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Lecture 6 Bootstraps Maximum Likelihood Methods

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Boostrapping A way to generate empirical probability distributions Very handy for making estimates of uncertainty

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100 realizations of a normal distribution p(y) with y=50 y =100

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What is the distribution of y est = i y i ? N 1

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We know this should be a Normal distribution with expectation=y=50 and variance= y / N=10 p(y) y p(y est ) y est

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Here’s an empirical way of determining the distribution called bootstrapping

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y1y2y3y4y5y6y7…yNy1y2y3y4y5y6y7…yN y’ 1 y ’ 2 y ’ 3 y ’ 4 y ’ 5 y ’ 6 y ’ 7 … y ’ N 4 3 7 11 4 1 9 … 6 N original data Random integers in the range 1-N N resampled data N 1 i y’ i Compute estimate Now repeat a gazillion times and examine the resulting distribution of estimates

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Note that we are doing random sampling with replacement of the original dataset y to create a new dataset y’ Note: the same datum, y i, may appear several times in the new dataset, y’

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pot of an infinite number of y’s with distribution p(y) cup of N y’s drawn from the pot Does a cup drawn from the pot capture the statistical behavior of what’s in the pot?

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More or less the same thing in the 2 pots ? Take 1 cup p(y) Duplicate cup an infinite number of times Pour into new pot p(y)

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Random sampling easy to code in MatLab yprime = y(unidrnd(N,N,1)); vector of N random integers between 1 and N original data resampled data

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The theoretical and bootstrap results match pretty well ! theoretical Bootstrap with 10 5 realizations

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Obviously bootstrapping is of limited utility when we know the theoretical distribution (as in the previous example)

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but it can be very useful when we don’t for example what’s the distribution of y est where ( y est ) 2 = 1/(N-1) i (y i -y est ) 2 and y est = (1/N) i y i ( Yes, I know a statistician would know it follows Student’s T-distribution …)

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To do the bootstrap we calculate y’ est = (1/N) i y’ i ( y’ est ) 2 = 1/(N-1) i (y’ i -y’ est ) 2 and y’ est = ( y’ est ) 2 many times – say 10 5 times

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Here’s the bootstrap result … Bootstrap with 10 5 realizations y true I numerically calculate an expected value of 92.8 and a variance of 6.2 Note that the distribution is not quite centered about the true value of 100 This is random variation. The original N=100 data are not quite representative of the an infinite ensemble of normally- distributed values p y est ) y est

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So we would be justified saying y 92.6 ± 12.4 that is, 2 6.2, the 95% confidence interval

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The Maximum Likelihood Distribution A way to fit parameterized probability distributions to data very handy when you have good reason to believe the data follow a particular distribution

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Likelihood Function, L The logarithm of the probable-ness of a given dataset

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N data y are all drawn from the same distribution p(y) the probable-ness of a single measurement y i is p(y i ) So the probable-ness of the whole dataset is p(y 1 ) p(y 2 ) … p(y N ) = i p(y i ) L = ln i p(y i ) = i ln p(y i )

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Now imagine that the distribution p(y) is known up to a vector m of unknown parameters write p(y; m) with semicolon as a reminder that its not a joint probabilty The L is a function of m L(m) = i ln p(y i ; m)

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The Principle of Maximum Likelihood Chose m so that it maximizes L(m) L/ m i = 0 the dataset that was in fact observed is the most probable one that could have been observed

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Example – normal distribution of unknown mean y and variance 2 p(y i ) = (2 ) -1/2 -1 exp{ -½ -2 (y i -y) 2 } L = i ln p(y i ) = -½Nln(2 ) –Nln( ) -½ -2 i (y i -y) 2 L/ y = 0 = -2 i (y i -y) L/ = 0 = - N -1 + -3 i (y i -y) 2 N’s arise because sum is from 1 to N

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Solving for y and 0 = -2 i (y i -y) y = N -1 i y i 0 = -N -1 + -3 i (y i -y) 2 2 = N -1 i (y i -y) 2

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y = N -1 i y i 2 = N -1 i (y i -y) 2 Sample mean is the maximum likelihood estimate of the expected value of the normal distribution Sample variance (more-or-less*) is the maximum likelihood estimate of the variance of the normal distribution * issue of N vs. N-1 in the formula Interpreting the results

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Example – 100 data drawn from a normal distribution true y=50 =100

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L(y, ) y max at y=62 =107

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Another Example – exponential distribution p(y i ) = ½ -1 exp{ - -1 |y i -y| } Check normalization … use z= y i -y p(y i )dy = ½ -1 - + exp{ - -1 |y i -y| } dy i = ½ -1 2 0 + exp{ - -1 z } dz = -1 (- ) exp{- -1 z}| 0 + = 1 Is this parameter really the expectation ? Is this parameter really variance ?

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Is y the expectation ? E(y i ) = - + y i ½ -1 exp{ - -1 |y i -y| } dy i use z= y i -y E(y i ) = ½ -1 - + (z+y) exp{ - -1 |z| } dz = ½ -1 2 y o + exp{ - -1 z } dz = - y exp{ - -1 z }| o + = y z exp(- -1 |z|) is odd function times even function so integral is zero YES !

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Is the variance ? var(y i ) = - + (y i -y) 2 ½ -1 exp{ - -1 |y i -y| } dy i use z= -1 (y i -y) E(y i ) = ½ -1 - + 2 z 2 exp{ -|z| } dz = 2 0 + z 2 exp{ -z } dz = 2 2 2 CRC Math Handbook gives this integral as equal to 2 Not Quite …

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Maximum likelihood estimate L = Nln(½) – Nln( ) - -1 i |y i -y| L/ y = 0 = - -1 i sgn (y i -y) L/ = 0 = - N -1 + -2 i |y i -y| y such that i sgn (y i -y) = 0 x |x| x d|x|/dx +1 Zero when half the y i ’s bigger than y, half of them smaller y is the median of the y i ’s

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Once y is known then … L/ = 0 = - N -1 + -2 i |y i -y| = N -1 i |y i -y| with y = median(y) Note that when N is even, y is not unique, but can be anything between the two middle values in a sorted list of y i ’s

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Comparison Normal distribution: best estimate of expected value is sample mean Exponential distribution best estimate of expected value is sample median

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Comparison Normal distribution: short tailed outlier extremely uncommon expected value should be chosen to make outliers have as small a deviation as possible Exponential distribution: relatively long-tailed outlier relatively common expected value should ignore actual value of outliers yiyi medianmean outlier yiyi medianmean

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another important distribution Gutenberg-Richter distribution (e.g. earthquake magnitudes) for earthquakes greater than some threshhold magnitude m 0, the probability that the earthquake will have a magnitude greater than m is –b (m-m 0 ) or P(m) = exp{ – log(10) b (m-m 0 ) } = exp{-b’ (m-m 0 ) } with b’= log(10) b P(m)=10

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This is a cumulative distribution, thus the probability that magnitude is greater than m 0 is unity P(m) = exp{ –b’ (m-m 0 ) } = exp{0} = 1 Probability density distribution is its derivative p(m) = b’ exp { –b’ (m-m 0 ) }

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Maximum likelihood estimate of b’ is L(m) = N log(b’) – b’ i (m i -m 0 ) L/ b’ = 0 = N/b’ - i (m i -m 0 ) b’ = N / i (m i -m 0 )

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Originally Gutenberg & Richter made a mistake … magnitude, m Log 10 P(m) slope = -b … by estimating slope, b using least-squares, and not the Maximum Likelihood formula least-squares fit

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yet another important distribution Fisher distribution on a sphere (e.g. paleomagnetic directions) given unit vectors x i that scatter around some mean direction x, the probability distribution for the angle between x i and x (that is, cos( )=x i x) is p( ) = sin( ) exp{ cos( ) } 2 sinh( ) is called the “precision parameter”

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Rationale for functional form p( ) exp{ cos( ) } For close to zero 1 – ½ 2 so p( ) exp{ cos( ) } = exp{ exp{ – ½ 2 } which is a gaussian

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I’ll let you figure out the maximum likelihood estimate of the central direction, x, and the precision parameter,

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