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{\large {\bf Adding Uniform Densities
}} 

    

March 17, 2019

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   Eric   Rasmusen  

{\it Abstract}

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\begin{small}
   A write-up of how to add uniform random variables. 

\noindent
Rasmusen: Dan R. and Catherine M. Dalton Professor, Department of Business Economics and Public Policy, Kelley School of Business, Indiana University.   1309 E. 10th Street, Bloomington, Indiana, 47405-1701. (812) 855-9219.  
\href{mailto:erasmuse@indiana.edu}{ erasmuse@indiana.edu}, \url{http://www.rasmusen.org}.  

{\small
  \noindent This paper:
\url{http://www.rasmusen.org/papers/adding-uniform-densities.pdf}. }
 
  
 \end{small}
 
  This is a rewrite of the excellent but poorly typeset
 ``Example:
Sum of Two Uniform Densities,''
 \url{ https://www.chem.purdue.edu/courses/chm621/text/stat/funcs/twovar/sum2uniform.htm}.   I've left out the first example, the standard one of adding two [0,1] random variables that you can find in lots of places. 
 
 Suppose you want to find the density of the sum $z = x+y$ of two independent uniform random variables with different supports, where
  $f(x) = 1$ on [0,1] and  and $g(y) =  .5$  on [0,2]. 
  
  First, note that $z$ will have the range [0,3], since its smallest value is where $x=y=0$ and its biggest is where $x=1$ and $y=2$. 
  
  Second, note that probably the density is going to look different over  [0,1], [1,2] and [2,3]. Over [0,1], both $x$ and $y$ have positive densities, though $x$ is going to be most important. Over [1,2], $y$ is going to get more important, though of course $x$ still adds on to the effect of $y$. Over [2,3], the importance of $x$ is dwindling.  So we'll look for what the density will be on each of these three intervals. Maybe two will turn out the same-- no harm done, then. 
  
  Note also that the basic idea of a {\bf convolution} and finding the density of the sum is that
  $$
  h(z) = \int f(x) g(z-x)dx. 
  $$
   That's because we can think of $h(x+y)$ as being made up of lots of combinations of f(x)g(y), for all the $x$ and $y$ that sum to $z$. To limit it to combinations that sum to $z$, we note that $f(x)g(y) = f(x) g(z-x)$ for those combinations. (We could equally well have used $f(z-y)g(y)$.) An integral is like summing up continuously all the combinations to get the grand total of the density of $h(z)$ for a particular $z$ from all the combinations of $x$ and $y$ such that $x+y=z$. 
   
   I didn't specify the bounds of integration yet. They will be different for each of our three ranges. For a given range, the bounds have to be set so that $x$ lies  inside of  [0,1] and $y$ lies inside of [0,2], though we don't need them to vary over their entire ranges. 
   
   First, consider $z$ in [0,1]. We can make our bounds of integration 0 and $z$. Then $x$ varies from 0 up to 1 (when $z=1$).  Also, then $z-x$ can't go negative (which it could if, say, we just specified an upper bound of .9 and z=.1, so $z-x = -.8$.)  So we have
 $$
For \; z \in [0,1],  h(z) = \int_0^z f(x) g(z-x)dx =  \int_0^z (1) (.5) dx = |_0^z .5x  = .5z - .5(0)  = z/2. 
 $$
   Thus, $h(z)$ starts off looking like a rising triangle density. 
  
Second, consider $z$ in [1,2]. We can make our bounds of integration 0 and $1$. Then $x$ varies from 0 up to 1 (when $z=1$). But $y$ is OK too, because $z$ has a minimum value of 1, so $z-x$ can't be any smaller than 0. In addition, $y$ doesn't become too big, because its maximum value if when $z=2$ and $x=0$, so  $y = z-x - 2$.  So we have
 $$
For \; z \in [1,2],  h(z) = \int_0^1 f(x) g(z-x)dx =  \int_0^1  (1) (.5) dx = |_0^1 .5x  = .5(1)  - .5(0)  = 1/2.
 $$
   Thus, $h(z)$  is a flat uniform density in the middle range.  
 
 We can check this for the extreme values of $z$ and a middle value. \\
When z = 1, the range of x is 0 to 1 and the range of y is 1 to 0.\\ When z = 1.5, the range of x is 0 to 1 and the range of y is 1.5 to 0.5.\\ 
When z = 2, the range of x is 0 to 1 and the range of y is 2 to 1.\\
It's a bit weird that since $y=z-x$, $y$ has a range from a bigger number to a smaller one as $x$ increases, but writing a range as [0,1] instead of [1,0] is just a convention after all.  


Third, consider $z$ in [2,3]. We can make our bounds of integration $z-2$ and $1$. How do you come up with those numbers? Well, we want $x$ to be in [0,1], and by varying $z$ we go from z-2=0 to z-2=1. 

  But $y$ is OK too. The minimum value of $z-x$ is going to be where $z=2$ and $x=1$ so $z-x=1.$  The maximum value of    $z-x$ is going to be where $z=3$ and $x=1$ so  $z-x=2$ or   $z=2$ and $x = z-2 =0$ so $z-x=2$. Note that $y$ is not varying all the way from 0 to 2, its entire range. That is fine, even good,  because to get $z>2$, we need $y$ to equal at least 1. 
  
   So we have
 $$
For \; z \in [2,3],  h(z) = \int_{2-z}^1 f(x) g(z-x)dx =  \int_{2-z}^1  (1) (.5) dx = |_{2-z}^1 .5x   = .5 -    (2-z)/2    = z/2 -.5
 $$
   Thus, $h(z)$  is a   downward triangle density for large values of $z$. 
 
 We can check this for the extreme values of $z$ and a middle value. \\  
When z = 2, the range of x is 0 to 1 and the range of y is 2 to 1.\\ When z = 2.5, the range of x is 0.5 to 1 and the range of y is 2 to 1.5.\\
 When z = 3, the range of x is 1 to 1 and the range of y is 2 to 2.

   The convolution of two uniform pdfs having different ranges of the random variable generates a symmetric trapezoid. The trapezoid's base will  equal to the sum of the  ranges. The top of the trapezoid   will   equal to the difference between the ranges of $x$ and $y$, with a height of   $|f(x) - g(y)|$. The two slants,  the up-ramp and down-ramp, will each  have a length equalt to the  range of the more constricted of the two variables we are summing.  
 
    It might be good to check that we actually have a density that integrates to 1. A nice low-tech way to do that is to calculate the area of the two  triangles and the rectangle that make up the density. Each triangle has area .5(1)(.5)= .25 each. The rectangle has areas (1)(.5) =.5. Thus, the areas do sum up to 1.  
    
    

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