Knowing that the density of a uniform random variable on $[0,1]$ is:

$f_{U}=\left\{\begin{matrix} 1 & x\in [0,1]\\ 0 & x\notin[0,1] \end{matrix}\right.$

How to determine the density of a uniformly distributed random variable on $[a,b]$ using change of variable?

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    $\begingroup$ Hint: If $U$ is a random variable on $[0, 1]$, then what distribution does $(b-a)U+a$ have? $\endgroup$ – angryavian Feb 18 at 4:07
  • $\begingroup$ I know how to do this by what you said, but does is use the change of variable? Uniform dist is pretty simple so you know the pdf is 1/(b-a) and make a random variable Y=(b-a)X+a and get this pdf. But what if you don't know the pdf? I know this problem itself is somehow weird $\endgroup$ – suntoto Feb 18 at 5:18
  • $\begingroup$ I understood. When constructing Y=(b-a)X+a, we consider it as a scaling and shifting of X, so Y is still a uniformly distributed RV and its domian is therefore [a,b], right? $\endgroup$ – suntoto Feb 18 at 5:24
  • $\begingroup$ You know the density of $U$, so you can use change of variables on $f_U$ to get the density of $(b-a)U +a$. $\endgroup$ – angryavian Feb 18 at 5:31

Denote the described variable by x. Construct new variable y as $y = (b-a)x+a$

For any value z, The cumulative density function $F(y \leq z) = F(x \leq \frac{z-a}{b-a}) = \begin{cases} 0, \, z< a\\ (z-a)/(b-a), \, z \in [a,b]\\ 1, \, z > b \end{cases} $

Take the derivative of cdf with respect to z, the pdf $f(y=z) = \begin{cases} 1/(b-a), z \in [a,b]\\ 0, otherwise \end{cases}$


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