## TS Module 15 Mean reversion intuition

 Author Message NEAS Supreme Being Group: Administrators Posts: 4.2K, Visits: 1.2K TS Module 15 Mean reversion intuition (The attached PDF file has better formatting.) A.     The time series courses emphasizes the intuition of ARIMA processes. B.     Exam problems use scenarios and probabilities to test the intuition.  Autoregressive processes have mean reversion.  Question 1.1: Mean reversion An AR(1) model of 90 day Treasury bill yields has a mean of 5.00%, φ1 of 0.60, and ó = 0.4% per month.  The values are for the first day of each month. The date now is January 15, 20X8.  Let  C.    AT = the absolute value of the difference between January’s Treasury bill yield and 5%.D.    AT+1 = absolute value of the difference between February’s Treasury bill yield and 5%.  For which of the following values of the January 20X8 90 day Treasury bill yield is the probability greatest that AT+1 > AT?  A.     4.3%B.     4.6%C.    4.9%D.    5.2%E.     5.5% Answer 1.1: C If the problem asks: “For which value of the January 20X8 90 day Treasury bill yield is the probability greatest that yT+1 > yT,” we choose the lowest value of yT (Choice A). For the absolute values, use the following reasoning.  If yt = ì, the difference now is zero. The time series is stochastic with continuous values, so the probability that the value will again be the mean next period is infinitesimally small (i.e., close to zero).  If the time series has discrete values, the probability is low. The absolute value of the difference from the mean increases, whether the actual interest rate increases or decreases.. The mean reversion is proportional, so the farther yt now is from the mean, the more it moves toward the mean. The stochastic term may still move it away from the mean.  But the stochastic term is constant for all values of the time series (since it is stationary). As the deterministic movement toward the mean increases, the less likely it will be offset by the stochastic movement. Stationary autoregressive processes show mean reversion.If the current value is not the mean, the forecasts eventually move toward the mean. Suppose the mean is ì and the most recent value of the time series is yT.             For an AR(1) model, the mean reversion is immediate. Next period’s forecast is between ì and yT.           For an AR(p) model, the mean reversion may be delayed p-1 periods. The p periods ahead forecast is between ì and yT.   Question 1.2: Mean reversion An AR(1) model of 90 day Treasury bill yields has a mean of 5.00%, φ1 of 0.60, and ó = 0.4% per month.  The values are for the first day of each month. The date now is January 15, 20X8.  Let             AT = the absolute value of the difference between January’s Treasury bill yield and 5%.           AT+1 = absolute value of the difference between February’s Treasury bill yield and 5%.  For which of the following values of the January 20X8 90 day Treasury bill yield is the probability smallest that AT+1 > AT?  A.     4.3%B.     4.6%C.    4.9%D.    5.2%E.     5.5% Answer 1.2: A To grasp the intuition, reason through the following sequence. If the January 20X8 Treasury bill yield is 5%, AT is zero. In this exercise, the Treasury bill yield is an ARMA process with a error term. The probability that the February 20X8 Treasury bill yield is also 5% is zero (with no rounding) and very small even with rounding. It is likely that AT+1 > AT. If the January 20X8 Treasury bill yield is 5.01%, AT is 0.01%. The forecast for February is 5.006%. The actual February yield has a normal distribution with a mean of 5.006% and a standard deviation of 0.4%. The 95% confidence interval for the February 20X8 Treasury bill yield is 4.206% to 5.806%. The probability that the February yield is between 4.99% and 5.01% is very small. We can compute the exact probability with a cumulative density function for the normal distribution. The exam problems do not ask for precise probabilities. If the January 20X8 Treasury bill yield is 10%, AT is 5%. The forecast for February is 8%. The actual February yield has a normal distribution with a mean of 8% and a standard deviation of 0.4%. The 95% confidence interval for the February 20X8 Treasury bill yield is 7.2% to 8.8%. The probability that the February yield is between 0% and 10% is almost 100%. This probability is the cdf for ±5 standard deviations from the mean.   For an MA(q) model, the mean reversion may be delayed q-1 periods. The q periods ahead forecast is between ì and yT.  Attachments TS Module 15 mean reversion intuition.pdf (1.5K views, 51.00 KB) minnie53053 Junior Member Group: Forum Members Posts: 11, Visits: 1 Hi NEAS,I can't understand what means the question 1.1 and 1.2, could you explain a little more?For which of the following values of the January 20X8 90 day Treasury bill yield is theprobability greatest that AT+1 > AT?if 4.9%, then AT=-0.1%,so P(AT+1>-0.1%) is greatest?After my comprehension for this question,AT+1 has a normal distribution with mean 0 and delta 0.4%. For other answers, respectively :A: P(AT+1>-0.7%) B: P(AT+1>-0.4%) C: P(AT+1>-0.1%)D: P(AT+1>0.2%) E: P(AT+1>0.5%) SO, absolutely A is greatest. where i made the mistake? smh1021 Forum Newbie Group: Forum Members Posts: 9, Visits: 1 In question 1.2, what is the equation used to calculate the February forecasts (5.006% and 8%)?
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