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Uniform Distribution Simplified Revision Notes

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16.1.4 Uniform Distribution

Continuous Uniform Distribution

The discrete uniform distribution models an event with n possible outcomes where each outcome has an equal probability of occurring. The continuous uniform distribution is similar in nature and is also referred to as the "rectangular distribution" for this reason:

(Graph showing a rectangle with height 1ba \frac{1}{b-a} between points a and bb , and the area under the curve is 11.)

image

f(x)={1ba,axb0,otherwisef(x) = \begin{cases} \frac{1}{b-a}, & a \leq x \leq b \\ 0, & \text{otherwise} \end{cases}

The height of the rectangle is such that h(ba)=1h(b - a) = 1 , therefore: h=1bah = \frac{1}{b - a}

(Table showing the continuous uniform distribution over [a,ba, b] , the probability density function (P.D.F.), expectation E(X) E(X) , and variance Var(X)\text{Var}(X)

image

p.d.f.=1ba,E(X)=12(a+b),Var(X)=112(ba)2\text{p.d.f.} = \frac{1}{b - a}, \quad E(X) = \frac{1}{2}(a + b), \quad \text{Var}(X) = \frac{1}{12}(b - a)^2

infoNote

Example Problem (Q5, Jan 2009, Q2):

The continuous random variable XX is uniformly distributed over the interval [2,7][-2, 7].

(a) Write down fully the probability density function f(x)f(x) of XX .

(b) Sketch the probability density function f(x)f(x) of XX .

Find:

(c) E(X2)E(X^2)

(d) P(0.2<X<0.6)P(-0.2 < X < 0.6)


(a) f(x)f(x) is given by:

f(x)=17(2)=19f(x)={19,2x70,otherwise f(x) = \frac{1}{7 - (-2)} = \frac{1}{9} \quad f(x) = \begin{cases} \frac{1}{9}, & -2 \leq x \leq 7 \\ 0, & \text{otherwise} \end{cases}

(b)


(c)

27x2f(x)dx=2719x2dx=[127x3]27\int_{-2}^{7} x^2 f(x) \, dx = \int_{-2}^{7} \frac{1}{9} x^2 \, dx = \left[ \frac{1}{27} x^3 \right]_{-2}^{7}

=(127(7)3)(127(2)3)=:success[13]= \left( \frac{1}{27} (7)^3 \right) - \left( \frac{1}{27} (-2)^3 \right) = :success[13]


(d)

p=0.8×19=:success[445]p = 0.8 \times \frac{1}{9} = :success[\frac{4}{45}]

The Uniform Discrete Distribution

This probability distribution describes a situation in which we play a game once and the game has n different outcomes. Each outcome is equally likely.

lightbulbExample

Example: A dice is thrown, and the number on the dice is noted. The following table details the probability distribution of this, where XX is "the score obtained":

x123456P(X=x)161616161616\begin{array}{c|c|c|c|c|c|c} x & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline P(X = x) & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} \end{array}

This is an example of a uniform distribution and can be more succinctly written as: XU(6)X \sim U(6)

lightbulbExample

Example: Put XU(2)X \sim U(2) in a table.

x12P(X=x)1212 \begin{array}{c|c|c} x & 1 & 2 \\ \hline P(X = x) & \frac{1}{2} & \frac{1}{2} \end{array}
lightbulbExample

Example : A fair spinner has outcomes 2, 4, 6, or 8. Find the expected mean score and expected variance after a large number of spins.

(Note: This situation is not quite a uniform distribution because the outcomes are not consecutive integers starting at 11. However, for now, let's pretend they are.)


XU(4) implies:X \sim U(4)\ implies:

E(X)=4+12=:success[2.5]E(X) = \frac{4 + 1}{2} = :success[2.5]

Var(X)=112(421)=:success[1.25]\text{Var}(X) = \frac{1}{12} (4^2 - 1) = :success[1.25]


However, this is not the mean and variance we need. If we consider E(2X)E(2X) and Var(2X)\text{Var}(2X) , we do get the quantities we need since the outcomes are double that of XU(4)X \sim U(4) .

E(2X)=2E(X)=:success[5]E(2X) = 2E(X) = :success[5]

Var(2X)=22Var(X)=:success[5]\text{Var}(2X) = 2^2 \text{Var}(X) = :success[5]

lightbulbExample

Example: A fair die is numbered 7, 10, 13, 16, 19, 22. Find the expected score per throw and the variance after a large number of throws.


XU(6) implies:X \sim U(6)\ implies:

E(X)=1+62=:success[72]E(X) = \frac{1 + 6}{2} = :success[\frac{7}{2}]

Var(X)=112(621)=:success[3512]\text{Var}(X) = \frac{1}{12} (6^2 - 1) = :success[\frac{35}{12}]

If XU(6)X \sim U(6) , then the outcomes are described by 3X+43X + 4 .

E(3X+4)=3E(X)+4=3×72+4=:success[292]E(3X + 4) = 3E(X) + 4 = 3 \times \frac{7}{2} + 4 = :success[\frac{29}{2}]

Var(3X+4)=32Var(X)=9×3512=:success[1054]\text{Var}(3X + 4) = 3^2 \text{Var}(X) = 9 \times \frac{35}{12} = :success[\frac{105}{4}]

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