问题如下图：
选项：
A. 
B. 
C. 
解释：
请问这道题为什么还要zhe折Y3的利率，三年的债券不是在二叉树上到T2就好了吗？
NO.PZ201602270200001904 问题如下 4. Baseon Exhibit 4 anusing Metho2, the correprifor BonX is closest to: A.97.2998. B.109.0085. C.115.0085. B is correct.The first step is to intify the cash flows:Next, calculate the cash flows for eayebeginning with Ye3 anmove backwar to Ye1:Ye3: 0.5×[(1061.06)+(1061.06)]+6=106.00000.5\times\lbrack(\frac{106}{1.06})+(\frac{106}{1.06})\rbrack+6=106.00000.5×[(1.06106)+(1.06106)]+6=106.00000.5×[(1061.05)+(1061.05)]+6=106.95240.5\times\lbrack(\frac{106}{1.05})+(\frac{106}{1.05})\rbrack+6=106.95240.5×[(1.05106)+(1.05106)]+6=106.95240.5×[(1061.03)+(1061.03)]+6=108.91260.5\times\lbrack(\frac{106}{1.03})+(\frac{106}{1.03})\rbrack+6=108.91260.5×[(1.03106)+(1.03106)]+6=108.9126Ye2:0.5×[(106.00001.04)+(106.95241.04)]+6=108.38100.5\times\lbrack(\frac{106.0000}{1.04})+(\frac{106.9524}{1.04})\rbrack+6=108.38100.5×[(1.04106.0000)+(1.04106.9524)]+6=108.38100.5×[(106.95241.02)+(108.91261.02)]+6=111.81620.5\times\lbrack(\frac{106.9524}{1.02})+(\frac{108.9126}{1.02})\rbrack+6=111.81620.5×[(1.02106.9524)+(1.02108.9126)]+6=111.8162Ye1:0.5×[(108.38101.01)+(111.81261.01)]=109.00850.5\times\lbrack(\frac{108.3810}{1.01})+(\frac{111.8126}{1.01})\rbrack=109.00850.5×[(1.01108.3810)+(1.01111.8126)]=109.0085A is incorrebecause the coupon payment is not accountefor eano calculation. C is incorrebecause it assumes tha coupon is paiin Ye1 (time zero) when no coupon payment is paitime zero. 可不可以用二叉树的图 来解析下这道题？
109.0085. 115.0085. B is correct. The first step is to intify the cash flows: Next, calculate the cash flows for eayebeginning with Ye3 anmove backwar to Ye1: Ye3: 0.5×[(1061.06)+(1061.06)]+6=106.00000.5\times\lbrack(\frac{106}{1.06})+(\frac{106}{1.06})\rbrack+6=106.00000.5×[(1.06106)+(1.06106)]+6=106.0000 0.5×[(1061.05)+(1061.05)]+6=106.95240.5\times\lbrack(\frac{106}{1.05})+(\frac{106}{1.05})\rbrack+6=106.95240.5×[(1.05106)+(1.05106)]+6=106.9524 0.5×[(1061.03)+(1061.03)]+6=108.91260.5\times\lbrack(\frac{106}{1.03})+(\frac{106}{1.03})\rbrack+6=108.91260.5×[(1.03106)+(1.03106)]+6=108.9126 Ye2: 0.5×[(106.00001.04)+(106.95241.04)]+6=108.38100.5\times\lbrack(\frac{106.0000}{1.04})+(\frac{106.9524}{1.04})\rbrack+6=108.38100.5×[(1.04106.0000)+(1.04106.9524)]+6=108.3810 0.5×[(106.95241.02)+(108.91261.02)]+6=111.81620.5\times\lbrack(\frac{106.9524}{1.02})+(\frac{108.9126}{1.02})\rbrack+6=111.81620.5×[(1.02106.9524)+(1.02108.9126)]+6=111.8162 Ye1: 0.5×[(108.38101.01)+(111.81261.01)]=109.00850.5\times\lbrack(\frac{108.3810}{1.01})+(\frac{111.8126}{1.01})\rbrack=109.00850.5×[(1.01108.3810)+(1.01111.8126)]=109.0085 A is incorrebecause the coupon payment is not accountefor eano calculation. C is incorrebecause it assumes tha coupon is paiin Ye1 (time zero) when no coupon payment is paitime zero.请问在从y3折现到y2的时候，不是应该是这样计算的吗？ （106/1.06+106/1.06）*2/1=100吗 为什么还要+6的coupon不是已经计算过一遍了吗？ 和最新版的基础班讲义第92页的标准算法有出入
没明白这道题里的forwarrate如何应用到折现？forwarrate不是用作中期利率来折现吗？为什么没有给出波动率？
请问这里的折现率为什么不用上一年的？能帮助整理下哪里的二叉树用上年的吗？
