Reference no: EM131237883
Q1. Do the following:
a. Show that for positive values of monthly mean return not greater than µ = .01 (1 percent), the approximation T · µ given by (1.21) under-estimates the exact result of equation (1.20) by less than 1 percent. An acceptable solution will be to write simple R code to show the result for a range of positive monthly mean returns µ less than or equal to 1 percent. A desirable analytic solution is to show the result for a 1 percent monthly mean return, and prove that the difference between the exact and approximate solution increases with increasing µ and correspondingly decreases with decreasing µ.
b. Show that for a windfall constant monthly mean return of 2% the exact formula gives 26.8% for annual mean return, whereas the approximate gives 24%, an under estimate of nearly 3%.
Q2. Derive the expression (1.22) for the variance of multi-period arithmetic returns under the assumption that the returns are independent with constant mean µ and constant variance σ2. Hint: Make use of the fact that the expectation of a product of independent random variables is the product of the expectations of the individual random variables.
Q3. Use the fact that the log is a concave function to prove the inequality (1.47). Hint: Use the fact that for any set of values x1, x2,··· , xn and any set of non-negative constants a1,a2,··· ,an such that i=1∑nai = 1, a concave function g(x) has the property that i=1∑nai g(xi) ≤ g (i=1∑nai xi) with equality if and only if the xi are all equal.
Q4. (Advanced) Prove the approximate formula (1.48) relating the geometric mean to the true mean and volatility.
Q5. Show that if f(r) is any probability density function that is symmetric about a location parameter µ and that has a finite third moment E(r3), and hence finite third central moment E(r-µ)3, then the coefficient of skewness is zero.
Q6. Show that if f(r) is a normal density with mean µ and variance σ2 the ∫(r -µ)4f(r)dr = 3·σ4, and so the (excess) kurtosis is zero.
Q7. Derive the equations (1.57) and (1.58) for the mean and variance of a two component normal mixture distribution.
Q8. Show that a kernel density estimator ˆf(r) given by (1.64) is a bona fide probability density, i.e., fˆ(r) is non-negative and integrates to one.
Attachment:- Assignment.rar
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