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December 6th, 2016, 06:36 AM  #1 
Senior Member Joined: Feb 2014 Posts: 112 Thanks: 1  Uniform distribution random numbers
hello, For generate Uniform distribution random number we have the below formula. a + (ba) * r and 0<=r<=1 How can proof the above formula? 
December 6th, 2016, 02:42 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,343 Thanks: 534 
You need to give more detail. a,b? Is this to get one number?

December 14th, 2016, 05:24 AM  #3 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,734 Thanks: 707 
You say "prove the above formula" (not "proof"). One "proves" a statement, not a formula. What do you want to prove about it? It is clear that, for any r between 0 and 1, b> a, this produces a number between a and b: 0<= r<= 1 so, multiplying by the positive number, b a, 0<= r(b a)<= b a Then adding a to each part, a<= a+ r(b a)<= b. Now, what more do you want to do? Do you want to prove that "if r is a random number distributed uniformly between 0 and 1 then a+ r(b a) is a random number uniformly distributed between a and b? Obviously that will depend on exactly what "uniformly distributed" means! What definition of "uniformly distributed" are you using? 
December 14th, 2016, 07:30 AM  #4 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,503 Thanks: 758 
I would guess that this question is given $r$ is a uniform random variate on $[0,1]$ show that $Y=a + (ba)r$ is a uniform random variate on $[a,b]$ this is pretty simple assume $b > a$ $F_Y(y) = P[Y < y] = P[a + (ba)r < y] = P[r < \dfrac{ya}{ba}]$ $F_Y(y) = \begin{cases} 0 &\dfrac{ya}{ba}<0 \\ \dfrac{ya}{ba} &0 \leq \dfrac{ya}{ba} \leq 1 \\ 1 &\dfrac{ya}{ba}>1 \end{cases}$ let's rewrite this a bit to make it clearer $F_Y(y) = \begin{cases} 0 &y < a \\ \dfrac{ya}{ba} &a \leq y \leq b \\ 1 &b < y \end{cases}$ to find the PDF of $Y$ we differentiate w/respect to $y$ $f_Y(y) = \begin{cases} 0 &y < a \\ \dfrac{1}{ba} &a \leq y \leq b \\ 0 &b < y \end{cases}$ This is exactly the PDF of a uniform random variate on $[a,b]$ and thus $Y = a + (ba)r \sim U[a,b]$ if $b < a$ then just flip $a$ and $b$ and you end up with $a + (ba)r \sim U[b,a]$ Last edited by romsek; December 14th, 2016 at 07:32 AM. 
December 14th, 2016, 07:34 AM  #5 
Senior Member Joined: Feb 2014 Posts: 112 Thanks: 1 
Excellent and very helpful.


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distribution, numbers, random, uniform 
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