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Section III Gaussian distribution Probability distributions (Binomial, Poisson)

Section III Gaussian distribution Probability distributions (Binomial, Poisson). Notation Statistic Sample Population mean Y μ Std deviation S or SD σ proportion P π

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Section III Gaussian distribution Probability distributions (Binomial, Poisson)

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  1. Section III Gaussian distribution Probability distributions (Binomial, Poisson)

  2. Notation Statistic Sample Population mean Y μ Std deviation S or SD σ proportion P π mean difference d δ Correlation coeff r ρ rate (regression) b β Num of obs n N

  3. Densities –PercentilesBMI=22 is the 88th percentile

  4. Standard Z scores Definition: Z = (Y – mean)/ SD Y = mean + Z SD Z is how many SD units Y is above or below mean. Mean & SD might be sample (Y, S) or population (μ,σ) values if population values are known.

  5. Survival data, mean=17.54, SD=11.68

  6. Standard Gaussian (Normal)a distribution model

  7. Selected Gaussian percentiles Z lower area (P <Z) -2.00 2.28% -1.96 2.50% -1.50 6.68% -1.00 15.87% 0.00 50.00% 1.00 84.13% 1.50 93.32% 1.96 97.50% 2.00 97.72%

  8. Gaussian percentiles

  9. Example- SAT Verbal Mean=μ=500, SD=σ=100 What is your percentile if Y=700? Z= (700-500)/100=2.0, area=0.977=97.7% What score is the 80th percentile, Z0.80=0.842 Y = 500 + 0.842 (100) = 584 What percent are between 450 and 500? For Y=450, Z=(450-500)/100=-.5, area=0.3085 For Y=500, Z=0, area=0.5000, so area between is 0.500-0.3085=0.1915=19%

  10. Example- Anesthesia Effective dose, μ=50 mg/kg, σ=10 mg/kg Lethal dose, μ=110 mg/kg, σ=20 mg/kg Q1= What dose with put 90% to sleep? Q2- What is the risk of death from this dose?

  11. Example- Anesthesia Effective dose, μ=50 mg/kg, σ=10 mg/kg Lethal dose, μ=110 mg/kg, σ=20 mg/kg Q1= What dose with put 90% to sleep? Z0.90=1.28, Y=50+1.28 (10) = 62.8 mg/kg Q2- What is the risk of death from this dose? Z=(62.8-110)/20= -2.36, area < 1%

  12. Prediction intervals (not CI) If μ and σ are known and the data is known to have a Gaussian distribution, the interval formed by (μ-Zσ, μ+Zσ) is the (2k-100th) prediction interval for the kth percentile Z (Z>0). Z=2, (μ-2σ, μ+2σ) is (approximately) the 95% prediction interval Implies SD ≈ range/4 (extremes excluded)

  13. Specificity & Sensitivity For serum Creatinine in normal adults  = 1.1 mg/dl  = 0.2 mg/dl In one type of renal disease  = 1.7 mg/dl  = 0.4 mg/dl If a cutoff value of 1.6 mg/dl is used Prob false pos= prob Y > 1.6 given normal Prob false neg = prob Y < 1.6 given disease

  14. Data transformations & logs Some continuous variables follow the Gaussian on a transformed scale, not the original scale. Statland implies that perhaps 80% of continuous lab test variables follow a Gaussian on either the original (50%) or a transformed scale, usually the log scale. (Clinical Decision Levels for lab Tests, 2nd ed, 1987, Med Econ)

  15. Example-Bilirubin Bilirubin umol/L Log Bilirubin, log10 umol/L Mean=64.3 Median=34.7 SD=104.3 n=216 Mean=1.55 Median=1.54 SD=0.456 n=216

  16. 95% prediction intervals Original scalelog 10 scale Mean 64.3 1.55 SD 104.3 0.456 2 SD 208.6 0.912 Lower -144.3 0.64 Upper 272.9 2.46 ******************************************* Geometric mean=101.55=35.5 mmol/L Prediction interval (100.64,102.46) or (4.3, 290)

  17. Normal probability plotBilirubin – original scale Data is Gaussian if plot is a straight line- above not Gaussian

  18. Normal probability plotBilirubin- log scale Data is Gaussian if plot is a straight line as above

  19. Log transformation (cont). The distribution of ratios is much closer to Gaussian on the log scale The “inverse” of 3/1 is 1/3. This is symmetric only on the log scale Original: 100/1, 10/1, 1/1, 1/10, 1/100 Log: 2, 1, 0, -1, -2 true for OR, RR and HR Measures of growth & proliferation have distribution closer to the Gaussian on the log scale

  20. Data distributions that tend to be Gaussian on the log scale Growth measures - bacterial CFU Ab or Ag titers (IgA, IgG, …) pH Neurological stimuli (dB, Snellen units) Steroids, hormones (Estrogen, Testosterone) Cytokines (IL-1, MCP-1, …) Liver function (Bilirubin, Creatinine) Hospital Length of stay (can be Poisson)

  21. Quick Probability Theory Mutually exclusive events: levels of one variable Blood type probability A 30% B 12% AB 8% O 50% Probability A or O = 30% + 50%=80%. Mutually exclusive probabilitiesadd. All (exhaustive) categories sum to 100%

  22. Probability-Independent events The probabilities of two independent events multiply. (two or more variables) If 5% of pregnant women have gestational diabetes If 8% of pregnant women have pre-eclampsia Probability of gest. diabetes and pre- eclampsia = 5% x 8% = 0.4% if independent.

  23. Conditional probability Probability of an event changes if made conditional on another event. Probability (prevalence) of TB is 0.1% in general population. In Vietnamese immigrants, TB probability is 4%. Conditional on being a Vietnamese immigrant, probability is 4%.

  24. Conditional Probability & Bayes n=1,000,000 B=TB+ n=1000 A=Vietnamese n=5000 A∩B N=200 Want prob TB|Vietnamese but can’t check all Vietnamese for TB

  25. Conditonal Prob & Bayes RuleWhat is TB prevalence in Orange Co Vietnamese population? Too hard to take census of all Vietnamese. Assume we know: P(A)=prop in Orange Co who are Viet=0.5% P(B)=prop in Orange Co who have TB = 0.1% P(A|B)=prop of those with TB who are Viet=20% Want P(B|A) = P(A|B) P(B)/ P(A) = (0.2 x 0.001)/(0.005) = 0.04=4%

  26. Bayes rule for conditional probability (formula) Probability of B given A = P(B|A)= Joint probability of A and B/Probability of A= P(A ∩ B)/P(A) = Probability of A given B x Probability of B Probability of A P(B|A)=[ P(A|B)P(B)] / P(A) If A and B are independent, P(B|A)=P(B) Also P(B) = ∑ P(B|Ai) (sum over all Ai)

  27. Example: Bayes rule A=Vietnamese, B=TB+ In pop of 1,000,000, 5000 (0.5%=0.005) are Vietnamese=P(A), 1000 (0.1%=0.001) have TB+ =P(B). Of 1000 with TB+, 200 (20%=0.20) are Vietnamese=P(A|B) Want prob. of TB given Vietnamese? =P(B|A). P(B|A)= 0.20 (0.001)/0.005 = 0.04=4%. =200/5000 Can’t test all Viet for TB+, can check all TB+ for Viet

  28. Bayes rule (graph) 1,000,000 pop B Conditional probability of TB+ given Vietnamese = 200/5000=4% B|A 1000 TB+ A A ∩ B 5000 Viet 200 Viet + TB+ Check all TB+ for Viet rather than check all Viet for TB

  29. BayesianvsFrequentist Bayesian computes Prob(hypothesis|data) = Prob(data|hypothesis) P(hypothesis) Prob(data) = DataLikelihood x prior probability If data (evidence) refutes a hypothesis Prob(data | hypothesis)=0 so Prob(hypothesis | data)=0 Frequentist computes Prob(data*|hypothesis)= p value * p value is prob of observed data or more extreme data

  30. Binomial distribution Population: Positive= π = 0.30, negative = 1- π = 0.70 Y= number of positive responses out of n trials n=1 Y probability 0 0.700 1 0.300 • n=2 • Y probability • 0 0.49=0.7 x 0.7 • 0.42= 0.7 x 0.3 x 2 • 2 0.09= 0.3 x 0.3

  31. Binomial (cont.) • n=3 • Y probability • 0 0.343 • 0.441 • 0.189 • 3 0.027 • n=4 • Y probability • 0 0.2401 • 0.4116 • 0.2646 • 0.0756 • 4 0.0810

  32. General binomial formula Probability of y positive out of n where π is prob of a single positive = n!/[y!(n-y)!] πy (1-π)(n-y) Mean=πn, SD=√nπ(1-π) Ex:Prob of y=5 herpes cases out n=50 teens if herpes incidence=π=4%=0.04 Prob=50!/(5! 45!)(0.04)5(0.96)45=3.4% Can compute using “=Binomdist(y,n,π,0)” in EXCEL

  33. Binomial-fair coin example for π=0.5, easy to compute y=number of “heads” (success) out of n prob y out of n = n!/[y!(n-y)!] / 2n Ex: n=3, flip 3 fair coins, 23=8 possibilities 0+0+0=0=y y freq prob 0+0+1=1=y 0 1 1/8 0+1+0=1=y 1 3 3/8 1+0+0=1=y 2 3 3/8 0+1+1=2=y 3 1 1/8 1+0+1=2=y total 8 8/8 1+1+0=2=y 1+1+1=3=y

  34. Pascal’s triangle n y: 0 to n “success” 2n - 1 1 1 1 2 2 1 2 1 4 3 1 3 3 1 8 4 1 4 6 4 1 16 5 1 5 10 10 5 1 32 For n=5, prob(y=2) is 10/32 prob(y≤2) is (1+5+10)/32=16/32

  35. Headache remedy success The “old” headache remedy was successful π=50% of the time, a true “population” value well established after years of study. A “new” remedy is tried in 10 persons and is successful in 7 of the 10 (70%).

  36. Hypothesis testing-Binomial How likely is y=7 success out of n=10 if π=0.5, prob = 10!/(7!3!) / 210 = 120/1024=0.1172 How likely y=7 or more? y probability 7 120/1024 = 0.1172 8 45/1024 = 0.0439 9 10/1024 = 0.0098 10 1/1024 = 0.0010 total 176/1024= 0.1719

  37. How likely is observing y=70 success out of n=100 if π=0.5 for each trial? Prob(y=70)=[100!/(70! 30!)] / 2100 = 2.32 x 10-5 How likely is it to observe 70 or more successes out of 100? pr(y=70) + pr(y=71) + …+pr(y=100) = 3.93 x 10-5 This is a simple example of hypothesis testing. The probability of observing y=70 or more successes out of n=100 under the “null hypothesis” that the true population π=0.5 is called a one sided p value.

  38. Gaussian approximation to Binomialok for large n, π not near 0 or 1 π =0.15, n=50, mean=0.15(50)=7.5, SD=√50(0.15)(0.85)=2.52 Actual 2.5th percentile is between 2 & 3, Gaussian 7.5-2(2.5)=2.5 Actual 97.5th percentile is between 12 and 13, Gaussian=7.5 +2(2.5)=12.5

  39. Poisson dist for count data For a patient, y is a positive integer: 0,1,2,3,… Probability of “y” responses (or events) given mean μ = (μy e-μ)/ (y!) (Note: μ0=1 by definition) For Poisson, if mean=μ then SD=√μ

  40. Poisson example: Q: If average num colds in a single winter is μ=1.9, what is the probability that a given patient will have 4 colds in one winter? A: (1.9)4e-1.9/4x3x2x1 = 0.0812 ≈ 8%. What is the probability of 4 or more (find for 0-3, subtract from 1), prob=12%

  41. Poisson distribution

  42. Poisson process Mean rate of events is h events/unit=h (Hazard rate). In T units, we expect μ=hT events on average. Can substitute into (μy e-μ)/ (y!) to get probability of “y” events in T units.

  43. Poisson process example Example: Cancer clusters Q: Given a cancer rate of h=3/1000 person-years, what is the expected number of cases in 2 years in a population of 1500? A: Rate in 2 years is 2 x (3/1000) =h= 6/1000. Expected is μ=hT= 6/1000 x 1500 = 9 cases. Q: What is the probability of observing exactly 15 cases? A: μ=9, Probability =(915 e-9)/15! = 0.019431≈ 2%. Q: What is the probability of observing 15 or more cases in 1500 persons? A: Plug in 0,1,2, …14 and add to get Q= probability of 14 or less. Probability is 1-Q = 1-0.958534 = 0.041466 ≈ 4%. Can compute with “=Poisson(y,μ,0)” in EXCEL.

  44. Summary: Descriptive stats for Normal, Binomial & Poisson n = sample size Distribution mean variance SD SE Normal µ σ2 σ σ/√n Binomial π π(1-π) √π(1-π) √π(1-π)/n Poisson µ µ √µ √µ/n SD = √variance, SE= SD/√n

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