please explain me the Q.15.3 in CT1-15:Stochastic Interest Rate Model What is the probability that Sn will take the value 1.02*1.04^(n-1)? Ans: There are n different years in which the 2% could fall and so: Probability=n*(1/3)*(1/3)^(n-1)=n/3^n
There are n places the interest rate 0.02 could be appear in the list of interest rates each year: - the 0.02 could be in the first year, then all the others are 0.04 - the first year could be 0.04, then the second year is 0.02, then all the others are 0.04 etc..etc - the first n-1 years could be 0.04, then the last year is 0.02. This gives n ways of getting 0.02 in one year and 0.04 in the other n-1 years. The probability of choosing 0.02 in a given year is 1/3. The probability of choosing 0.04 in n-1 years is (1/3)^(n-1). Putting it all together the overall probability is n*(1/3)^n.
