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Probability and Statistics Grinshpan 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.