Continuous Uniform Distribution

Section 6.3 Continuous Uniform Distribution

Modeling the idea of "equally-likely" in a continuous world requires a slightly different perspective since there are obviously infinitely many outcomes to consider. Instead, you should consider requiring that intervals in the domain which are of equal width should have the same probability regardless of where they are in that domain. This behaviour suggests \(P(u \lt X \lt v) = P(u + w \lt X \lt v + w)\text{.}\) In integral notation you obtain the following:

\begin{gather*} \int_u^v f(x) dx = \int_{u+w}^{v+w} f(x) dx\\ F(v)-F(u) = F(v+w)-F(u+w)\\ F(u+w)-F(u) = F(v+w)-F(v) \end{gather*}

which is true regardless of w so long as you stay in the domain of interest. This only happens if F is linear and therefore f must be constant. Say, f(x)=c. In many situations, the space of X will be a single interval with \(R = [a,b]\text{.}\) Unless otherwise noted, this will be our assumption as well.

Theorem 6.3.1 Properties of the Continuous Uniform Probability Function

\(f(x) = \frac{1}{b-a}\) satisfies the properties of a probability function over R = [a,b].
\(\mu = \frac{a+b}{2}\)
\(\sigma^2 = \frac{b^2-a^2}{12}\)
\(\gamma_1 = 0\)
\(\gamma_2 = \frac{9 \, {\left(a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}\right)} {\left(a - b\right)}}{5 \, {\left(a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right)}^{2}}\)

Example 6.3.2 Occurence of exactly one event randomly in a given interval

Suppose you know that only one person showed up at the counter of a local business in a given 30 minute interval of time. Then, R=[0,30] given f(x) = 1/30.

Further, the probability that the person arrived within the first 6 minutes would be \(\int_0^6 \frac{1}{30} dx = 0.2\text{.}\)

Theorem 6.3.3 Distribution Function for Continuous Uniform

For \(x \in [a,b], F(x) = \frac{x-a}{b-a}\)

Proof

For x in this range,

\begin{equation*} F(x) = \int_a^x \frac{1}{b-a} du = \frac{u}{b-a} \big |_a^x = \frac{x-a}{b-a}. \end{equation*}