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    Probability primer

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    1/08/13   Probability Primer Principles of Econometrics‚ 4th Edition Probability Primer Page 1 !   Announcement: !   Please make sure you know who your tutor is and remember their names. This will save confusion and embarrassment later. !   Kai Du (David) !   Ngoc Thien Anh Pham (Anh) !   Zara Bomi Shroff Principles of Econometrics‚ 4th Edition Probability Primer Page 2 Chapter Contents ¡  P.1 Random Variables ¡  P.2 Probability Distributions ¡  P.3 Joint

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    Sta304

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    STA304 H1 S/1003 H S Winter 2013 Dragan Banjevic (I) Note: A lot of material will be used from Internet‚ some with reference‚ some without. 2 CITY OF TORONTO NEIGHBOURHOODS 1 West Humber-Clairville 19 Long Branch 36 Newtonbrook West 54 O’Connor-Parkview 2 Mount Olive-SilverstoneJamestown 20 Alderwood 37 Willowdale West 55 Thorncliffe Park 3 Thistletown-Beaumond Heights 21 Humber Summit 38 Lansing-Westgate 56 Leaside-Bennington 4 Rexdale-Kipling

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    Quantitative Techniques

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    increases 3 points‚ the median will become __. a. 21 b. 21.5 c. 24 d. Cannot be determined without additional information. e. none of these 3. If you are told a population has a mean of 25 and a variance of 0‚ what must you conclude? a. Someone has made a mistake. b. There is only one element in the population. c. There are no elements in the population. d. All the elements in the population are 25. e

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    Partner s healthcare

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    24.32% 3.15% Foreign Equity 14.44% 12.42% 30.87% 3.83% Bonds 11.10% 5.40% 10.83% 0.58% REITs 13.54% 9.44% 9.91% 0.94% Commodities 18.43% 10.05% 24.07% 2.42% Total 100.00% 10.92% 0.815960738 Portfolio Mean Return 10.92% Portoflio Variance 0.90% Portfolio S.D 9.46% Calculation of Covariance (Do Not Alter Formula)                               Correlation Matrix             US Equity Foreign Equity Bonds REITs Commodities US Equity 1 0.62 0.25 0.56 -0.02 Foreign

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    Risk & Return

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    stocks. b. Calculate the variance and standard deviation of the small company returns and large company common returns. 5. The table below provides a probability distribution for the returns on stocks A and B State Probability Return On Stock A Return OnStock B 1 20% 5% 50% 2 30% 10% 30% 3 30% 15% 10% 4 20% 20% -10% a. Given a probability distribution of returns‚ calculate the expected return‚ variance‚ standard deviation of Stock

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    Bond Valuation

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    Assignment for Week -2 Chapter 5 (5 - 9) Bond Valuation and Interest Rate Risk Bond L Bond S INS = $100 INS = $100 M = $1‚000 M = $1‚000 N = 15 Years N = 1 Year a) 1) rd = 5% VBL = INT/ (1 + rd)t + M/ (1 + rd)N =INT [1/rd – 1/ rd(1 + rd)N ] + M/ (1 + rd)N =$100 [1/0.05 – 1/ 0.05(1 + 0.05)15] + $1‚000/ (1 + 0.05)15 =$1040 + $480.77 = $1518.98

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    weweqeweq

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    TUTORIAL 1 -95253936900 Each of the following processes involves sampling from a population. Define the population‚ and state whether it is tangible or conceptual. A shipment of bolts is received from a vendor. To check whether the shipment is acceptable with regard to shear strength‚ an engineer reaches into the container and selects 10 bolts‚ one by one to test. The resistance of a certain resistor is measured 5 times with the same ohmmeter 8 welds are made with the same process‚ and the

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    Problems on Risk and Return 1) Using the following returns‚ calculate the arithmetic average returns‚ the variances and the standard deviations for X and Y. Year X Y 1 8% 16% 2 21 38 3 17 14 4 -16 -21 5 9 26 2) You bought one of the Great White Shark Repellant Co’s 8 per cent coupon bonds one year ago for $1030. These bonds make annual payments and mature six years from now. Suppose you decide to sell your bonds today ‚when the required return on the bonds is 7 per cent

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    Normal Distribution

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    A population of measurements is approximately normally distributed with mean of 25 and a variance of 9. Find the probability that a measurement selected at random will be between 19 and 31. Solution: The values 19 and 31 must be transformed into the corresponding z values and then the area between the two z values found. Using the transformation formula from X to z (where µ = 25 and σ √9 = 3)‚ we have z19 = (19 – 25) / 3 = -2 and z31 = (31 - 25) / 3 = +2 From the area between z =±2 is 2(0

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    Answer Key

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    FALSE For a random variable X‚ let µ = E (X). The variance of X can be expressed as: V ar(X) = E X 2 − µ2 7. TRUE For random variables Y and X‚ the variance of Y conditional on X = x is given by: V ar(Y |X = x) = E Y 2 |x − [E (Y |x)]2 8. TRUE An estimator‚ W ‚ of θ is an unbiased estimator if E (W ) = θ for all possible values of θ. 9. FALSE The central limit theorem states that the average from a random sample for any population (with finite variance) when it is standardized‚ by subtracting the mean

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