Definition D1
Fix a starting point a and a width h, and call Iⱼ = [ a + (j−1)h, a + jh ) the j-th bin. Here 0 < h, and j runs far enough to cover all the data.
Proof.
(1)[u, v) denotes the x with u ≤ x < v. The left endpoint is included and the right endpoint is not.convention
(2)Since 0 < h, no bin is empty as an interval, and adjacent bins do not overlap.A2, convention
(3)Therefore each data point falls in exactly one bin.(2), A1
(4)a and h are set by the person counting, not by the data. The data single out neither of the two numbers.convention
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Definition D2
The number of data points falling in bin Iⱼ is written fⱼ, and (f₁, f₂, …, f_k) is called the frequency sequence.
Proof.
(1)Each data point falls in exactly one bin, so f₁ + f₂ + ⋯ + f_k = n.D1, A1
(2)This equation holds however a and h are chosen.(1), D1
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Definition D3
A position in the frequency sequence larger than its left neighbor and larger than its right neighbor is called a peak. At either end, only the existing neighbor is examined. One peak makes the sequence unimodal; two or more make it bimodal.
Proof.
A peak is read only from the frequency sequence, not directly from the data.D2, convention
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Theorem T1
Counting the data 0, 1, 2, 2, 2, 3, 3, 3, 5, 6, 6, 7, 7, 7 (n = 14) with a = 0, h = 2 gives the frequency sequence (2, 6, 1, 5), which has two peaks.
Proof.
(1)The bins are [0,2), [2,4), [4,6), [6,8).assumption, D1
(2)The data falling in [0,2) are 0, 1, so f₁ = 2(1), D2, N
(3)The data falling in [2,4) are 2, 2, 2, 3, 3, 3, so f₂ = 6(1), D2, N
(4)The only datum falling in [4,6) is 5, so f₃ = 1(1), D2, N
(5)The data falling in [6,8) are 6, 6, 7, 7, 7, so f₄ = 5(1), D2, N
(6)2 + 6 + 1 + 5 = 14, so the total equals n.(2), (3), (4), (5), D2, N
(7)f₁ = 2 < 6 = f₂ and f₂ = 6 > 1 = f₃, so the second position is a peak.(2), (3), (4), D3, N
(8)f₃ = 1 < 5 = f₄ and f₄ is the last position, so the fourth position is also a peak.(4), (5), D3, N
(9)The first and third positions are smaller than their right neighbors, so they are not peaks. There are two peaks, and the frequency sequence is bimodal.(7), (8), D3, N
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Theorem T2
Counting the same data with different bins makes it unimodal.
Proof.
(1)Recount the same data with a = 0, h = 4.K, D1
(2)The bins are [0,4) and [4,8), containing 0, 1, 2, 2, 2, 3, 3, 3 and 5, 6, 6, 7, 7, 7 respectively.(1), D1, N
(3)Hence the frequency sequence is (8, 6), and 8 + 6 = 14.(2), D2, N
(4)f₁ = 8 > 6 = f₂ and f₁ is the first position, so the only peak is the first position. Unimodal.(3), D3, N
(5)This time keep the width and shift only the starting point: count with a = −1, h = 2.K, D1
(6)The bins are [−1,1), [1,3), [3,5), [5,7), [7,9), containing 0 / 1, 2, 2, 2 / 3, 3, 3 / 5, 6, 6 / 7, 7, 7 respectively.(5), D1, N
(7)Hence the frequency sequence is (1, 4, 3, 3, 3), and 1 + 4 + 3 + 3 + 3 = 14.(6), D2, N
(8)1 < 4 and 4 > 3, so the second position is a peak. The last position equals its left neighbor 3, so it is not a peak, and neither are the remaining positions. Unimodal.(7), D3, N
(9)Whether the width is changed or the width is kept and only the starting point is shifted, the number of peaks changes from two to one.(4), (8), T1
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Theorem T3
The data do not determine the shape of the frequency distribution.
Proof.
(1)Consider the proposition 'once the data are fixed, the number of peaks is fixed'.D3
(2)One and the same data set produced both a two-peak count and a one-peak count.T1, T2
(3)Therefore this single data set makes the proposition false.(1), (2), A3
(4)For the shape to be determined, a and h are needed in addition to the data, and both come from outside the data.(3), D1
(5)Therefore the number of peaks read from a histogram is not a property of the data, but a joint product of the data and the choice of bins.(4), D3
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Corollary C
What the data determine is the total of the frequencies, not the shape.
Proof.
(1)Whatever a and h are chosen, the total of the frequencies is the same n = 14.D2, A1
(2)Indeed the totals of (2,6,1,5), (8,6), and (1,4,3,3,3) are all 14.T1, T2, N
(3)But the number of peaks is sometimes two and sometimes one.T1, T2
(4)Therefore, to say from a histogram that 'these data split into two groups', one must also state how the bins were chosen.(3), T3
(5)Narrow bins raise the number of peaks and wide bins lower it, not because the data changed, but because the counting changed.(1), (4), D1
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