\fAre you good at finance or computational finance staff? Can you provide me your personal e-mail, then I can pay you to do homework for me. Coursehero would take 60% of the price I provide. If you work directly with me, we don't have to worry about it. You can send me an attached file including your personal e-mail in it. f ( a ) is minimised with respect to 1. a) At the point that So, I proceed to establish the value of Here, a a , the gradient of f ( a )=0 . that meets the following criteria: f ' ( a ) =0. f ' ( a ) is the gradient. Expanding f (a ) : n Eq. 1 n f a xi a xi a xi a 2 i 1 i 1 n xi xi a a xi a i 1 n xi2 2axi a 2 i 1 Applying the summation operator inside the bracket gives the following: n Eq. 2 n f a x 2a xi na i 1 2 i 2 i 1 Note the following about Eq. 2. The summation operator does not affect coefficients. Thus: n n i 1 i 1 2axi 2a xi Summing a constant n times is equal to that constant multiplied by n. Thus: n a 2 na 2 i 1 Differentiating Eq. 2 with respect to a: Eq. 3 n df a f a 2 xi 2na da i 1 To find the point that minimises f ( a ) we equate Eq. 3 to zero. n f a 2 xi 2na 0 i 1 n a x i 1 i n x b) Using the same logic as in 1 a), we proceed as follows: Expanding f (b) : Eq. 4 f b E x b 2 E x 2 b2 2bx E x 2 b2 2bE x Note the following about Eq. 4. The expectation operator does not affect coefficients. Thus: E 2bx 2bE x The expected value of a constant is that constant. Thus: E b2 b 2 Differentiating Eq. 4 with respect to b: Eq. 5 f ' b 2b 2 E x To find the point that minimises f ( b ) we equate Eq. 5 to zero. f ' b 2b 2 E x 0 b E x c) Using the same logic as in 1 a), we proceed as follows: Expanding f () : Eq. 6 f E y x 2 E y 2 2 x 2 2 xy E y 2 2 E x 2 2 E xy Differentiating Eq. 6 with respect to : Eq. 7 f ' 2 E x 2 2 E xy f ( ) we equate Eq. 7 to zero. To find the point that minimises f ' 2 E x 2 E xy 0 E xy E x 2. a) Let L denote the likelihood function. n 1 L e i 1 x Eq. 8 1 1 exp n Since the value of 1 Log L nLog i 1 Log L x i 1 n i i that maximises Log(L). Eq. 9 n x i 1 i with respect to dLog L n 1 n 2 xi 0 d i 1 n x that maximises L is the same as that which maximises Log(L), we can proceed to find the value of Differentiating n and equating to zero: Eq. 10 Note, in Eq. 10, denotes the maximum likelihood estimator of . b) Let OI denote the observed information. It is calculated below. n xi d 2 Log ( L) n OI 2 2 i 1 3 2 d c) The 95% confidence interval (symmetric) is: 2 n 2 n 2 2 0.975,2 n 0.025,2 n 3.We have the following statistics from the sample: Item Sample mean (m) Sample count (n) Sample variance Sample standard deviation (s) Value 0.58 5 0.88 0.94 Since the sample size is small, we use the t-distribution to obtain the confidence interval. Degrees of freedom for t-distribution = n - 1 = 4. Alpha level of t-distribution = (1 - 0.95) 2 = 2.5%. Critical value of t-distribution (from table) = 2.776. Therefore, the 95% confidence interval is: x t s n 0.58 2.776 0.59;1.75 0.94 5 4.d) and e). > > > > library(quantmod) library(MASS) set.seed(123) getSymbols("B", from = "2012-01-01", to = "2012-12-31") [1] "B" Warning message: In download.file(paste(yahoo.URL, "s=", Symbols.name, "&a=", from.m, : downloaded length 14917 != reported length 200 > daily <- as.numeric(dailyReturn(Cl(B)))[-1] > fitdistr(daily,"t") m s df -0.00051 0.01374 7.54471 ( 0.00097) ( 0.00100) ( 3.34505) There were 17 warnings (use warnings() to see them) > tq <- qt(ppoints(daily), df = 7.54471) > qqplot(daily,tq) f ( a ) is minimised with respect to a , the gradient of f ( a )=0 . 1. a) At the point that So, I proceed to establish the value of Here, a that meets the following criteria: f ' ( a ) =0. f ' ( a ) is the gradient. f (a ) : Expanding n Eq. 1 n f a xi a xi a xi a 2 i 1 i 1 n xi xi a a xi a i 1 n xi2 2axi a 2 i 1 Applying the summation operator inside the bracket gives the following: n n i 1 Eq. 2 i 1 f a xi2 2a xi na 2 Note the following about Eq. 2. The summation operator does not affect coefficients. Thus: n n i 1 i 1 2axi 2a xi Summing a constant n times is equal to that constant multiplied by n. Thus: n a na 2 2 i 1 Differentiating Eq. 2 with respect to a: Eq. 3 n df a f a 2 xi 2na da i 1 To find the point that minimises n f a 2 xi 2na 0 i 1 n a x i 1 n i x f ( a ) we equate Eq. 3 to zero. b) Using the same logic as in 1 a), we proceed as follows: Expanding f (b) : Eq. 4 2 f b E x b E x 2 b2 2bx E x 2 b2 2bE x Note the following about Eq. 4. The expectation operator does not affect coefficients. Thus: E 2bx 2bE x The expected value of a constant is that constant. Thus: E b2 b 2 Differentiating Eq. 4 with respect to b: Eq. 5 f ' b 2b 2 E x To find the point that minimises f ( b ) we equate Eq. 5 to zero. f ' b 2b 2 E x 0 b E x c) Using the same logic as in 1 a), we proceed as follows: Expanding f () : Eq. 6 2 f E y x E y 2 2 x 2 2 xy E y 2 2 E x 2 2 E xy Differentiating Eq. 6 with respect to : Eq. 7 f ' 2 E x 2 E xy 2 To find the point that minimises f ( ) we equate Eq. 7 to zero. f ' 2 E x 2 E xy 0 E xy E x 2. a) Let L denote the likelihood function. n 1 L e i 1 x Eq. 8 1 1 exp n Since the value of x i 1 i that maximises L is the same as that which maximises Log(L), we can proceed to find the value of 1 Log L nLog Differentiating n Log L that maximises Log(L). Eq. 9 n x i 1 i with respect to and equating to zero: Eq. 10 dLog L n 1 n 2 xi 0 d i 1 n x i 1 i n Note, in Eq. 10, denotes the maximum likelihood estimator of b) Let OI denote the observed information. It is calculated below. n xi d 2 Log ( L) n OI 2 2 i 1 3 2 d c) The 95% confidence interval (symmetric) is: 2 n 2 n 2 2 0.975,2 n 0.025,2 n . 3.We have the following statistics from the sample: Item Sample mean (m) Sample count (n) Sample variance Sample standard deviation (s) Value 0.58 5 0.88 0.94 Since the sample size is small, we use the t-distribution to obtain the confidence interval. Degrees of freedom for t-distribution = n - 1 = 4. Alpha level of t-distribution = (1 - 0.95) 2 = 2.5%. Critical value of t-distribution (from table) = 2.776. Therefore, the 95% confidence interval is: x t s n 0.58 2.776 0.59;1.75 0.94 5 4.d) and e). manojkumar1165@gmail.com