## Continued Uniform Distributions |
## October 2nd, 2017 |

Today we are going to work on an interesting probability distribution that I spent some time thinking about recently. The simplest distribution is the uniform distribution. The uniform distribution has equal probability for all numbers in a range. For example, if you have a uniform distribution on $\{1,2,3\}$, then you have a $\frac{1}{3}$ probability of getting each number in that range. It is a little confusing to think about continuous distributions, because each number actually has a 0 percent chance of occurring. If we have a uniform distribution on $[0,5]$, then each individual point has a probability of 0 of occurring, but the probability that we get a number *near* each number is $\frac{1}{5}$. Make sense? A uniform distribution on $[0,a]$ has probability density function, denoted $f_x$, given by

\[

f_x=

\begin{cases}

\frac{1}{a} & 0\leq x\leq a \\

0 & \text{else} \\

\end{cases}

\]

**I. Our Distribution**

So, let’s say we want to generate a number $x$ from a uniform distribution on $[0,a]$, and *then* we want to generate a number $y$ from a uniform distribution on $[0,x]$. Let’s write down what we are given; we know that $x$ is uniformly distributed on $[0,a]$ and is independent of $y$

\[

f_x=

\begin{cases}

\frac{1}{a} & 0\leq x\leq a \\

0 & \text{else} \\

\end{cases}

\]

Additionally, if we are given $x$, we know what the distrubution for $y$ is — it’s uniform on $[0,x]$. We denote this as $f_{y|x}$, the probability of $y$ *given* $x$.

\[

f_{y|x}=

\begin{cases}

\frac{1}{x} & 0\leq y\leq x \\

0 & \text{else} \\

\end{cases}

\]

**II. Manipulation**

We start by finding $f_{xy}$, the probability that *both* $x$ and $y$ occur. We can use the formula

$$ f_{xy} = f_{y|x}\cdot f_{x}$$

This gives us

$$f_{xy} = \frac{1}{ax}$$

We know that $x$ will never be more than $a$ or less than 0, and that $y$ will never be more than $x$ or less than 0. So, we can properly define our probability function.

\[f_{xy}=

\begin{cases}

\frac{1}{ax} & 0\leq y\leq x \hspace{3mm}\text{and}\hspace{3mm} 0\leq x\leq a\\

0 & \text{else} \\

\end{cases}

\]

This area is shown in the picture below

To find $f_y$, we need to integrate x to remove it from the equation. If we wanted to integrate over the entire region, we would do

$$\int_0^a\int_y^a f_{xy}\hspace{2mm}\text{d}x\text{d}y \hspace{6mm}\text{or}\hspace{6mm}\int_0^a\int_0^x f_{xy}\hspace{2mm}\text{d}y\text{d}x$$

Since we want to eliminate $x$, we need $x$ to be the interior variable of integration. This will ensure that we have no $x$ term in the integrand for the exterior integral. Therefore, to eliminate $x$, we will use

$$f_y = \int_y^a f_{xy}\hspace{2mm}\text{d}x$$

$$ = \frac{1}{a}\int_y^a \frac{1}{x}\hspace{2mm}\text{d}x$$

$$ = \frac{1}{a}\ln{x}\hspace{2mm}\Big|_{x=y}^{x=a}$$

$$ = \frac{1}{a}(\ln{a}-\ln{y})$$

$$ = \frac{1}{a}\ln{\frac{a}{y}}$$

$$ \bbox[15px,border:1px solid black]{f_y = \frac{\ln\frac{a}{y}}{a} \hspace{7mm} 0\leq y\leq a}$$

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