问题如下图:
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解释:
这里的Mean loss是Expected shortfall的意思吗
NO.PZ2016070202000002 问题如下 A large, internationbank ha trang book whose size pen on the opportunities perceiveits trars. The market risk manager estimates the one-y VAR, the 95% confinlevel, to US50 million. You are asketo evaluate how gooa job the manager is ing in estimating the one-y VAR. Whiof the following woulthe most convincing evinththe manager is ing a poor joassuming thlosses are inticaninpenntly stribute(i.i.)? Over the past 250 ys, there are eight exceptions. Over the past 250 ys, the largest loss is US500 million. Over the past 250 ys, the meloss is US60 million. Over the past 250 ys, there is no exception. is correct. We shoulexpe(1−95%)250=12.5{(1-95\%)}250=12.5(1−95%)250=12.5 exceptions on average. Having eight exceptions is too few, but the fferencoule to luck. Having zero exceptions, however, woulvery unusual, with a probability of 95%250, whiis very low. This means ththe risk manager is proving Vestimates thare mutoo high. Otherwise, the largest or melosses are not rectly useful without more information on the stribution of profits. 如题
NO.PZ2016070202000002 问题如下 A large, internationbank ha trang book whose size pen on the opportunities perceiveits trars. The market risk manager estimates the one-y VAR, the 95% confinlevel, to US50 million. You are asketo evaluate how gooa job the manager is ing in estimating the one-y VAR. Whiof the following woulthe most convincing evinththe manager is ing a poor joassuming thlosses are inticaninpenntly stribute(i.i.)? Over the past 250 ys, there are eight exceptions. Over the past 250 ys, the largest loss is US500 million. Over the past 250 ys, the meloss is US60 million. Over the past 250 ys, there is no exception. is correct. We shoulexpe(1−95%)250=12.5{(1-95\%)}250=12.5(1−95%)250=12.5 exceptions on average. Having eight exceptions is too few, but the fferencoule to luck. Having zero exceptions, however, woulvery unusual, with a probability of 95%250, whiis very low. This means ththe risk manager is proving Vestimates thare mutoo high. Otherwise, the largest or melosses are not rectly useful without more information on the stribution of profits. 这题根据置信区间计算平均例外天数是12.5天,如果实际只有0天出现例外,不是说明基金经理表现很好吗
NO.PZ2016070202000002 Over the past 250 ys, the largest loss is US500 million. Over the past 250 ys, the meloss is US60 million. Over the past 250 ys, there is no exception. is correct. We shoulexpe(1−95%)250=12.5{(1-95\%)}250=12.5(1−95%)250=12.5 exceptions on average. Having eight exceptions is too few, but the fferencoule to luck. Having zero exceptions, however, woulvery unusual, with a probability of 95%250, whiis very low. This means ththe risk manager is proving Vestimates thare mutoo high. Otherwise, the largest or melosses are not rectly useful without more information on the stribution of profits. 老师好,本题解答用的mean=12.5天,作为判断标准;为什么不是用统计量作为判断标准,例如95%的置信区间,统计量算出应该是19.2天(cutoff)。用mean判断和用t统计量判断的区别是什么?谢谢!
NO.PZ2016070202000002 如果说mean值超过了60m 那var值是50m 肯定是说明var低估了呀 那这个模型就不准确了呀