Econ 424

ECON 424

SPRING 2015

Lab#2  

Lei, Tao #1263559

Exercises

1. X is normally distributed random variable, with mean 0.05, variance c(0.10)2.

• Pr(X>0.10): 1-pnorm(0.10, 0.05, 0.10)= 0.3085
• Pr(X<-0.10): pnorm (0.10, 0.05, 0.10)= 0.06681
• Pr(-0.05
• 1% quantile: qnorm (0.01, 0.05, 0.10)= -0.1826
• 5% quantile: qnorm (0.05, 0.05, 0.10)= -0.1145
• 95% quantile: qnorm (0.95, 0.05, 0.10)= 0.2145
• 99% quantile: qnorm (0.99, 0.05, 0.10)= 0.2826

2. X ~ N(0.05, (0.10)2) and Y ~ N(0.025, (0.05)2).

(a) Plot the normal graph[pic 1]

(b) Comment on risk-return trade-offs for the two stocks.

[pic 2]

According to the graph, the Microsoft stocks has a higher mean and a higher standard deviation. This tells us that Microsoft stocks possibly has a higher expected return, and at the same time, higher risk.

3. Value at Risk Calculations with simple monthly return

(a) 1% quantile:

w0 = 100000

w0*qnorm (0.01, 0.04, 0.09) = -16937.13

(b) 5% quantile:

w0 = 100000

w0*qnorm (0.05, 0.04, 0.09) = -10803.68

There’s 1% probability that there’ll be a loss of $16,937 next month, and there’s 5% probability that the loss will be $10,803 next month from the initial investment.

4. Value at Risk Calculations with cc return

(a) Using Excel functions, 1% quantile: $-15,580.46.

     5% quantile: $-10,240.55.

When cc return is normally distributed, there’s 1% probability that there’ll be a loss of $15,580 next month; there’s 5% probability that there’ll be a loss of $10,241.

(b) If annually invest with cc returns, we just times the mean and the standard deviation with 12 months.

Mean= 0.04*12= 0.48

Standard deviation= [pic 3]

Apply the two changes in Excel functions,

1% quantile: -$21,751.89

5% quantile: -$3,228.34.

There’s 1% probability that $21,752 or more next year from our initial investment, 5% probability that the loss will be $3,228.

5. Chi-square and student’s t-distribution

• Using r-studio, code as follows:

x.vals = seq(0, 20, length=100)

ub = max( c(dchisq(0, df=c(1,2,5,10)), dnorm(0)) )

plot(x.vals, dchisq(x.vals, df=1), type="l", lwd=2, ylim=c(0,1), ylab="pdf", xlab="x")