justify it completely? Thanks! $\endgroup$
    – Ben B
    Apr 15, 2017 at 10:56
  • $\begingroup$ Yes, you can prove it by showing they have the identical distribution function function or density function. $\endgroup$
    – User
    Apr 15, 2017 at 10:58
  • $\begingroup$ Could you complete your solution please so I can compare with what I have. (This isn't a homework task or anything just for my learning). $\endgroup$
    – Ben B
    Apr 15, 2017 at 10:59
  • $\begingroup$ the rest is follow the definition, plugging in the value of $F_X(x)$ for different $x$ ~ $\endgroup$
    – User
    Apr 15, 2017 at 11:00
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You are right, there is a theorem stating that If $U\sim U(0,1)$ then $V = (b — a)U \sim U(a,b)$ where $a < b$.

The proof is straightforward with the pdf: $f(x) = \frac{1}{b-a}$ if $a<x<b$ and $0$ otherwise.

Thus, if we multiply the inequality by $2$ we have $2a<2x<2b$ and setting $a' =a, b'=2b$ and thus $f(2x) = \frac{1}{b'-a'}$ if $a'<2x<b'$ for $a=0$, $b=1$ and $x=u$ you get your answer as $b'=2$,$a'=0$ and $v=2u$

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