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  #161  
Old 05-29-2017, 07:06 PM
BelayMeMaybe BelayMeMaybe is offline
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Wow, Adapt doesn't think much of dg approximation:


Has anyone come across questions like this already? I'm guessing the zero difficulty score means either a) too few responses to assign a difficulty b) no one's missed it yet c) bug
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  #162  
Old 05-31-2017, 02:37 PM
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Quote:
Originally Posted by BelayMeMaybe View Post
Using adapt but haven't started doing full length exams yet. For me, Practice exams are too long to be effective studying and it's too easy to skip review. I'll do 4 or 5, but only once the theory is crystal clear.

From the March 2017 thread it looks like some relatively high adapt scores (5+) failed because adapt doesn't (can't) hit theory as hard as the exam.

Computationally challenging adapt problems don't translate well to conceptually difficult exam questions.

Also, the removal of difficult material (stochastic calculus) and replacement with easy material (derivatives) scares me because to keep the pass-rate around 45% they'll have to ask harder questions on easier concepts, which means that past exams might not reflect the actual difficulty.



Thank you InTheDark! The formula for symmetry on currency options was exactly what I needed!
or they could just increase pass mark to something higher like 80%
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  #163  
Old 05-31-2017, 03:38 PM
Levi Levi is offline
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The stochastic calculus/IRM questions only appear scary before you actually attempt them. Almost all of my real exam questions on these topics were Ito's Lemma, S^a, and general theory (trivially simple multiple choice) which were all relatively quick and straightforward material. The hardest questions I had were actually Black Scholes pricing questions due to the lengthy algebra. I don't see the pass mark doing anything crazy - I passed under the old syllabus with a 6 and likely would have failed under the new.
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  #164  
Old 05-31-2017, 04:34 PM
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The Disreputable Dog The Disreputable Dog is offline
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I'm completely lost on something with BDT trees. Can anyone help?

Quote:
Originally Posted by ASM pg. 456
Now let's verify the lognormal yield volatility after one year of 3-year bonds issued at time 0, or 2-year bonds issued at time 1. In figure 22.6, we see that the prices of 2-year bonds at time 1 are 0.84027 and 0.86625. The yields of these zero-coupon bonds are the square roots of the reciprocals of the prices, or (1/0.84027)^0.5 -1 = 0.09091 and (1/0.86625)^0.5 -1 = 0.07443, so...
I'm with everything so far.

Quote:
Originally Posted by ASM pg. 456
...the volatility is 0.5ln(0.09091/0.07443) = 0.1
But this jump loses me.

In the tree being discussed each branch is one year in length. The general BDT material talks about the yield volatility of 1-year bonds (σh) being a parameter necessary for constructing the tree. Any two vertically-adjacent nodes must have a ratio of e^2σh√h. It looks like this example is maybe saying that the same relationship exists here? Even though we're looking at bonds that are two periods in length?

I'm missing something crucial I think.
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  #165  
Old 06-01-2017, 11:01 AM
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ALivelySedative ALivelySedative is offline
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Quote:
Originally Posted by The Disreputable Dog View Post
It looks like this example is maybe saying that the same relationship exists here? Even though we're looking at bonds that are two periods in length?

I'm missing something crucial I think.
Your definition of 'h' is incorrect. They are 2-year bonds, but the time period from node to node is only 1 year (this is h; "the price of 2 year bonds at time 1). Thus Sqrt(h) = 1 and the adjacent vertical nodes only differ by a factor of e^(2*sigma).
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  #166  
Old 06-01-2017, 12:55 PM
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The Disreputable Dog The Disreputable Dog is offline
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Interesting. Thank you.

The other problem I was having was just inferring how far I could take that relationship between nodes.

So is the following correct? If I have a tree like the following with 1-year nodes and p* = 0.50, I could determine the yield volatility of two-year bonds sold at time 4 using just the 1-year rates in the red squares?

[0.5/A + 0.5/B]/W = [0.5/B + 0.5/C]/X * e^2σ√1 (then solve for σ)

I would also get the same yield volatility of two-year bonds sold at time 4 if I used only the blue squares?

[0.5/D + 0.5/E]/Y = [0.5/E + 0.5/F]/Z * e^2σ√1 (then solve for σ)

Or if I really wanted to, I could get the same yield volatility of two-year bonds sold at time 4 with non-adjacent nodes?

[0.5/A + 0.5/B]/W = [0.5/E + 0.5/F]/Z * e^8σ√1 (then solve for σ)

And just to clarify, this is the same as the yield volatility of 6-year bonds sold at time zero after 4 years?
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  #167  
Old 06-03-2017, 11:25 AM
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I clearly only have a tenuous grasp of some of the rules of BDT trees.

In problem 22.14 of the ASM manual we're given a 3-period BDT tree that has 3-month branches. The yields shown in the tree are all annual yields. We're asked to find the "yield volatility for 6-month zero-coupon bonds issued at the end of 3 months".

I did this problem wrong, differently, on several attempts, but these are the lessons I've taken away from it. I was hoping someone could confirm them for me.

1. To construct anything using a BDT tree you need to start with yields which correspond to the length of the branches.

So in this case we need to take the annual yields shown in the tree, and we need to convert them to quarterly yields. This, as opposed to taking the annual yields shown at 3 months, and splitting them into semi-annual yields at three months.

2. When we're asked for yield volatilities we're necessarily talking about annualized yields.

Once we compute the prices of 6-month bonds at the up and down nodes at 3 months, the solution says that we need to annualize their 6-month yields before backing out the yield volatility.

I'm afraid I've worded these poorly, but the solution has us following these steps:

1. break the annual yields shown in the tree into 3-month rates.
2. construct 6-month bond prices at time 3-months out of the 3-month rates using the normal process. I.e., Pu = (0.5/ruu + 0.5/rud)/ru; Pd = (0.5/rud + 0.5/rdd)/rd.
3. Annualize the yields of these bond prices. yu = (1/Pu)^2 -1; yd = (1/Pd)^2 -1.
4. Extract the yield volatility as ln(yu/yd)/(2√0.25).

And my takeaways were: 1) you have to work with yields that match the length of each period, and 2) the yield volatilities require us to annualize the yields of whatever we're looking at.

Does that sound right?

Last edited by The Disreputable Dog; 06-03-2017 at 11:32 AM.. Reason: a words
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  #168  
Old 06-03-2017, 02:24 PM
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Are you working problems at the end of sections from the Mahler manual? Any other problems?
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  #169  
Old 06-04-2017, 01:24 PM
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I've been doing all of the ASM problems, and supplementing with ADAPT quizzes if I wanted additional practice. The only Mahler questions I really sat down and worked were on option Greeks.
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  #170  
Old 06-08-2017, 09:00 AM
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