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# FEC 512.04

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• 1. Introduction to Sampling Distributions and Estimating Population Values Istanbul Bilgi University FEC 512 Financial Econometrics-I Dr. Orhan Erdem
• 2. Unbiasedness ˆ A point estimator θ is said to be an unbiased estimator of the parameter θ if the expected value, or mean, of the ˆ sampling distribution of θ is θ, ˆ E(θ) = θ Examples: The sample mean is an unbiased estimator of µ The sample variance is an unbiased estimator of σ2 Lecture 4- 2 FEC 512
• 3. Unbiasedness (continued) ˆis ˆ θ1 an unbiased estimator, θ 2 is biased: ˆ ˆ θ1 θ2 ˆ θ θ Lecture 4- 3 FEC 512
• 4. Bias ˆ Let θ be an estimator of θ ˆ The bias in θ is defined as the difference between its mean and θ ˆ ˆ Bias(θ) = E(θ) − θ The bias of an unbiased estimator is 0 Lecture 4- 4 FEC 512
• 5. Consistency ˆ Let θ be an estimator of θ ˆ θ is a consistent estimator of θ if the ˆ difference between the expected value of θ and θ decreases as the sample size increases Consistency is desired when unbiased estimators cannot be obtained Lecture 4- 5 FEC 512
• 6. Most Efficient Estimator Suppose there are several unbiased estimators of θ The most efficient estimator or the minimum variance unbiased estimator of θ is the unbiased estimator with the smallest variance ˆ ˆ Let θ1 and θ2 be two unbiased estimators of θ, based on the same number of sample observations. Then, ˆ ˆ θ1 is said to be more efficient than θ 2 if ˆ ˆ Var(θ ) < Var(θ ) 1 2 Lecture 4- 6 FEC 512
• 7. Sampling Distribution A sampling distribution is a distribution of the possible values of a statistic for a given size sample selected from a population Lecture 4- 7 FEC 512
• 8. Developing a Sampling Distribution Assume there is a population … D Population size N=4 C A B Random variable, X, is age of individuals Values of X: 18, 20, 22, 24 (years) Lecture 4- 8 FEC 512
• 9. Developing a Sampling Distribution (continued) Summary Measures for the Population Distribution: ∑X P(x) µ= i N .25 18 + 20 + 22 + 24 = = 21 4 0 ∑ (X − µ) 2 x 18 20 22 24 σ= = 2.236 i A B C D N Uniform Distribution Lecture 4- 9 FEC 512
• 10. Developing a Sampling Distribution (continued) Now consider all possible samples of size n = 2 1st 2nd Observation 16 Sample Obs 18 20 22 24 Means 18 18,18 18,20 18,22 18,24 1st 2nd Observation 20 20,18 20,20 20,22 20,24 Obs 18 20 22 24 22 22,18 22,20 22,22 22,24 18 18 19 20 21 24 24,18 24,20 24,22 24,24 20 19 20 21 22 22 20 21 22 23 16 possible samples (sampling with 24 21 22 23 24 replacement) Lecture 4- 10 FEC 512
• 11. Developing a Sampling Distribution (continued) Sampling Distribution of All Sample Means Sample Means 16 Sample Means Distribution 1st 2nd Observation _ P(X) Obs 18 20 22 24 .3 18 18 19 20 21 .2 20 19 20 21 22 .1 22 20 21 22 23 _ 0 24 21 22 23 24 18 19 20 21 22 23 24 X (no longer uniform) 4- 11 Lecture FEC 512
• 12. Developing a Sampling Distribution (continued) Summary Measures of this Sampling Distribution: ∑X 18 + 19 + 21+ L + 24 E(X) = = = 21 = µ i N 16 ∑ ( Xi − µ)2 σX = N (18 - 21)2 + (19 - 21)2 + L + (24 - 21)2 = = 1.58 16 Lecture 4- 12 FEC 512
• 13. Comparing the Population with its Sampling Distribution Population Sample Means Distribution n=2 N=4 µX = 21 σ X = 1.58 µ = 21 σ = 2.236 _ P(X) P(X) .3 .3 .2 .2 .1 .1 _ 0 0 X 18 19 20 21 22 23 24 18 20 22 24 X A B C D Lecture 4- 13 FEC 512
• 14. Sampling in Excel Tools/Data Analysis/Sampling Lecture 4- 14 FEC 512
• 15. Histogram of 500 Sample Means from Sample Size n=10 Mean of the Sample Means is 2.41 with 0.421 St.Dev. σ 1.507 = = 0.477 where n 10 Lecture 4- 15 FEC 512
• 16. Mean of the Sample Means is 2.53 with 0.376 St.Dev. σ 1.507 = = 0.337 where n 20 Lecture 4- 16 FEC 512
• 17. Properties of a Sampling Distribution For any population, the average value of all possible sample means computed from all possible random samples of a given size from the population is equal to the population mean: µx = µ Theorem 1 The standard deviation of the possible sample means computed from all random samples of size n is equal to the population standard deviation divided by the square root of the sample size: σ σx = Theorem 2 n Lecture 4- 17 FEC 512
• 18. If the Population is Normal If a population is normal with mean µ and standard deviation σ, the sampling distribution x of is also normally distributed with σ σx = µx = µ and n Theorem 3 Lecture 4- 18 FEC 512
• 19. z-value for Sampling Distribution of x Z-value for the sampling distribution of x : ( x − µ) z= σ n x = sample mean where: µ = population mean σ = population standard deviation n = sample size Lecture 4- 19 FEC 512
• 20. Sampling Distribution Properties The sample mean is an unbiased estimator Normal Population Distribution x µ µx = µ Normal Sampling Distribution (has the same mean) µx x Lecture 4- 20 FEC 512
• 21. Sampling Distribution Properties (continued) The sample mean is a consistent estimator (the value of x becomes closer to µ as n Population increases): x Small sample size As n increases, x σ x = σ/ n Larger sample size decreases x µ Lecture 4- 21 FEC 512
• 22. If the Population is not Normal We can apply the Central Limit Theorem: Even if the population is not normal, …sample means from the population will be approximately normal as long as the sample size is large enough …and the sampling distribution will have σ σx = µx = µ n and Theorem 4 Lecture 4- 22 FEC 512
• 23. Central Limit Theorem the sampling As the n↑ distribution sample becomes size gets almost normal large regardless of enough… shape of population x Lecture 4- 23 FEC 512
• 24. If the Population is not Normal (continued) Population Distribution Sampling distribution properties: Central Tendency µx = µ µ x Sampling Distribution Variation σ (becomes normal as n increases) σx = Larger n Smaller sample sample size size (Sampling with replacement) x µx Lecture 4- 24 FEC 512
• 25. How Large is Large Enough? For most distributions, n > 25 will give a sampling distribution that is nearly normal For fairly symmetric distributions, n > 15 is sufficient For normal population distributions, the sampling distribution of the mean is always normally distributed Lecture 4- 25 FEC 512
• 26. Example Suppose a population has mean µ = 8 and standard deviation σ = 3. Suppose a random sample of size n = 36 is selected. What is the probability that the sample mean is between 7.8 and 8.2? Lecture 4- 26 FEC 512
• 27. Example (continued) Solution: Even if the population is not normally distributed, the central limit theorem can be used (n > 30) … so the sampling distribution of x is approximately normal µx = µ = 8 … with mean σ 3 σx = = = 0.5 …and standard deviation n 36 Lecture 4- 27 FEC 512
• 28. Example (continued) Solution (continued) -- find z-scores:    7.8 - 8 8.2 - 8  µx -µ P(7.8 < µ x < 8.2) = P < <  3 σ 3    36  36 n = P(-0.4 < z < 0.4) = 0.3108 Population Sampling Standard Normal Distribution Distribution Distribution .1554 ??? +.1554 ? ?? ? ?? Sample Standardize ? ? ? -0.4 0.4 µz = 0 7.8 8.2 z x x µ=8 µx = 8 Lecture 4- 28 FEC 512
• 29. Suppose that Y1, Y2 Y3... Yn are i.i.d., and let µx and σx2 denote the mean and the variance of Yi. n 1 ∑ E (Yi ) = µ Y E (Y ) = n i =1 1n V a r (Y ) = V a r ( ∑ Yi ) n i =1 1n 1n n = 2 ∑ V a r (Yi ) + 2 ∑ ∑ C o v (YiY j ) n i =1 n i =1 j = 1, j ≠ i σ Y2 = n Lecture 4- 29 FEC 512
• 30. Point and Interval Estimates A point estimate is a single number, a confidence interval provides additional information about variability Upper Lower Confidence Confidence Point Estimate Limit Limit Width of confidence interval Lecture 4- 30 FEC 512
• 31. Confidence Intervals How much uncertainty is associated with a point estimate of a population parameter? An interval estimate provides more information about a population characteristic than does a point estimate Such interval estimates are called confidence intervals Never 100% sure: “The surer we want to be, the less we have to be sure of” -Freund and Williams(1977)- Lecture 4- 31 FEC 512
• 32. Estimation Process I am 95% Random Sample confident that µ is between Population Mean 40 & 60. (mean, µ, is x = 50 unknown) Sample Lecture 4- 32 FEC 512
• 33. General Formula The general formula for all confidence intervals is: Point Estimate ± (Critical Value)(Standard Error) Lecture 4- 33 FEC 512
• 34. Confidence Level, (1-α) (continued) Suppose confidence level = 95% Also written (1 - α) = .95 A relative frequency interpretation: In the long run, 95% of all the confidence intervals that can be constructed will contain the unknown true parameter A specific interval either will contain or will not contain the true parameter No probability involved in a specific interval Lecture 4- 34 FEC 512
• 35. Confidence Interval for µ (σ Known) Assumptions Population standard deviation σ is known Population is normally distributed If population is not normal, use large sample σ x ± z α/2 Confidence interval estimate n Lecture 4- 35 FEC 512
• 36. Finding the Critical Value Consider a 95% confidence interval: z α/2 = ± 1.96 1 − α = .95 α α = .025 = .025 2 2 z.025= -1.96 z.025= 1.96 z units: 0 Lower Upper x units: Point Estimate Confidence Confidence Limit Limit Lecture 4- 36 FEC 512
• 37. Common Levels of Confidence Commonly used confidence levels are 90%, 95%, and 99% Confidence Confidence z value, Coefficient, z α/2 Level 1− α 80% .80 1.28 90% .90 1.645 95% .95 1.96 98% .98 2.33 99% .99 2.57 99.8% .998 3.08 99.9% .999 3.27 Lecture 4- 37 FEC 512
• 38. Interval and Level of Confidence Sampling Distribution of the Mean 1− α α/2 α/2 x Intervals µx = µ extend from x1 100(1-α)% σ x2 x + z α/2 of intervals n constructed to contain µ; σ x − z α/2 100α% do not. n Confidence Intervals Lecture 4- 38 FEC 512
• 39. Margin of Error Margin of Error (e): the amount added and subtracted to the point estimate to form the confidence interval Example: Margin of error for estimating µ, σ known: σ σ x ± z α/2 e = z α/2 n n Lecture 4- 39 FEC 512
• 40. Factors Affecting Margin of Error σ e = z α/2 n Data variation, σ : e as σ Sample size, n : e as n Level of confidence, 1 - α : if 1 - α e Lecture 4- 40 FEC 512
• 41. Confidence Interval for µ (σ Unknown) If the population standard deviation σ is unknown, we can substitute the sample standard deviation, s This introduces extra uncertainty, since s is variable from sample to sample So we use the t distribution instead of the normal distribution Lecture 4- 41 FEC 512
• 42. Confidence Interval for µ (σ Unknown) (continued) Assumptions Population standard deviation is unknown Population is normally distributed If population is not normal, use large sample (if you have a large sample you can still use z dist.) Use Student’s t Distribution s Confidence Interval Estimate x ± t α/2 n Lecture 4- 42 FEC 512
• 43. Student’s t Distribution The t is a family of distributions The t value depends on degrees of freedom (d.f.) Number of observations that are free to vary after sample mean has been calculated d.f. = n - 1 Lecture 4- 43 FEC 512
• 44. Degrees of Freedom (df) Idea: Number of observations that are free to vary after sample mean has been calculated Example: Suppose the mean of 3 numbers is 8.0 Let x1 = 7 If the mean of these three Let x2 = 8 values is 8.0, What is x3? then x3 must be 9 (i.e., x3 is not free to vary) Here, n = 3, so degrees of freedom = n -1 = 3 – 1 = 2 (2 values can be any numbers, but the third is not free to vary for a given mean) Lecture 4- 44 FEC 512
• 45. Student’s t Distribution Note: t z as n increases Standard Normal (t with df = ∞) t (df = 13) t-distributions are bell- shaped and symmetric, but have ‘fatter’ tails than the t (df = 5) normal t 0 Lecture 4- 45 FEC 512
• 46. Student’s t Table Upper Tail Area Let: n = 3 df = n - 1 = 2 df .25 .10 .05 α = .10 α/2 =.05 1 1.000 3.078 6.314 2 0.817 1.886 2.920 α/2 = .05 3 0.765 1.638 2.353 The body of the table 0 2.920 t contains t values, not probabilities Lecture 4- 46 FEC 512
• 47. t distribution values With comparison to the z value Confidence t t t z Level (10 d.f.) (20 d.f.) (30 d.f.) ____ .80 1.372 1.325 1.310 1.28 .90 1.812 1.725 1.697 1.64 .95 2.228 2.086 2.042 1.96 .99 3.169 2.845 2.750 2.57 Note: t z as n increases Lecture 4- 47 FEC 512
• 48. Example A random sample of n = 25 has x = 50 and s = 8. Form a 95% confidence interval for µ d.f. = n – 1 = 24, so t n−1,α/2 = t 24,.025 = 2.0639 The confidence interval is S S x − t n-1,α/2 < µ < x + t n-1,α/2 n n 8 8 50 − (2.0639) < µ < 50 + (2.0639) 25 25 46.698 < µ < 53.302 Lecture 4- 48 FEC 512
• 49. Example A money manager wants to obtain a 95% CI for fund inflows and outflows over the future. He calls a random sample of 10 clients enquiring about their planned additions to and withdrawals from the fund. He computes that there will be an average of 5.5m cash inflows with 10m standard deviation. A histogram of past data looks fairly normal. Calculate a 95% CI for the population mean. Lecture 4- 49 FEC 512
• 50. Solution s 10 x ± t0.025 = 5.5 ± 2.262 = 5.5 ± 7.15% n 10 The CI for the population means spans -1.65m to 12.65m. The manager can be confident at the 95% level that this range includes the population mean Lecture 4- 50 FEC 512
• 51. Approximation for Large Samples Since t approaches z as the sample size increases, an approximation is sometimes used when n ≥ 30: Technically Approximation correct for large n s s x ± t α/2 x ± z α/2 n n Lecture 4- 51 FEC 512
• 52. Example: Sharpe Ratio Suppose an investment advisor takes a random sample of stock funds and calculates the average Sharpe ratio(excess return/st.dev). The sample size is 100, and has a standard deviation of 0.30. If the average Sharpe ratio is 0.45, determine a 90% confidence interval for the true population mean of Sharpe ratio. Lecture 4- 52 FEC 512
• 53. Solution: (continued) s 0.30 x ± z0.025 = 0.45 ± 1.645 = 0.45 ± 1.645 * 0.03 n 100 = 0.401 : 0.499 Lecture 4- 53 FEC 512
• 54. Value At Risk Lecture 4- 54 FEC 512