题目是想问什么

NO.PZ2022071105000002 问题如下 analyst investment bank uses interest-rate trees to forecast short-term interest rates. The analystapplies the following mol for estimating monthly changes in a short-term interest rate tree: = λ(t)* + σ(t)*In this process, λ(t) represents the ift in month t, σ(t) represents the volatility in month t, is the timeintervmeasurein years, an is a normally stributeranm variable with a meof zero anastanrviation of the square root of . The analyst uses the following information to make thecalculations:• Current level of short-term interest rate: 3.1%• ift in month 1 (λ(1)): 0.0024• ift in month 2 (λ(2)): 0.0036• Annualizevolatility of the interest rate in month 1 (σ(1)): 0.0060• Annualizevolatility of the interest rate in month 2 (σ(2)): 0.0080• Probability of upwaror wnwarmovement in interest rates: 0.5Whis the volatility component of the change in interest rate from the upper no of month 1 to the upperno of month 2? A.23 bps B.26 bps C.40 bps 45 bps 中文解析A是正确的。在日期1和日期2之间的利率变化的波动性对日期2上的任何节点的影响都是相同的。由于的标准差为√，所以收益率的标准差为σ(t)*=σ(t)*√。因此，利率变化的波动性成分是0.0080√1/12=0.0023,23 bps，上升或下降。B是不正确的。这个答案的选择包括在日期2向上节点的漂移和波动。0.0036/12+0.0080√1/12=0.0003+0.0023=0.0026。C是不正确的。这个答案包括在日期1和日期2时的波动性影响。0.0060√1/12+0.0080√1/12=0.0040。正确。这个答案包括从日期2的初始利率到上节点的漂移和波动性影响。A is correct. The impaof volatility to the change in the interest rate between te 1 anate 2 will the same any no on te 2. Sinthe stanrviation of is √,the stanrviation of the rate change is σ(t)* = σ(t)*√. So, the volatilitycomponent of the change in interest rate is 0.0080√1/12 = 0.0023, 23 bps, up or wn.B is incorrect. This answer choiinclus ift anvolatility to the upper no te 2.0.0036/12 + 0.0080√1/12 = 0.0003 + 0.0023 = 0.0026C is incorrect. This inclus the volatility impate 1 ante 2. 0.0060√1/12 +0.0080√1/12 = 0.0040is incorrect. This inclus the ift anvolatility impafrom the initirate to the upperno te 2. 这里波动率是年化波动率，为啥不先通过平方根成月度的波动率呢？

NO.PZ2022071105000002问题如下 analyst investment bank uses interest-rate trees to forecast short-term interest rates. The analystapplies the following mol for estimating monthly changes in a short-term interest rate tree: = λ(t)* + σ(t)*In this process, λ(t) represents the ift in month t, σ(t) represents the volatility in month t, is the timeintervmeasurein years, an is a normally stributeranm variable with a meof zero anastanrviation of the square root of . The analyst uses the following information to make thecalculations:• Current level of short-term interest rate: 3.1%• ift in month 1 (λ(1)): 0.0024• ift in month 2 (λ(2)): 0.0036• Annualizevolatility of the interest rate in month 1 (σ(1)): 0.0060• Annualizevolatility of the interest rate in month 2 (σ(2)): 0.0080• Probability of upwaror wnwarmovement in interest rates: 0.5Whis the volatility component of the change in interest rate from the upper no of month 1 to the upperno of month 2? A.23 bpsB.26 bpsC.40 bps45 bps 中文解析A是正确的。在日期1和日期2之间的利率变化的波动性对日期2上的任何节点的影响都是相同的。由于的标准差为√，所以收益率的标准差为σ(t)*=σ(t)*√。因此，利率变化的波动性成分是0.0080√1/12=0.0023,23 bps，上升或下降。B是不正确的。这个答案的选择包括在日期2向上节点的漂移和波动。0.0036/12+0.0080√1/12=0.0003+0.0023=0.0026。C是不正确的。这个答案包括在日期1和日期2时的波动性影响。0.0060√1/12+0.0080√1/12=0.0040。正确。这个答案包括从日期2的初始利率到上节点的漂移和波动性影响。A is correct. The impaof volatility to the change in the interest rate between te 1 anate 2 will the same any no on te 2. Sinthe stanrviation of is √,the stanrviation of the rate change is σ(t)* = σ(t)*√. So, the volatilitycomponent of the change in interest rate is 0.0080√1/12 = 0.0023, 23 bps, up or wn.B is incorrect. This answer choiinclus ift anvolatility to the upper no te 2.0.0036/12 + 0.0080√1/12 = 0.0003 + 0.0023 = 0.0026C is incorrect. This inclus the volatility impate 1 ante 2. 0.0060√1/12 +0.0080√1/12 = 0.0040is incorrect. This inclus the ift anvolatility impafrom the initirate to the upperno te 2. 如题。。。。。。。。。。