Q1. Using 625 trading days of data, you estimated the daily log return follows a normal distribution with a mean of 5 bps and and a stdev of 125 bps. Q1a. based on information above, what is the...

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Q1. Using 625 trading days of data, you estimated the daily log return follows a normal distribution with a mean of 5 bps and and a stdev of 125 bps.
Q1a. based on information above, what is the probability of true daily log return average is 0? can you reject the true mean is 0? can you reject the true mean is 10 bps?
Q1b. what is the 90, 95, and 99% confidence interval for your mean return estimate?
Q1c. what is the mean log return and stdev of log return over one year period and four year period (assuming 252 trading days per year)?
Q1d. based on Q1c what is the probably of losing money (negative log return) or doubling your money (total log return = ln(2)) over 1 year and 4 year period?
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Q2. You collected 500 weeks of data (2500 days total). Based on that you find Tuesday's mean return is 12 bps. Mean return of all days is 2 bps. Stdev across all days is 100 bps. There is no noticeable difference b/w Tuesday stdev vs other weekdays' stdev.
Q2a. What is the mean return for weekdays other than Tuesday?
Q2b. What is the z-score (t-stat) for the difference b/w Tuesday's mean return and other weekdays's mean return?
Q2c. What is the stdev for the difference above? What is the 95% Confidence interval?
Q2d. Based on 95% CI, should you reject or accept that Tuesday's mean return is not different from other weekdays?
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Q3. You observed a stock moves 1% stdev each day usually, except for days of earning announcement. In the latter case, it has a stdev of 5%. Today is a Friday. You are interested in an option that expires in 2 weeks. Earning announcement will be on the next Friday.
Q3a. What should be annualized IV now?
Q3b. On next Thurday at close, what should be IV?
Q3b. On Friday at close, just after ER, what should be IV?
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Q4. You conduct a poll comparing two candidates. In a poll of 400 people, 190 prefer candidate A, and 210 prefer B.
Q4a. What is the point estimate for the probability that a voter will choose candidate A?
Q4b. What is the stdev for that point estimate?
Q4c. What is the 95% Confidence interval?
Q4d. What is the probability that candidate B loses in the actual election?
Q4e. If you want to shrink one stdev error to 1%, how many people do you need to survey?
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Below Q5-Q6 for HW4.
Q5. You are considering two investment choices: a. 1 year CD that pays 2% for sure; b. investing in SP500 with 8% expected and 20% stdev.
Q5a. If you put 50% in each of the two, what in the mean and stdev for your portfolio return?
Q5b. You want to maximize your expected return for the portfolio, as long as the stdev risk is no more than 10% per year. What is the allocation to CD and to stocks? What is the mean return of the portfolio?
Q5c. You want to reach 15% mean return per year as a minimum. What is the least amount you need to allocate to stocks? What is the stdev of the portfolio?
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Q6. In addition to two instruments in Q4, you also can invest in a long term bond with 4% mean return and 10% stdev annually. Stock and bond funds have a correlation of -0.4 (negative 0.4).
Q6a. What is the mean return and stdev for a portfolio that is 50% in stock and 50% in bond?
Q6b. What is the mean return and stdev for a portfolio that is 25% in stock and 25% in bond and 50% in 1 year CD?
Q5c. What is the mean return and stdev for a portfolio that is 75% in stock and 75% in bond and -50% (negative 50%, ie borrow money) in 1 year CD?
Answered Same DayApr 07, 2021

Answer To: Q1. Using 625 trading days of data, you estimated the daily log return follows a normal distribution...

Pooja answered on Apr 08 2021
141 Votes
Table of Contents
Q1)    2
Q1) a)    2
Q1) b)    3
Q1) c)    3
Q1) d)    4
Q2)    5
Q4)    6
Q1)
Q1) a)
P(Xbar=0) = 0 ; as Xbar f
ollows N(u, sigma^2/n) which is a continuous distribution. And hence probability at a point is zero.
(i)
Ho: u=0 v/s h1: u =/=0
mean =     5.00
sd = sqrt(var) =     11.180
u = 0
n= 625.00
alpha=    5%
critical value, z(a/2) = z(0.05/2) = 1.960
test statistic, z = (mean-u)/(sd/sqrt(n))
= (5-0)/(11.18/sqrt(625))
=11.1803
Since |z| > z(a/2), i reject Ho at 5% level of significance and conclude that u =/=0
(ii)
Ho: u=10 v/s h1: u =/=10
mean=     5.00
sd= sqrt(var) =     11.180
u=     10.00
n=     625.00
alpha=    5%
critical value, z(a/2) = z(0.05/2) = 1.960
test statistic, z = (mean-u)/(sd/sqrt(n))
z = (5-10)/(11.1803398874989/sqrt(625))
z = -11.1803
Since |z| > z(a/2), i reject HO at 5% level of significance and conclude that u =/= 10
Q1) b)
(i) 90% CI
critical value, z(a/2) = z(0.1/2) = 1.645
CI = mean +- z(a/2,n-1)*(sd/sqrt(n))        
lower    = 5 - 1.645*(11.18/sqrt(625))=    4.26
upper    = 5 + 1.645*(11.18/sqrt(625))=    5.74
(ii) 95% CI
critical value, z(a/2) = z(0.05/2) = 1.960
CI = mean +- z(a/2,n-1)*(sd/sqrt(n))        
lower    = 5 -...
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