Question

1.3 A portfolio consists of 10 shares of stock A and 8 shares of stock B. The price of A has a mean of 10 and variance of 16, while the price of B has a mean of 12 and a variance of 9 The correlation between prices is 0.3. What are the mean and variance of the portfolio value? 2.1 2.1.a State the definition of the sampling distribution of the sample mean 21h Let's assume that we have a Bernoulli random variable X-1 with probability 0.6, X-0 with probability 0.4 mean when the sample size is 3 Develop the sampling distribution of the sample the mean and variance for the sample mean in question 2.1.b? 2.1.c What are 2.2 Let the random variable X be a random variable with a of u and a standard deviation mean of o. Let X1 be the mean of a sample of 36 observations randomly chosen from this population, population and be the mean of a sample of 49 observations randomly chosen from the same Evaluate the statement P(u- 0.2a <1<u+0.20)< P(u-0.2a<2<u+0.20) as to whether (Hint: which theorem should you use?) for it is true or false. Please explain in detail the reasons your evaluation results.

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Answer 1

1.3 Let P be random variable denoting Portfolio value Then P 10X 8Y where X random variable denoting price of A Y = random variable denoting price of B Then E(P) of portfolio value 10E(X) 8E(Y) mean = 10 10 + 8 12 196 Var (P) 102Var(X)82VarY) - 102Var(X) 160Cov(X, Y) 8Var (Y)+ 160ρνVar(X ) Var (Y) = 102 * 16 8 9160 0.3 V16* 9 2752 where p correlation coefficient between prices 0.3 2.1 The sampling distributionof a sample based on a mean is a probability distribution of sample mean large number of samples of size from a given population

2.1 S X1X2 + X3 where Xs SNBinomial (3, 0.6) X1X2X3 independent and Xi 1(0) with probability 0.6(0.4) there are sample mean 3 The sampling distribution of sample P(X 0) P(S 0) (0.4)3 0.064 mean is _ P(X 1/3) P(S 1 (0.6)(0.4)2 0.288 P(S 2) (0.6)2(0.4)0.432 2 P(X 2/3 P(X 1) P(S 3) (0.6)3 0.216 P(X ) 0 otherwise 1 E(S) 3 1 Var (S) 32 1 2.1c. E(X) x 3 x 0.6 0.6 1 Var (X x 3 x 0.6 x 0.4 0.08 2.2 We know if sufficiently large then n is Vn(X H) N(0, 1) (using central limit theorem) 6(X1 -) 1.2 P(u0.2 X1 <0.20) -1.2 < 24 (1.2) 10.7699 7(X2 - X2 < 0.2a) P 1.4< P(u -0.2 1.4 2(1.4) 1 0.8385 Hence true,