An airline company is concerned about potential financial losses due to currency movements in various countries. An analyst has been hired to model daily currency movements in these economies. The objective is to understand the likelihood of extreme currency movements that may have major financial consequences for the airline. The analyst has suggested fitting several types of standard statistical distributions to the entire datasets, using maximum likelihood estimation, including Normal, Gamma, Pareto and lognormal. Estimates can then be made regarding probabilities of extreme currency movements based on the chosen model. extreme daily losses. For this analysis, currency losses were treated as positive, and gains were treated as negative. A threshold was set at daily losses of +2.2%, resulting in a data subset of 150 ‘extreme’ values from the 3,000 original data points to be used in the extreme value analysis. The following results were obtained from the subsequent analysis: Beta = 0.009785. Gamma = 5.417 (shape). (ii) Calculate G(0.02) using the above parameter values, where G is the distribution function of the GPD, and write down its meaning. (iii) Estimate the probability that the currency losses tomorrow exceed 5%, based on this GPD model. The probability that a loss is less than 5% given that it is greater than 2.2% is G(0.028)= 0.8995. The probability that a loss is greater than 5% given that it is greater than 2.2% is therefore 1 – 0.8995 = 0.1005 The estimated probability that a loss is greater than 5% is therefore 0.1005 * 150 / 3,000 = 0.0050 For(ii), the solution is The formula for the GPD CDF is G(x) = 1 – (1 + x / (gamma * beta)) ^ (-gamma). Hence G(0.02) = 1 – (1 + 0.02 / (5.417 * 0.009785)) ^ (-5.417)= 0.823 This is the probability that a loss is less than 4.2% given that it is greater than 2.2% Where is 4.2% come from? For(iii) the solution is The probability that a loss is less than 5% given that it is greater than 2.2% is G(0.028)= 0.8995. The probability that a loss is greater than 5% given that it is greater than 2.2% is therefore 1 – 0.8995 = 0.1005 The estimated probability that a loss is greater than 5% is therefore 0.1005 * 150 / 3,000= 0.0050 Why we use 0.028 to represent "less than 5%"? A Normal distribution was also fitted to the data. The best fitting Normal distribution was determined to be N(0, 0.014^2). However, currency returns are generally considered to exhibit leptokurtic behaviour. (iv) Calculate the probability that daily losses exceed 5% using this Normal distribution, and comment on the results. The solution is, The number of standard deviations is 0.05 / 0.014 = 3.57 From the Tables, the required probability is therefore 1 – 0.99982 = 0.00018 Hence the probability of a loss exceeding 5% is significantly lower under the Normal distribution than under the GPD distribution Given the expectation of leptokurtic behaviour, there is reason to believe that the GPD distribution is more appropriate than the Normal distribution Using the Normal distribution would therefore be expected to significantly underestimate the risk of extreme currency movements I don't know why we use 0.05 div by 0.014.What is this action meaning? In fact,I can't get any clue from GPD section of CH16 of CMP CS2. For this question,the only one I can know is how to calculate GPD of (ii)
Hello 1. Remember with the GPD that we are working with a threshold exceedance amount. So, if X is the underlying currency loss then we have W = X - u | X > u, which is the amount by which the loss exceeds some threshold. The threshold here is 2.2% (as we're told in the question) so we have: W = X - 0.022 | X > 0.022 G(0.02) = P(W <= 0.02) = P(X - 0.022 <= 0.02 | X > 0.022) = P(X <= 0.042 | X > 0.022). So, this is the probability that a loss is less than 4.2%, given it is larger than 2.2%. 2. When considering 5%, we have: P(X > 0.05) = P(X > 0.05 | X > 0.022) P(X > 0.022) Now: P(X > 0.05 | X > 0.022) = P(X - 0.022 > 0.028 | X > 0.022) = 1 - P(X - 0.022 <= 0.028 | X > 0.022) = 1 - G(0.028) and we estimate P(X > 0.022) using the data as 150/3000. 3. Here we are standardising. We are saying that the best fitting distribution is X ~ N(0, 0.014^2). So, according to this distribution: P(X > 0.05) = P([X - 0] / 0.014 > [0.05 - 0] / 0.014) = P(Z > 3.57) where Z ~ N(0,1). Hope this helps! Andy
