1. Suppose the following quantities related to the size of the ”jump” of a stock

Question

1. Suppose the following quantities related to the size of the ”jump” of a stock price. Tracking the size of jumps is important as it helps assess the volatility of the market.

## [1] 2.61 12.13 1.31 3.20 2.14 4.86 0.56 1.20 3.97 5.61 0.43

## [12] 0.92 4.20 2.66 3.15 6.05 4.48 2.07 1.81 0.32

Assuming these are representative of a larger population of jumps, I want to make some probabilistic statements. To answer the following questions, you may use R or SAS.

(a) I assume a normal distribution. My estimate for µ is 3.184. My estimate for 2 is 7.39. What is the probability that a stock from this larger population experiences a jump greater than 9?

(b) Now I’ve decided to instead assume a gamma distribution. My estimate for is 0.99. My estimate for is 3.02. What is the probability that a stock from this larger population experiences a jump greater than 9?

(c) Suppose I am tracking 100 stocks. What is the probability at least 10 of them have a jump size larger than 9 assuming the normal model?

(d) Suppose I am tracking 100 stocks. What is the probability at least 10 of them have a jump size larger than 9 assuming the gamma model?

(e) Which model seems more appropriate? Can you back this up by referencing the original data values?

Explanation / Answer

I have used R to perform the calculations.

(a)

Assuming normal model, the probability that a stock from this larger population experiences a jump greater than 9 is 0.01619961

pnorm(9,3.184,sqrt(7.39),lower.tail = FALSE)

(b)

Assuming Gamma model, the probability that a stock from this larger population experiences a jump greater than 9 is 0.04980999

pgamma(9, shape = 0.99, scale = 3.02, lower.tail = FALSE)

(c)

Usig normal approximation of binomial distribution where the probability that a stock experiences a jump greater than 9 is p=0.01619961 and number of trials is n = 100.

The mean of normal approximated model is np = 100 * 0.01619961 = 1.619961

Standard deviation of sqrt(np(1-p)) = sqrt(100*0.01619961*(1-0.01619961)) = 1.262425

So, the probability at least 10 of them have a jump size larger than 9 assuming the normal mode is 2.518839e-09

pnorm(9, mean = 1.619961, sd = 1.262425, lower.tail = FALSE)

(d)

Usig normal approximation of binomial distribution where the probability that a stock experiences a jump greater than 9 is p=0.04980999 and number of trials is n = 100.

The mean of normal approximated model is np = 100 * 0.04980999 = 4.980999

Standard deviation of sqrt(np(1-p)) = sqrt(100*0.04980999*(1-0.04980999)) = 2.175522

So, the probability at least 10 of them have a jump size larger than 9 assuming the normal mode is 0.03234654

pnorm(9, mean = 4.980999, sd = 2.175522, lower.tail = FALSE)

(e)

From the data, we see that there is one occurence of data (12.13) out of 20 which is greater than 9. So, the probability that a stock from this larger population experiences a jump greater than 9 is 1/20 = 0.05.

Comparing with the results of part (a) and (b), we see that Gamma model is more appropriate.

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.