My Math Forum Uniform distribution random numbers

 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 + (b-a) * r and 0<=r<=1 How can proof the above formula?
 December 6th, 2016, 02:42 PM #2 Global Moderator   Joined: May 2007 Posts: 6,494 Thanks: 578 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: 3,092 Thanks: 845 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: USA Posts: 1,857 Thanks: 962 I would guess that this question is given $r$ is a uniform random variate on $[0,1]$ show that $Y=a + (b-a)r$ is a uniform random variate on $[a,b]$ this is pretty simple assume $b > a$ $F_Y(y) = P[Y < y] = P[a + (b-a)r < y] = P[r < \dfrac{y-a}{b-a}]$ $F_Y(y) = \begin{cases} 0 &\dfrac{y-a}{b-a}<0 \\ \dfrac{y-a}{b-a} &0 \leq \dfrac{y-a}{b-a} \leq 1 \\ 1 &\dfrac{y-a}{b-a}>1 \end{cases}$ let's rewrite this a bit to make it clearer $F_Y(y) = \begin{cases} 0 &y < a \\ \dfrac{y-a}{b-a} &a \leq y \leq b \\ 1 &b < y \end{cases}$ to find the PDF of $Y$ we differentiate w/respect to $y$ $f_Y(y) = \begin{cases} 0 &y < a \\ \dfrac{1}{b-a} &a \leq y \leq b \\ 0 &b < y \end{cases}$ This is exactly the PDF of a uniform random variate on $[a,b]$ and thus $Y = a + (b-a)r \sim U[a,b]$ if $b < a$ then just flip $a$ and $b$ and you end up with $a + (b-a)r \sim U[b,a]$ Thanks from life24 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.

 Tags distribution, numbers, random, uniform

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post mia6 Advanced Statistics 1 February 23rd, 2011 01:11 PM Advanced Statistics 0 February 3rd, 2011 06:24 PM begyu85 Algebra 1 March 28th, 2008 04:31 PM JesusGumbau Algebra 9 December 21st, 2007 04:01 AM Mathematics19 Advanced Statistics 1 April 30th, 2007 05:06 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top