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Uniform distribution: continuous case
Probability as length. Consider an experiment of randomly drawing a number from [0, 1].
Suppose that every possible choice is treated equally, in the sense that from any two subintervals of equal
length a number is drawn with the same probability.
Let X represent the result of our experiment. It is a random variable taking values in [0, 1].
The probability of any particular outcome X = x is zero.
The events 0 ≤ X ≤ 1/2 and 1/2 ≤ X ≤ 1 have the same probability of 1/2.
The events 0 ≤ X ≤ 1/3, 1/3 ≤ X ≤ 2/3, and 2/3 ≤ X ≤ 1 have the same probability of 1/3.
And so on.
In fact, the probability of X falling into [x1 , x2 ] agrees with the length of [x1 , x2 ],
P (x1 ≤ X ≤ x2 ) = x2 − x1 ,
0 ≤ x1 ≤ x2 ≤ 1.
Probability as mass. Let X be a random variable taking values in the interval [0, 100].
Suppose, as above, that in any two equal subintervals of [0, 100] X occurs with the same probability.
Let us endow our sample interval with some physical characteristics. For instance, view it as a cylindrical
rod of length 100 feet with a small circular cross-section of area 1/1000 square feet and a total mass of
1 pound, which is distributed uniformly. Then the mass density of the rod is constant, 10 pounds per
cubic foot. The mass of any portion of the rod of length ∆x feet is the same,
1
1
· ∆x =
∆x (lb/ft3 · ft2 · ft = lb),
1000
100
and proportional to ∆x. We may, therefore, interpret probability as mass,
∆m = 10 ·
P (x1 ≤ X ≤ x2 ) = mass([x1 , x2 ]),
0 ≤ x1 ≤ x2 ≤ 100.
Uniform distribution on an interval. A random variable X taking values in the interval [a, b]
is uniformly distributed if in any two equal subintervals of [a, b] X occurs with the same probability.
It follows that the probability is proportional to the length,
x2 − x1
P (x1 ≤ X ≤ x2 ) =
,
a ≤ x1 ≤ x2 ≤ b.
b−a
The cumulative distribution function FX (x) = P (X ≤ x)

0,


x−a
,
FX (x) =

b−a


1,
of X is
x ≤ a,
a ≤ x ≤ b,
x ≥ b.
The probability density function (concentration) of X is
1
fX (x) =
,
a ≤ x ≤ b,
b−a
and zero outside the interval.
Probability as area. Consider an experiment of randomly choosing a point from the square
[0, 1] × [0, 1]. Assume that each point (x, y), 0 ≤ x, y ≤ 1, is treated equally, i.e., from any two
rectangles in [0, 1] × [0, 1] of equal area a point is chosen with the same probability.
Let (X, Y ) represent the result of our experiment, it is a random variable ranging in the unit square.
Both X and Y are random variables ranging in [0, 1]. The probability of any particular outcome
(X, Y ) = (x, y) is zero. We expect the probability that (X, Y ) falls into the rectangle
[x1 , x2 ] × [y1 , y2 ] to be proportional to the area of [x1 , x2 ] × [y1 , y2 ].
Since the total area of the sample square is 1, we have
P (x1 ≤ X ≤ x2 , y1 ≤ Y ≤ y2 ) = (x2 − x1 )(y2 − y1 ),
0 ≤ x1 ≤ x2 ≤ 1, 0 ≤ y1 ≤ y2 ≤ 1.