Dear All, Pls refer to this qu and answer >> a)Where do they get 1.10 in the answer( I can see only 1.08, 1.06 and 1.04. b) Is it correct to say that in t=0 it is 1.06 c) Why only 4 possibilities >> Is it to do with the "time 0 will accumulate to more than 1.2 at time 3?" Pls help out Question Let it denote the effective rate of interest in the year t to t + 1. It is assumed that, for t = 0, 1, 2, ... i(t)+1 = i(t) + .02 with probability 0.25 i(t)+1 = i(t) with probability 0.5 i(t)+1 = i(t) − .02 with probability 0.25 Given that i(0) = 0.06, which of the following gives the probability that an investment of 1 at time 0 will accumulate to more than 1.2 at time 3? Suggested answer 1. Note that 1.063 = 1.191016 < 1.2. Relevant possible combinations of interest rates are therefore 1.06 × 1.06 × 1.08 ! probability 0.5 × 0.25 1.06 × 1.08 × 1.06 ! probability 0.25 × 0.25 1.06 × 1.08 × 1.08 ! probability 0.25 × 0.5 1.06 × 1.08 × 1.10 ! probability 0.25 × 0.25 Total probability = 0.375 Hence B is correct.
First of all, when posting threads on the forum please make it clear (preferably in the header) where a question comes from. This makes it much easier for other users of the forum to help out or find an answer to their query. Also I think where you've typed i(t)+1 in the question you mean i(t+1). Answering your questions in turn: a) The question is effectively telling us that each year the interest rate is the same as that in the previous year, or 2% higher than that in the previous year, or 2% lower than that in the previous year. So since it is 6% in the first year, it can be 8% in the second year and the 10% in the third year (if each year it increases by 2%). b) Yes - i(0) = 0.06 means that the interest rate in the first year is 6%, so the relevant accumulation factor is 1.06 for the year. c) Yes - other combinations of possible interest rates give values less than 1.2 so are excluded e.g. (1.06)^3 = 1.191... Try drawing a tree diagram including all the different possibilities for the interest rates over the three years to prove this to yourself.
