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Quantitative Analysis - Chapter 1- Probabilities - FRM May 2015

Dr. Jayanthi Sankaran

Well-Known Member
Hi David,

I notice that Chapter 1 - Probabilities (Quantitative Analysis) - FRM Part I 2015, has been scaled down, somewhat, over the corresponding Chapter I of FRM Part 1 2014. It is not a substantial one - I must admit. Your notes are quite exhaustive.....pulled down from various books, I can see. Do your study notes cover previous FRM Exams? Basically, the question is whether we need to delve this deeply......

Thanks!
 

brian.field

Well-Known Member
Subscriber
Jayanthi,

BT does indeed retain material from year to year where they deem it appropriate, particularly when the retained material is fundamental to the FRM program, as is the case with the Chapter you identified. Is it required? Perhaps not! But if you have trouble with the elementary probability topics of chapter 1, you may have a hard time with subsequent material, so I would suggest that you spend a little time on it, even if it is review only!

Err on the side of preparing too much!

Good luck!

Brian
 

Dr. Jayanthi Sankaran

Well-Known Member
Thanks, Brian - appreciate your good advice and input. Will do! It's just that I am trying to race against time, given the voluminous coverage of material - must stop getting myself into a panic mode!

Thanks, also, for your good wishes:)

Jayanthi
 

Dr. Jayanthi Sankaran

Well-Known Member
Hi Brian,

In David's study notes for P1.T2 Quantitative Analysis, Study Notes Reading 10, page 5 - do you know what pmf means?

Thanks!
Jayanthi
 

brian.field

Well-Known Member
Subscriber
PMF stands for "probability mass function". It is equivalent to PDF, or "probability distribution function", but is specific to discrete distributions.

To reiterate, discrete distributions are not typically described as PDFs; rather, discrete distributions are called PMFs whereas continuous distributions are called PDFs..
 
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Dr. Jayanthi Sankaran

Well-Known Member
Hi David,

In your Study Notes P1.T2 Quantitative Analysis Reading 10:

  • In page 20, the answer for Prob [Passage|Constant] works out to 4.44% (as against 22.22%), according to the Bayes Theorem:
*P[Passage|Constant] = {P[Constant|Passage] x P[Passage]}/P[Constant] = {(.2 x .3) x (.2)}/.27 = 4.44% as against P[Passage|Constant] = Joint P(Passage, Constant)/P(Constant) = 6%/27% = 22.22%
*There is a typo - instead of Prob(Neutral|Constant) it should read Prob(Passage|Constant)​
  • In Page 19, a minor typo under He estimates:
    • If the state of the economy is POOR, the probability that the stock price increases is 15%, and the probability that the stock price decreases is 70%
  • In Page 13, in the Stock and Watson example, P[M=2] = 0.06, instead of 0.03 - just a typo....
  • In Page 7, under EVT Distributions, what do GPD and GEV stand for? Is it General Probability Distribution and General Extreme Value Distributions? - Just a guess...
Thanks!
 
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brian.field

Well-Known Member
Subscriber
Hello Jayanthi.

I can't look at the notes at the moment but looking at your questions, I have a couple thoughts. On 1) remember that Bayes says,

P(A|B) = P(B|A)•P(A) / (P(B|A)•P(A) + P(B|A')•P(A'))

where A' refers to the compliment of A.

Regarding EVT, GEV, and GPD, they refer to Extreme Value Theory, Generalized Extreme Value, and Generalized Pareto Distribution, respectively. The GEV is used (in part 2) for the block maxima approach whereas the GPD is used (part 2 again) for the Points Over Threshold approach. I don't think EVT is on part 1 but I may be mistaken.

Best,

Brian
 

Dr. Jayanthi Sankaran

Well-Known Member
Hi Brian,

Thanks a tonne! I took the mathematical expression for Bayes Theorem straight out of Chapter 1 - Probabilities, GARP FRM 2014 Material. It is given as follows:

  • P[A|B] = {P[B|A].P[A]}/P ..............................Equation (1.20)


The definition you have given above is from David's Notes.

  • EVT, GPD and GEV were stated as examples of Continuous Distributions - was just curious to know, what they stood for! - Thanks for the clarification
All the best for FRM Part 2!

Best
Jayanthi
 

brian.field

Well-Known Member
Subscriber
I mentioned that Bayes example to emphasize that the denomonator should reflect the entire probability space. I hope that makes sense.

Brian
 

Dr. Jayanthi Sankaran

Well-Known Member
Hi Brian,
There is a typo in my remark above. Bayes Theorem as illustrated in the 2014 GARP FRM Material, is expressed as:
  • P[A|B] = {P[B|A].P[A]}/P..............................Equation (1.20)

Yes, your definition (as given in David's notes) is correct, in that it explicity defines the Probability space P.
However, my question was with regard to application in the numerical example, given above. Thanks!
Jayanthi
 

ShaktiRathore

Well-Known Member
Subscriber
Hi jayanthi,
P[Constant|Passage] x P[Passage] is same as Joint P(Passage, Constant) so how u getting different answers?
Gpd,gev and evt are distributions used to categorize the tail. These are distributions if the data points of the tail. These are useful to find Var and ES with their respective formulas and terminoligies.
Thanks
 

Dr. Jayanthi Sankaran

Well-Known Member
Hi Brian and Shakti,

Yes, discovered my mistake:

*P[Passage|Constant] = {P[Constant|Passage] x P[Passage]}/P[Constant] = {(.2 x .3) x (.2) }/.27 = 4.44 - there was an extra 0.2 in the numerator. Wonder where I got that from! So, the answer does come to 22.22% according to Bayes Theorem.
* Sorry, Brian for bugging you with that.......
* Shakti - thanks for explaining GPD, GEV and EVT. What does ES stand for?

Thanks!
 

Dr. Jayanthi Sankaran

Well-Known Member
Thanks, Shakti! Another question, on page 25 of the Quantitative Analysis Study Notes:
  • For f(x) = x/8 - 0.75, F(x) = xsquare/16 - 0.75x + c
  • I thought integration of f(x) = xsquare/16 + c. How do we get the term 0.75x?
Thanks!
 

ShaktiRathore

Well-Known Member
Subscriber
Hi
Integration of f(x)=x/8-.75=x^2/16-.75x+c
Integrstion of x/8=x^2/16
Integration of .75 which is a constant is .75x its from here that .75x is cmg from(integration of any constant c is cx).
After intgrn of any function we add a constant c.
Differentiate above integration of f(x) we again get function f(x)
Thanks
 

brian.field

Well-Known Member
Subscriber
Jayanthi! You did not "bug me" with anything! Please continue to ask questions - it is good for all of us to have an open dialogue.

Regarding your question on the 0.75.....this is a very basic concept in integration (or calculus for that matter.) If the exam relied an a significant amount of calculus, I would recommend that you review a calculus textbook carefully. However, since the exam is not too focused on calculus, you should probably still give a quick glance of the first few chapters on differentiation and integration in a standard calculus textbook, but don't spend too much time on it.

Looking forward to chatting with you over the next few months!

Brian
 

Dr. Jayanthi Sankaran

Well-Known Member
Hi Brian,

Thanks - glad to know that I am not 'bugging' you - yes completely agree that we should flesh out all our doubts, in order to have a constructive dialogue and resolution, so that we are well prepared. It challenges us and contributes to learning....

I looked up some Youtube integration videos, recommended by David. Also, looked up Khan Academy. Have become somewhat rusty with integral calculus, although it is all coming back....

Looking forward to chatting with you too, over the next few months!

Jayanthi
 
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