Essentially, the example would make the formula look like "Value at Node = .5[((VH+C)/…))] + C"
When I compare 2 separate sources, its seems like the methods are different, and the second source doesn’t add the extra cash flow to the bond value. Which was is correct? Are they actually similar,and I am just missing a subtle nuance between the two?
Both sources cite the same formula in the text: “Value at Node = .5 * [((VH+C)/(1+i))+((VL+C)/(1+i))]”
However, when using backward induction in the practice examples, one set of materials seems to value the bond at each node as show above (which I expected), whereas the “standard” source of materials is showing this:
Value at Node = (.5 * [((VH+C)/(1+i))+((VL+C)/(1+i))] ) + C (where C = coupon)
Why, in practice, does one text add an additional coupon amount to the ending bond value, and the other materials do not? This is only done at nodes beyond T=0. I’m really confused as to whether one source is incorrect, or if I am missing some subtle difference in the approaches.
There is no way you’d leave a coupon payment undiscounted in calculating present values. Unless the coupon is recieved today. In which case, it would be a future value, and it’s present value already included in the node before.
The second formula is not famillar with me, I guess it’s incorrect.
That makes sense, but it’s not useful for calculating the PV at n0.
Not to mention that a bond holder is unlikely to include the coupon payment he recieved today when selling it beyond issue date, only the present value of future payments.
You take the principal and coupon at time 6, discount them to time 5, and add the coupon at time 5. Then you discount that to time 4 and add the coupon at time 4. And so on. At time 1 you’ve discounted the value from time 2 and added the coupon at time 1. You discount that to time zero and voilà! You have the value at time 0.
For 6 years I valued mortgage-backed securities using binomial trees; this is how it’s done.
I get what you’re saying. What I meant is, it doesn’t mean anything leaving it undiscounted as part of the value at the node. You only add the coupon at T+1 when discounting to today as the value of N at T. Which I guess what the OP was trying to say in deriving the present value of a node other than N0.
Hi all, thanks for the replies. The sources are Schweser vs. Curriculum. As far as I can tell, the binomial tree examples in the two texts are the same question, but it seems they use different methods and arrive at different answers. I don’t know which one is correct.
Would ike to bump this… i know exactly what you’re talking about.
Schweser does it one way & CFAI curriculum does it another and they lead to different answers… Do you or do you not add the current node’s non discounted coupon to calculate the value at said node? (in addition to the discounted coupon from the t+1 year) And if you believe so, then you are implying that schweser is incorrect. Schweser does not do so.
I always thought the 2nd formula (CFAI) explains the cash flow for a certain node and not the value for that certain node.
While in Schweser you have the value of the bond for each node, and will add the coupon in order to discount it further towards present.
But I’m not sure my interpretation is correct. I think I saw a question somewhere asking the value of a bond for a node other than zero and the answer was value without adding the coupon to receive at that point in time. But I’m afraid it was a Schweser question, I can’t recall where I saw it.
say it is a 1 year bond, no trees involved. Do you discount principal with or without the coupon? As you receive the coupon at the end of the year, you need to discount it.
Two formulas work well. You will have the same value at time 0.
The difference in tree nodes between two formulas come from the different convention. The first one evaluate the bond just after receive the coupon, the second one (with coupon at the end of formula) evaluate the bond just before receive the coupon.
At time 0, because there is no coupon, the two methods (from two formulas) give the same answer.