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Question on Kaplan's question on Binomial Trees
An example of a standard tree used by FIData is given in Figure 2.
Figure 2: Binomial Interest Rate TreeYear 0 | Year 1 |
---|
4.5749% | 7.1826% |
| 5.3210% |
FIData's website uses rates in Figure 2 to value a two-year, 5% annual-pay coupon bond with a par value of $1,000 using the backward induction method.
The question is: Using the backward induction method, the value of the 5% annual-pay bond using the interest rate tree given in the three bonds in Figure 2 is closest to:
$945.
C)$993.
So I averaged node U2 (1005/1.071826) and node L2 (1005/1.05321) to come up with 943.6846. Then I discounted that by the year 0 rate of 4.5749% and came up with 907.1819. I answered A.
Kaplan says the answer is C. Their explanation is: The value of the 5%, two-year annual pay $1000 par bond is $992.88.
I find that explanation underwhelming. Can someone help me out here? Where did I screw up?
0 ·
Comments
I've ignored the $ value of the bond for now, and work on a % basis (of par) to simplify things:
Using your definition of nodes:
U2 = 105/1.071826 = 97.9637
L2 = 105/105321 = 99.6952
Then fair value of bond today, V0 = 0.5 * [(97.9637+5)/1.045749] + 0.5 * [(99.6952+5)/1.045749]
= 0.5 * [98.4593+100.1150]
= 99.2872
So 99.2872% of a par value of $1,000 should be $993, i.e. answer C.
Hope this helps!