1) (5 pts) Assume that a stock'slogreturns at any time scale have normal distribution. Suppose that its average annuallog return is 100%, and its annual standard deviation ("volatility") of log returns is 200%. What areits average (mu) and standard deviation (sigma) ofdailylog returns, assuming a year has 250 trading days?

2) (10 pts) Simulate 250 instances of thedaily log returns described in 1) with random seed set.seed(2015). Compute thenetreturns of these instances, and compute their average and standard deviation. Are the average (m) and standard deviation (s) of net returns same as the average and standard deviation of log returns computed in 1)?

3) (10 pts) Compute the quantitym

s

2

2

. How does this compare with the average log return mu computedin part 1)? Their equality can be proven analytically through amathematical theorem called Ito's Lemma that lies at the foundation of Black-Scholes options pricing formula. Their numerical equality is not that good here because Ito's Lemma assumes we can divide a period into infinitesimally small sub-periods. So divide ayearinto 25,000 sub-periods (think of these sub-periods roughly as minutes) instead, and compare the new mu (average log return per minute) with the newm

s

2

2

(m is now the average net return per minute). Also compare the new sigma (standard deviation of log returns per minute) with the new s (standard deviation of netreturn per minute).

4) (10 pts) If we assume thatthestock's initial price is $1,what is the expected value of its logprice log(P(t)) aftert minutesexpressed in terms of mu? And what is the expected value of its price P(t) expressed in terms of mu and sigma?(Hint: the answer depends on a careful reading of section A.9.4 on lognormal distribution of the textbook.) Finally, express these expected values in terms of m and s instead.

5) (5 pts)The continuously compounded rate of growth of a stock is log(P(t))/t. What is the expected continuously compounded rate of growth of the stock in part 4?

A keyfinancial concept embedded in this exercise:

The expected value of the exponential function of a normal variable is not equal to the exponential function of the expected value of that normal variable, orE(exp(x)) !=exp(E(x)). So while the expected value of P(t) does increase with an exponent of m, the expected continuously compounded rate of growthis mu=m-(s^2)/2. Inplain language: risk decreases your expected compounded rate of growth!