History of the Normal Distribution - University of Utahkenkel/normaldistributiontalk.pdf · Normal...
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History of theNormal
Distribution
Jenny Kenkel
History of the Normal Distribution
Jenny Kenkel
November 8th, 2016
History of theNormal
Distribution
Jenny Kenkel
Probabiltiy Distributions
A probability distribution is a function f (x) so that
P(a < X < b) =
∫ b
af (x)dx
History of theNormal
Distribution
Jenny Kenkel
The Normal Distribution
The normal distribution is afamily of distributions, given by
f (x) =1√
2πσ2e−
(x−µ)2
2σ2
The Standard Normal has µ =0 and σ = 1, i.e.
f (x) =1√2π
ex2
2
Changing µ changes the loca-tion of the curve, and chang-ing σ changes the spread of thecurve
History of theNormal
Distribution
Jenny Kenkel
Arbuthnot: “An argument for Divine Providence, taken from
the constant Regularity observ’d in the Births of both Sexes”
Argued the probabilitythat more males thanfemales were born everyyear from 1629 to 1710,
82 years, was(
12
)82
But the fact that notmany more males thanfemales were born everyyear was proof polygamy iscontrary to the laws ofnature
History of theNormal
Distribution
Jenny Kenkel
Arbuthnot: “An argument for Divine Providence, taken from
the constant Regularity observ’d in the Births of both Sexes”
Argued the probabilitythat more males thanfemales were born everyyear from 1629 to 1710,
82 years, was(
12
)82
But the fact that notmany more males thanfemales were born everyyear was proof polygamy iscontrary to the laws ofnature
History of theNormal
Distribution
Jenny Kenkel
‘sGravesande
Felt evidence for divineprovidence was evenstronger than Arbuthnothad argued
History of theNormal
Distribution
Jenny Kenkel
Bernoulli Trials
A Bernoulli Trial is an experiment with only two possibleoutcomes. A Binomial Experiment is a series of repeatedBernoulli trials.
The probability, p, of success stays constant as more trialsare performed
The probability of k successes in n trials is(n
k
)pk(1− p)n−k
History of theNormal
Distribution
Jenny Kenkel
Bernoulli Trials
A Bernoulli Trial is an experiment with only two possibleoutcomes. A Binomial Experiment is a series of repeatedBernoulli trials.
The probability, p, of success stays constant as more trialsare performed
The probability of k successes in n trials is(n
k
)pk(1− p)n−k
History of theNormal
Distribution
Jenny Kenkel
Bernoulli Trials
A Bernoulli Trial is an experiment with only two possibleoutcomes. A Binomial Experiment is a series of repeatedBernoulli trials.
The probability, p, of success stays constant as more trialsare performed
The probability of k successes in n trials is(n
k
)pk(1− p)n−k
History of theNormal
Distribution
Jenny Kenkel
‘sGravesande was very patient
‘sGravesande let n = 11, 429, p = 0.5
The lowest ratio, which occurred in 1703, was .5027,which corresponded to about 5745 out of 11429
The highest ratio, which occurred in 1661, was .5362,which corresponded to about 6128 out of 11429
‘sGravesande calculated
P[5745 ≤ x ≤ 6128] =6128∑5745
(11429
x
)(1
2
)11429
≈ 0.292
Thus the probability of this difference in birth ratesrecurring 82 years in a row is .29282
History of theNormal
Distribution
Jenny Kenkel
‘sGravesande was very patient
‘sGravesande let n = 11, 429, p = 0.5
The lowest ratio, which occurred in 1703, was .5027,which corresponded to about 5745 out of 11429
The highest ratio, which occurred in 1661, was .5362,which corresponded to about 6128 out of 11429
‘sGravesande calculated
P[5745 ≤ x ≤ 6128] =6128∑5745
(11429
x
)(1
2
)11429
≈ 0.292
Thus the probability of this difference in birth ratesrecurring 82 years in a row is .29282
History of theNormal
Distribution
Jenny Kenkel
‘sGravesande was very patient
‘sGravesande let n = 11, 429, p = 0.5
The lowest ratio, which occurred in 1703, was .5027,which corresponded to about 5745 out of 11429
The highest ratio, which occurred in 1661, was .5362,which corresponded to about 6128 out of 11429
‘sGravesande calculated
P[5745 ≤ x ≤ 6128] =6128∑5745
(11429
x
)(1
2
)11429
≈ 0.292
Thus the probability of this difference in birth ratesrecurring 82 years in a row is .29282
History of theNormal
Distribution
Jenny Kenkel
‘sGravesande was very patient
‘sGravesande let n = 11, 429, p = 0.5
The lowest ratio, which occurred in 1703, was .5027,which corresponded to about 5745 out of 11429
The highest ratio, which occurred in 1661, was .5362,which corresponded to about 6128 out of 11429
‘sGravesande calculated
P[5745 ≤ x ≤ 6128] =6128∑5745
(11429
x
)(1
2
)11429
≈ 0.292
Thus the probability of this difference in birth ratesrecurring 82 years in a row is .29282
History of theNormal
Distribution
Jenny Kenkel
‘sGravesande was very patient
‘sGravesande let n = 11, 429, p = 0.5
The lowest ratio, which occurred in 1703, was .5027,which corresponded to about 5745 out of 11429
The highest ratio, which occurred in 1661, was .5362,which corresponded to about 6128 out of 11429
‘sGravesande calculated
P[5745 ≤ x ≤ 6128] =6128∑5745
(11429
x
)(1
2
)11429
≈ 0.292
Thus the probability of this difference in birth ratesrecurring 82 years in a row is .29282
History of theNormal
Distribution
Jenny Kenkel
DeMoivre and Stirling’s Approximation
In 1733, in a second supplement that only appeared withcertain editions of his book[3], De Moivre noted that:
n! ∼ Bnn+ 12 e−n
James Stirling determined that B =√
2π
The probability of k successes in n trials is(n
k
)pk(1− p)n−k ≈ 1√
2πnp(1− p)e− (k−np)2
2np(1−p)
Let σ2 = np(1− p) and µ = np, and you have(n
k
)pk(1− p)n−k ≈ 1√
2πσ2e−
(k−µ)2
σ2
History of theNormal
Distribution
Jenny Kenkel
DeMoivre and Stirling’s Approximation
In 1733, in a second supplement that only appeared withcertain editions of his book[3], De Moivre noted that:
n! ∼ Bnn+ 12 e−n
James Stirling determined that B =√
2π
The probability of k successes in n trials is(n
k
)pk(1− p)n−k ≈ 1√
2πnp(1− p)e− (k−np)2
2np(1−p)
Let σ2 = np(1− p) and µ = np, and you have(n
k
)pk(1− p)n−k ≈ 1√
2πσ2e−
(k−µ)2
σ2
History of theNormal
Distribution
Jenny Kenkel
DeMoivre and Stirling’s Approximation
In 1733, in a second supplement that only appeared withcertain editions of his book[3], De Moivre noted that:
n! ∼ Bnn+ 12 e−n
James Stirling determined that B =√
2π
The probability of k successes in n trials is(n
k
)pk(1− p)n−k ≈ 1√
2πnp(1− p)e− (k−np)2
2np(1−p)
Let σ2 = np(1− p) and µ = np, and you have(n
k
)pk(1− p)n−k ≈ 1√
2πσ2e−
(k−µ)2
σ2
History of theNormal
Distribution
Jenny Kenkel
DeMoivre and Stirling’s Approximation
In 1733, in a second supplement that only appeared withcertain editions of his book[3], De Moivre noted that:
n! ∼ Bnn+ 12 e−n
James Stirling determined that B =√
2π
The probability of k successes in n trials is(n
k
)pk(1− p)n−k ≈ 1√
2πnp(1− p)e− (k−np)2
2np(1−p)
Let σ2 = np(1− p) and µ = np, and you have(n
k
)pk(1− p)n−k ≈ 1√
2πσ2e−
(k−µ)2
σ2
History of theNormal
Distribution
Jenny Kenkel
From Binomials to Errors
In 1733, DeMoivre first used the Normal distribution as anapproximation for probabilities of binomial experiments where nis very large.About 70 years later, it would be used as the probabilitydistribution of random errors.
History of theNormal
Distribution
Jenny Kenkel
The beginning of the average
Astronomy called for accurate measurements
In the 1600s, Tycho Brahe advocated incorporatingrepeated measurements but did not specify how to userepeated measurements
Sometimes astronomers used median observations,sometimes mean observations, some times ????
History of theNormal
Distribution
Jenny Kenkel
The beginning of the average
Astronomy called for accurate measurements
In the 1600s, Tycho Brahe advocated incorporatingrepeated measurements but did not specify how to userepeated measurementsSometimes astronomers used median observations,sometimes mean observations, some times ????
History of theNormal
Distribution
Jenny Kenkel
The beginning of the average
Astronomy called for accurate measurements
In the 1600s, Tycho Brahe advocated incorporatingrepeated measurements but did not specify how to userepeated measurementsSometimes astronomers used median observations,sometimes mean observations, some times ????
History of theNormal
Distribution
Jenny Kenkel
Early reasoning about the probability distributionsof errors
In 1632, Galileo reasoned that
1 There is a true distance, which is only one number
2 All observations have errors
3 The errors are symmetric around the true value
4 Small errors are more common than large ones
History of theNormal
Distribution
Jenny Kenkel
Early reasoning about the probability distributionsof errors
In 1632, Galileo reasoned that
1 There is a true distance, which is only one number
2 All observations have errors
3 The errors are symmetric around the true value
4 Small errors are more common than large ones
History of theNormal
Distribution
Jenny Kenkel
Early reasoning about the probability distributionsof errors
In 1632, Galileo reasoned that
1 There is a true distance, which is only one number
2 All observations have errors
3 The errors are symmetric around the true value
4 Small errors are more common than large ones
History of theNormal
Distribution
Jenny Kenkel
Early reasoning about the probability distributionsof errors
In 1632, Galileo reasoned that
1 There is a true distance, which is only one number
2 All observations have errors
3 The errors are symmetric around the true value
4 Small errors are more common than large ones
History of theNormal
Distribution
Jenny Kenkel
The average didn’t take on immediately
In the 1660s, Robert Boyleargued that, rather thanaveraging at all, astronomersshould just focus on one verycareful experiment
In 1756, Simpson wrote that:“some persons, of considerablenote, have been of opinion, andeven publickly maintained, thatone single observation, takenwith due care, was as much tobe relied on as the Mean of agreat number ”
History of theNormal
Distribution
Jenny Kenkel
The average didn’t take on immediately
In the 1660s, Robert Boyleargued that, rather thanaveraging at all, astronomersshould just focus on one verycareful experiment
In 1756, Simpson wrote that:“some persons, of considerablenote, have been of opinion, andeven publickly maintained, thatone single observation, takenwith due care, was as much tobe relied on as the Mean of agreat number ”
History of theNormal
Distribution
Jenny Kenkel
Simpson’s probability distributions of errors!
Also in Simpson’s 1756 paper was the notion of probabilitydistributions of errors or error curve
Simpson computed the probabilities when the errors couldtake on values
−v , . . . ,−2,−1, 0, 1, 2, . . . , v
to have probabilities proportional to either
r−v , . . . , r−2, r−1, r0, r1, r2, . . . , r4
or
r−v , 2r1−v , 3r2−v . . . , (v + 1)r0, . . . , 2r v−1, r4
History of theNormal
Distribution
Jenny Kenkel
Simpson’s probability distributions of errors!
Also in Simpson’s 1756 paper was the notion of probabilitydistributions of errors or error curveSimpson computed the probabilities when the errors couldtake on values
−v , . . . ,−2,−1, 0, 1, 2, . . . , v
to have probabilities proportional to either
r−v , . . . , r−2, r−1, r0, r1, r2, . . . , r4
or
r−v , 2r1−v , 3r2−v . . . , (v + 1)r0, . . . , 2r v−1, r4
History of theNormal
Distribution
Jenny Kenkel
Laplace’s Error Curves
In 1774, Laplace proposed the first of his probabilitydistributions for errors .
Probability distribution, φ, must be symmetric about 0
Probability distribution, φ, must be monotone decreasingwith x“as we have no reason to suppose a different law for theordinates than for their differences...”, assumedφ(x) ∼ dφ(x)
dx ,
Thus φ(x) = m2 e−m|x |
Recall:
f (x) =1√
2πσ2e−
(x−µ)2
2σ2
History of theNormal
Distribution
Jenny Kenkel
Laplace’s Error Curves
In 1774, Laplace proposed the first of his probabilitydistributions for errors .
Probability distribution, φ, must be symmetric about 0Probability distribution, φ, must be monotone decreasingwith x
“as we have no reason to suppose a different law for theordinates than for their differences...”, assumedφ(x) ∼ dφ(x)
dx ,
Thus φ(x) = m2 e−m|x |
Recall:
f (x) =1√
2πσ2e−
(x−µ)2
2σ2
History of theNormal
Distribution
Jenny Kenkel
Laplace’s Error Curves
In 1774, Laplace proposed the first of his probabilitydistributions for errors .
Probability distribution, φ, must be symmetric about 0Probability distribution, φ, must be monotone decreasingwith x“as we have no reason to suppose a different law for theordinates than for their differences...”, assumedφ(x) ∼ dφ(x)
dx ,
Thus φ(x) = m2 e−m|x |
Recall:
f (x) =1√
2πσ2e−
(x−µ)2
2σ2
History of theNormal
Distribution
Jenny Kenkel
Laplace’s Error Curves
In 1774, Laplace proposed the first of his probabilitydistributions for errors .
Probability distribution, φ, must be symmetric about 0Probability distribution, φ, must be monotone decreasingwith x“as we have no reason to suppose a different law for theordinates than for their differences...”, assumedφ(x) ∼ dφ(x)
dx ,
Thus φ(x) = m2 e−m|x |
Recall:
f (x) =1√
2πσ2e−
(x−µ)2
2σ2
History of theNormal
Distribution
Jenny Kenkel
Gauss gets involved
On January 1, 1801, Giuseppe Piazzi spotted something hebelieved to be a new planet and named Ceres.
History of theNormal
Distribution
Jenny Kenkel
Gauss gets involved
Six weeks later, he lost it!
History of theNormal
Distribution
Jenny Kenkel
Gauss gets involved
Gauss suggested looking in a different area of the sky thanmost other astronomers... and he was right!
History of theNormal
Distribution
Jenny Kenkel
Gauss’ Proof: Assumptions and Notation
Gauss concluded that the probability density for the error is:
φ(x) =h√πe−h
2x2
Let p be the true but unknown error
Let M1, . . . ,Mn be estimates of p
Let φ(x) be the probability density function of the randomerror(recall: P(a < x < b) <
∫ b
aφ(x) )
φ(x) has a maximum at x = 0
φ(−x) = φ(x)
The average, M = 1n
∑ni=1 Mi , is the most likely value of p
History of theNormal
Distribution
Jenny Kenkel
Gauss’ Proof: Part 1 of 4
Let f (x) = φ′(x)φ(x)
Then f (−x) = −f (x)
If p is the true value and Mi is the i th estimate, thenMi − p is the i th error
If the errors are independent, then their joint probabilitydistribution is
Ω = φ(M1 − p)φ(M2 − p) · · ·φ(Mn − p)
If M = 1n
∑ni=1 Mi is the most likely value of p, then M
should maximize Ω
That is, dΩdp |M = 0
History of theNormal
Distribution
Jenny Kenkel
Gauss’ Proof: Part 1 of 4
Let f (x) = φ′(x)φ(x)
Then f (−x) = −f (x)
If p is the true value and Mi is the i th estimate, thenMi − p is the i th error
If the errors are independent, then their joint probabilitydistribution is
Ω = φ(M1 − p)φ(M2 − p) · · ·φ(Mn − p)
If M = 1n
∑ni=1 Mi is the most likely value of p, then M
should maximize Ω
That is, dΩdp |M = 0
History of theNormal
Distribution
Jenny Kenkel
Gauss’ Proof: Part 1 of 4
Let f (x) = φ′(x)φ(x)
Then f (−x) = −f (x)
If p is the true value and Mi is the i th estimate, thenMi − p is the i th error
If the errors are independent, then their joint probabilitydistribution is
Ω = φ(M1 − p)φ(M2 − p) · · ·φ(Mn − p)
If M = 1n
∑ni=1 Mi is the most likely value of p, then M
should maximize Ω
That is, dΩdp |M = 0
History of theNormal
Distribution
Jenny Kenkel
Gauss’ Proof: Part 1 of 4
Let f (x) = φ′(x)φ(x)
Then f (−x) = −f (x)
If p is the true value and Mi is the i th estimate, thenMi − p is the i th error
If the errors are independent, then their joint probabilitydistribution is
Ω = φ(M1 − p)φ(M2 − p) · · ·φ(Mn − p)
If M = 1n
∑ni=1 Mi is the most likely value of p, then M
should maximize Ω
That is, dΩdp |M = 0
History of theNormal
Distribution
Jenny Kenkel
Gauss’ Proof: Part 1 of 4
Let f (x) = φ′(x)φ(x)
Then f (−x) = −f (x)
If p is the true value and Mi is the i th estimate, thenMi − p is the i th error
If the errors are independent, then their joint probabilitydistribution is
Ω = φ(M1 − p)φ(M2 − p) · · ·φ(Mn − p)
If M = 1n
∑ni=1 Mi is the most likely value of p, then M
should maximize Ω
That is, dΩdp |M = 0
History of theNormal
Distribution
Jenny Kenkel
Gauss’ Proof: Part 1 of 4
Let f (x) = φ′(x)φ(x)
Then f (−x) = −f (x)
If p is the true value and Mi is the i th estimate, thenMi − p is the i th error
If the errors are independent, then their joint probabilitydistribution is
Ω = φ(M1 − p)φ(M2 − p) · · ·φ(Mn − p)
If M = 1n
∑ni=1 Mi is the most likely value of p, then M
should maximize Ω
That is, dΩdp |M = 0
History of theNormal
Distribution
Jenny Kenkel
Gauss’ Proof: Part 2 of 4
Ω = φ(M1 − p)φ(M2 − p) · · ·φ(Mn − p)
so 0 = dΩdp |p=M =
= −φ′(M1 −M)φ(M2 −M) · · ·φ(Mn −M)
− φ(M1 −M)φ′(M2 − p) · · ·φ(Mn −M)
− . . .− φ(M1 −M)φ(M2 −M) · · ·φ′(Mn −M)
=
(φ′(M1 −M)
φ(M1 −M) +
φ′(M2 −M)
φ(M2 −M) + . . .+
φ′(Mn −M)
φ(Mn −M)
)Ω
History of theNormal
Distribution
Jenny Kenkel
Gauss’ Proof: Part 2 of 4
Ω = φ(M1 − p)φ(M2 − p) · · ·φ(Mn − p)
so 0 = dΩdp |p=M =
= −φ′(M1 −M)φ(M2 −M) · · ·φ(Mn −M)
− φ(M1 −M)φ′(M2 − p) · · ·φ(Mn −M)
− . . .− φ(M1 −M)φ(M2 −M) · · ·φ′(Mn −M)
=
(φ′(M1 −M)
φ(M1 −M) +
φ′(M2 −M)
φ(M2 −M) + . . .+
φ′(Mn −M)
φ(Mn −M)
)Ω
History of theNormal
Distribution
Jenny Kenkel
Gauss’ Proof: Part 2 of 4
Ω = φ(M1 − p)φ(M2 − p) · · ·φ(Mn − p)
so 0 = dΩdp |p=M =
= −φ′(M1 −M)φ(M2 −M) · · ·φ(Mn −M)
− φ(M1 −M)φ′(M2 − p) · · ·φ(Mn −M)
− . . .− φ(M1 −M)φ(M2 −M) · · ·φ′(Mn −M)
=
(φ′(M1 −M)
φ(M1 −M) +
φ′(M2 −M)
φ(M2 −M) + . . .+
φ′(Mn −M)
φ(Mn −M)
)Ω
History of theNormal
Distribution
Jenny Kenkel
Gauss’ Proof: Part 3 of 4
0 =
(φ′(M1 −M)
φ(M1 −M) +
φ′(M2 −M)
φ(M2 −M) + . . .+
φ′(Mn −M)
φ(Mn −M)
)Ω
Recall, we defined f (x) = φ′(x)φ(x) , so
0 = f (M1 −M) + . . .+ f (Mn −M)
For arbitrary real numbers M and N, let M1 = M, andM2 = M3 = . . . = Mn = M − nN.Then M = M − (n − 1)N
0 = f ((n− 1)N) + (n− 1)f (−N) or f ((n− 1)N) = (n− 1)f (N)
History of theNormal
Distribution
Jenny Kenkel
Gauss’ Proof: Part 3 of 4
0 =
(φ′(M1 −M)
φ(M1 −M) +
φ′(M2 −M)
φ(M2 −M) + . . .+
φ′(Mn −M)
φ(Mn −M)
)Ω
Recall, we defined f (x) = φ′(x)φ(x) , so
0 = f (M1 −M) + . . .+ f (Mn −M)
For arbitrary real numbers M and N, let M1 = M, andM2 = M3 = . . . = Mn = M − nN.Then M = M − (n − 1)N
0 = f ((n− 1)N) + (n− 1)f (−N) or f ((n− 1)N) = (n− 1)f (N)
History of theNormal
Distribution
Jenny Kenkel
Gauss’ Proof: Part 3 of 4
0 =
(φ′(M1 −M)
φ(M1 −M) +
φ′(M2 −M)
φ(M2 −M) + . . .+
φ′(Mn −M)
φ(Mn −M)
)Ω
Recall, we defined f (x) = φ′(x)φ(x) , so
0 = f (M1 −M) + . . .+ f (Mn −M)
For arbitrary real numbers M and N, let M1 = M, andM2 = M3 = . . . = Mn = M − nN.
Then M = M − (n − 1)N
0 = f ((n− 1)N) + (n− 1)f (−N) or f ((n− 1)N) = (n− 1)f (N)
History of theNormal
Distribution
Jenny Kenkel
Gauss’ Proof: Part 3 of 4
0 =
(φ′(M1 −M)
φ(M1 −M) +
φ′(M2 −M)
φ(M2 −M) + . . .+
φ′(Mn −M)
φ(Mn −M)
)Ω
Recall, we defined f (x) = φ′(x)φ(x) , so
0 = f (M1 −M) + . . .+ f (Mn −M)
For arbitrary real numbers M and N, let M1 = M, andM2 = M3 = . . . = Mn = M − nN.Then M = M − (n − 1)N
0 = f ((n− 1)N) + (n− 1)f (−N) or f ((n− 1)N) = (n− 1)f (N)
History of theNormal
Distribution
Jenny Kenkel
Gauss’ Proof: Part 4
f ((n − 1)N) = (n − 1)f (N)
If f is continuous, then f (x) = kx
Thus, φ′(x)φ(x) = kx
Inegration with respect to x gives us
ln(φ(x)) =k
2x2 + c
φ(x) = Aekx2
2
In order for φ(x) to have a maximum at x = 0, we must havethat k is negative, so set k = −h2.We require that
∫∞−∞ Ae−h
2x2dx = 1, so A = h√
π
Thus φ(x) = h√πekx
2
History of theNormal
Distribution
Jenny Kenkel
Gauss’ Proof: Part 4
f ((n − 1)N) = (n − 1)f (N)
If f is continuous, then f (x) = kx
Thus, φ′(x)φ(x) = kx
Inegration with respect to x gives us
ln(φ(x)) =k
2x2 + c
φ(x) = Aekx2
2
In order for φ(x) to have a maximum at x = 0, we must havethat k is negative, so set k = −h2.We require that
∫∞−∞ Ae−h
2x2dx = 1, so A = h√
π
Thus φ(x) = h√πekx
2
History of theNormal
Distribution
Jenny Kenkel
Gauss’ Proof: Part 4
f ((n − 1)N) = (n − 1)f (N)
If f is continuous, then f (x) = kx
Thus, φ′(x)φ(x) = kx
Inegration with respect to x gives us
ln(φ(x)) =k
2x2 + c
φ(x) = Aekx2
2
In order for φ(x) to have a maximum at x = 0, we must havethat k is negative, so set k = −h2.We require that
∫∞−∞ Ae−h
2x2dx = 1, so A = h√
π
Thus φ(x) = h√πekx
2
History of theNormal
Distribution
Jenny Kenkel
Gauss’ Proof: Part 4
f ((n − 1)N) = (n − 1)f (N)
If f is continuous, then f (x) = kx
Thus, φ′(x)φ(x) = kx
Inegration with respect to x gives us
ln(φ(x)) =k
2x2 + c
φ(x) = Aekx2
2
In order for φ(x) to have a maximum at x = 0, we must havethat k is negative, so set k = −h2.We require that
∫∞−∞ Ae−h
2x2dx = 1, so A = h√
π
Thus φ(x) = h√πekx
2
History of theNormal
Distribution
Jenny Kenkel
Gauss’ Proof: Part 4
f ((n − 1)N) = (n − 1)f (N)
If f is continuous, then f (x) = kx
Thus, φ′(x)φ(x) = kx
Inegration with respect to x gives us
ln(φ(x)) =k
2x2 + c
φ(x) = Aekx2
2
In order for φ(x) to have a maximum at x = 0, we must havethat k is negative, so set k = −h2.We require that
∫∞−∞ Ae−h
2x2dx = 1, so A = h√
π
Thus φ(x) = h√πekx
2
History of theNormal
Distribution
Jenny Kenkel
Gauss’ Proof: Part 4
f ((n − 1)N) = (n − 1)f (N)
If f is continuous, then f (x) = kx
Thus, φ′(x)φ(x) = kx
Inegration with respect to x gives us
ln(φ(x)) =k
2x2 + c
φ(x) = Aekx2
2
In order for φ(x) to have a maximum at x = 0, we must havethat k is negative, so set k = −h2.
We require that∫∞−∞ Ae−h
2x2dx = 1, so A = h√
π
Thus φ(x) = h√πekx
2
History of theNormal
Distribution
Jenny Kenkel
Gauss’ Proof: Part 4
f ((n − 1)N) = (n − 1)f (N)
If f is continuous, then f (x) = kx
Thus, φ′(x)φ(x) = kx
Inegration with respect to x gives us
ln(φ(x)) =k
2x2 + c
φ(x) = Aekx2
2
In order for φ(x) to have a maximum at x = 0, we must havethat k is negative, so set k = −h2.We require that
∫∞−∞ Ae−h
2x2dx = 1, so A = h√
π
Thus φ(x) = h√πekx
2
History of theNormal
Distribution
Jenny Kenkel
Gauss’ Proof: Part 4
f ((n − 1)N) = (n − 1)f (N)
If f is continuous, then f (x) = kx
Thus, φ′(x)φ(x) = kx
Inegration with respect to x gives us
ln(φ(x)) =k
2x2 + c
φ(x) = Aekx2
2
In order for φ(x) to have a maximum at x = 0, we must havethat k is negative, so set k = −h2.We require that
∫∞−∞ Ae−h
2x2dx = 1, so A = h√
π
Thus φ(x) = h√πekx
2
History of theNormal
Distribution
Jenny Kenkel
Not Just For Errors Anymore
In 1846, Adolphe Quetelet examined chest measurementsof Scottish soldiers from a medical journal.
He realized that measurements of different objects of thesame type would also often take on the NormalDistribution.
His idea was that individuals were errors from some true,abstract form: the average man.
History of theNormal
Distribution
Jenny Kenkel
The Average Man
“The more knowledge becomes widespread, the moredo the deviations from the average diminish, and themore do we tend to approach that which is beautifuland that which is good.”
Quetelet, 1835
History of theNormal
Distribution
Jenny Kenkel
Florence Nightingale
Mentored by Quetelet,translated one of his worksto English
Popularized the pie chart,invented the circularhistogram
Was the first femalemember of the RoyalStatistical Society
History of theNormal
Distribution
Jenny Kenkel
Galton
Quetelet’s discoverybecame very popular inthe social sciences
Galton noticed the normalcurve appearing in heights,chest measurements, examscores, weight of sweetpeas
Galton was one of the firststatisticians to use theterm “normal” to describethe curve outlined by DeMoivre and Gauss
History of theNormal
Distribution
Jenny Kenkel
Galton
History of theNormal
Distribution
Jenny Kenkel
Pearson
In 1893, Pearson called the curve’ the normal curve(rather than the Gaussian).
He was trying to give some credit to Laplace, rather thanGauss- not even realizing that DeMoivre had discovered itearlier!
In 1900, Pearson developed a method of testing goodnessof fit
History of theNormal
Distribution
Jenny Kenkel
Pearson
In 1893, Pearson called the curve’ the normal curve(rather than the Gaussian).
He was trying to give some credit to Laplace, rather thanGauss- not even realizing that DeMoivre had discovered itearlier!
In 1900, Pearson developed a method of testing goodnessof fit
History of theNormal
Distribution
Jenny Kenkel
Pearson
In 1893, Pearson called the curve’ the normal curve(rather than the Gaussian).
He was trying to give some credit to Laplace, rather thanGauss- not even realizing that DeMoivre had discovered itearlier!
In 1900, Pearson developed a method of testing goodnessof fit
History of theNormal
Distribution
Jenny Kenkel
Value at Risk Model
Value at Risk is a way of quantifying risk of investmentinto a single number
For example, if a VaR of 5% is $1 million, that means theprobability the portfolio value will decrease more than $ 1million is 5%
The original VaR measurement used the normaldistribution to model amount a portfolio would gain or lose
This assumption contributed to the housing crisis.
History of theNormal
Distribution
Jenny Kenkel
Everyone Loves The Normal Distribution
History of theNormal
Distribution
Jenny Kenkel
Anders Hald. History of Probability and Statistics andTheir Applications before 1750. John Wiley & Sons, Inc.,New Jersey, 2003.
Anders Hald. A History of Mathematical Statistics From1750 to 1930. John Wiley & Sons, Inc., New York, 1998.
Norman L. Johnson and Samual Kotz, editors. LeadingPersonalities in Statistical Sciences from teh seventeenthcentury to the present. John Wiley & Sons, Inc., NewYork, 1997.
Stephen M. Stigler. The History of Statistics: TheMeasurement of Uncertainty before 1900. The BelknapPress of Harvard University Press, Massachusetts. 1986.
Saul Stahl. The Evolution of the Normal Distribution.Mathematics Magazine, 96-113.