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22S6 - Numerical and data analysis techniques Mike Peardon School of Mathematics Trinity College Dublin Hilary Term 2012 Mike Peardon (TCD) 22S6 - Data analysis Hilary Term 2012 1 / 14 Sampling Mike Peardon (TCD) 22S6 - Data analysis Hilary Term 2012 2 / 14 Sample mean Sample mean For a sequence of n random numbers, {X1 , X2 , X3 , . . . Xn }. The sample mean is n 1X X̄(n) = Xi n i=1 X̄(n) is also a random number. If all entries have the same mean, μX then E[X̄(n) ] = n 1X n i=1 E[Xi ] = μX If all entries are independent and identically distributed then 1 σX̄2(n) = σX2 n Mike Peardon (TCD) 22S6 - Data analysis Hilary Term 2012 3 / 14 The law of large numbers Jakob Bernoulli: “Even the stupidest man — by some instinct of nature per se and by no previous instruction (this is truly amazing) — knows for sure the the more observations that are taken, the less the danger will be of straying from the mark”(Ars Conjectandi - 1713). But the strong law of large numbers was only proved in the 20th century (Kolmogorov, Chebyshev, Markov, Borel, Cantelli, . . . ). The strong law of large numbers If X̄(n) is the sample mean of n independent, identically distributed random numbers with well-defined expected value μX and variance, then X̄(n) converges almost surely to μX . P( lim X̄(n) = μX ) = 1 n→∞ Mike Peardon (TCD) 22S6 - Data analysis Hilary Term 2012 4 / 14 Example: exponential random numbers X 0.299921 1.539283 1.084130 1.129681 0.001301 1.238275 4.597920 0.679552 0.528081 1.275064 0.873661 1.018920 0.980259 1.115647 1.664513 0.340858 X̄(2) X̄(4) X̄(8) X̄(16) 0.919602 1.013254 1.106906 1.321258 0.619788 1.629262 2.638736 1.147942 0.901572 0.923931 0.946290 0.974625 1.047953 1.025319 1.002685 Mike Peardon (TCD) 22S6 - Data analysis Hilary Term 2012 5 / 14 The central limit theorem As the sample size n grows, the sample mean looks more and more like a normally distributed p random number with mean μX and standard deviation σX / n The central limit theorem (de Moivre, Laplace, Lyapunov,. . . ) The sample mean of n independent, identically distributed random numbers, each drawn from a distribution with expected value μX and standard deviation σX obeys Za −aσ +aσ 1 2 (n) lim P( p < X̄ − μX < p ) = p e−x / 2 dx n→∞ n n 2π −a Mike Peardon (TCD) 22S6 - Data analysis Hilary Term 2012 6 / 14 The central limit theorem (2) The law of large numbers tells us we can find the expected value of a random number by repeated sampling The central limit theorem tells us how to estimate the uncertainty in our determination when we use a finite (but large) sampling. The uncertainty falls with increasing sample size like Mike Peardon (TCD) 22S6 - Data analysis Hilary Term 2012 1 p n 7 / 14 The central limit theorem An example: means of bigger sample averages of a random number X with n = 1, 2, 5, 50 14 14 12 12 n=1 10 8 6 6 4 4 2 2 0 0 0.2 0.4 0.6 0.8 1 14 0 0.2 0.4 0.6 0.8 1 12 n=5 10 8 6 6 4 4 2 2 0 n=50 10 8 Mike Peardon (TCD) 0 14 12 0 n=2 10 8 0.2 0.4 0.6 0.8 1 0 0 0.2 0.4 0.6 0.8 22S6 - Data analysis 1 Hilary Term 2012 8 / 14 Confidence intervals The central limit theorem tells us that for sufficiently large sample sizes, all sample means are normally distributed. We can use this to estimate probabilities that the true expected value of a random number lies in a range. One sigma What is the probability a sample mean X̄ is more than one p standard deviation σX̄ = σX / n from the expected value μX ? If n is large, we have 1 P(−σX̄ < X̄ − μX < σX̄ ) = p 2π 1 Z e−x 2/ 2 dx = 68.3% −1 These ranges define confidence intervals . Most commonly seen are the 95% and 99% intervals Mike Peardon (TCD) 22S6 - Data analysis Hilary Term 2012 9 / 14 Confidence intervals (2) Most commonly seen are the 95%(2σ) and 99%(3σ) intervals. P −σX̄ P −2σX̄ P −3σX̄ P −4σX̄ P −5σX̄ P −10σX̄ < X̄ − μX < X̄ − μX < X̄ − μX < X̄ − μX < X̄ − μX < X̄ − μX < σX̄ < 2σX̄ < 3σX̄ < 4σX̄ < 5σX̄ < 10σX̄ 68.2% 95.4% 99.7% 99.994% 99.99994% 99.9999999999999999999985% The standard deviation is usually measured from the sample variance. Beware - the “variance of the variance” is usually large. Five-sigma events have been known ... Mike Peardon (TCD) 22S6 - Data analysis Hilary Term 2012 10 / 14 Sample variance With data alone, we need a way to estimate the variance of a distribution. This can be estimated by measuring the sample variance: Sample variance For n > 1 independent, identically distributed samples of a random number X, with sample mean X̄, the sample variance is n 1 X σ̄X2 = (Xi − X̄)2 n − 1 i=1 Now we quantify fluctuations without reference to (or without knowing) the expected value, μX . Note the n − 1 factor. One “degree of freedom” is absorbed into “guessing” the expected value of X Mike Peardon (TCD) 22S6 - Data analysis Hilary Term 2012 11 / 14 Student’s t-distribution In 1908, William Gosset, while working for Guinness in St.James’ Gate published under the pseudonym “Student” Computes the scaling to define a confidence interval when the variance and mean of the underlying distribution are unknown and have been estimated Student’s t-distribution fT (t) = p Γ( 2n ) π(n − 1)Γ( n−1 ) 2 t2 1+ −n/ 2 n−1 Used to find the scaling factor c(γ, n) to compute the γ confidence interval for the sample mean P(−cσ̄ < μX < cσ̄) = γ For n > 10, the t-distribution looks very similar to the normal distribution Mike Peardon (TCD) 22S6 - Data analysis Hilary Term 2012 12 / 14 Student’s t-distribution (2) fX(x) 0.4 0.2 0 -3 -2 -1 0 x 1 2 3 blue - normal distribution red - Student t with n = 2. Mike Peardon (TCD) 22S6 - Data analysis Hilary Term 2012 13 / 14 Student’s t-distribution (3) For example, with just 2 samples, the sample mean and variance can be computed but now the confidence levels are: P −σ̄X < X̄ − μX < σ̄X 50% P −2σ̄X < X̄ − μX < 2σ̄X 70.5% P −3σ̄X < X̄ − μX < 3σ̄X 79.5% P −4σ̄X < X̄ − μX < 4σ̄X 84.4% P −5σ̄X < X̄ − μX < 5σ̄X 87.4% P −10σ̄X < X̄ − μX < 10σ̄X 93.7% “Confidences” are much lower because variance is very poorly determined with only two samples. Mike Peardon (TCD) 22S6 - Data analysis Hilary Term 2012 14 / 14