The post Stamp Duty Changes Calculator (Autumn Statement 2014) appeared first on Quantitative Finance Courses.

]]>My wife and I are enjoying our first property we bought back in June 2014. I can tell you we weren’t very happy about the 4% tax rate we had to pay on the whole of the purchase price considering it was not much above the 4% threshold. Stamp duty being a slab tax means when the purchase price crosses a threshold the new rate of tax applies to the entire purchase price not just the amount over the threshold.

Watching the autumn statement closely, we were glad to hear the Chancellor, George Osborne, has announced sweeping reforms to this unfair tax. It will now be a graduated tax so that when the purchase price crosses a threshold the new tax rate only applies to the amount in excess of the threshold.

The new rates of tax are

£000,000 – £125,000 0%

£125,000 – £250,000 2%

£250,000 – £925,000 5%

£925,000 – £1,500,000 10%

£1,500,000 upwards 12%

My joy over the reform was tempered somewhat after realising if we bought our flat today we would have saved more than £4,000!

You can use the calculator below to calculate the stamp duty under the old and new system to see if you are better or worse off. Simply enter the price of your property.

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]]>The post Pricing Financial Instruments – The Finite Difference Method appeared first on Quantitative Finance Courses.

]]>This book explains how to price derivatives with the finite difference technique. It is aimed at practitioners full of many different examples, such as pricing convertible bonds, American options, Barrier options and Parisian options. It also has a nice introduction to stability analysis using the matrix approach and the fourier approach.

The book covers pretty much all you need to know for solving 1d pde’s in finance. It pays special attention to issues such as discontinuties in the payoff and how to deal with critical pricing points such as strikes and barrier points with coordinate transformations. The chapter on discrete sampling for pricing Asian and Parisian options is very useful and is something i have implemented in practice. It works very well.

Rating 4.5/5

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]]>The post Financial Calculus appeared first on Quantitative Finance Courses.

]]>Financial Calculus is rather an old book now, first being published in 1996. Despite its age it has (as far as i know) a unique place in the literature in that it attempts to teach continuous time stochastic calculus, without requiring a knowledge of (nor teaching) measure theoretic probability. It does in fact succeed to do this difficult task! By using the binomial tree approach to pricing derivatives it introduces difficult concepts such as measure, risk neutral pricing and change of measure. Ito Calculus is explained using the derivative of \( W_t^2\) to illustrate that higher order terms are needed in the Taylor expansion in order for the differential to have the right expectation. Using the tools of the Martingale Representation theorem and the Cameron Martin Girsanov Change of Measure theorem the Black Scholes formula is derived. This is quite a feet considering that the word sigma algebra isn’t mentioned once in the book! The later part of the book is devoted to pricing interest rate derivatives. It provides a nice introduction to the Heath Jarrow Morton interest rate model framework for the forward rate curve but for a detailed approach you should probably look elsewhere. There is also a brief mention of the Libor market model (BJM) but again you should look elsewhere for more details as you could write (and some authors have!) a whole book on the subject.

Rating 4/5

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]]>The post What is Quantitative Finance? appeared first on Quantitative Finance Courses.

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]]>The post FRM Practice Questions appeared first on Quantitative Finance Courses.

]]>FRM Part 1 Practice Test – 10 questions in 24 minutes

This is a random selection of questions from our database. I am continually adding more questions so feel free to try the sample FRM exam again in future. Please bear in mind the questions are on the harder side of the spectrum so don’t be disheartened if you don’t get a high score! The actual part 1 exam is a 4 hours long multiple choice exam featuring 100 questions. This means you have 2.4 minutes per question. This is a tough exam due to time pressure as well as the actual difficulty of the questions.

Here are some tips on taking the exam

- Some questions can be answered very quickly in 10 to 20 seconds and some other questions take much longer than 2.4 minutes. Since each question is weighted the same skip the really long questions and revisit them at the end. If you skip a question make sure you skip it in the answer sheet otherwise your subsequent answers will be out of line!!
- Never leave a question unanswered when you hand your paper in, if you have to guess you still have a 25% chance of being correct. If you can rule out 2 wrong answers you increase your chances of guessing correctly to 50%!
- Make sure you have practised using your calculator including the special bond calculation functions. You don’t have time to figure this out in the exam!
- Check how many questions you have done at least every hour. You should be aiming for around 25 questions an hour.

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]]>The post Futures Delta and Forward Delta appeared first on Quantitative Finance Courses.

]]>The payoff from a long position in a forward contract on an asset expiring at time \( T \) is \( S_T – K\) where \(S_T\) is the final price of the asset and \( K \) is the agreed purchase or strike price. Assuming constant interest rates, using elementary pricing theory the present value of this payoff at time t \( V_t \) is

\begin{align*}

V_t = e^{-r (T – t)}\; (\mathbb{E}[S_T|\mathcal F_t] – K) = S_t – e^{-r(T-t)} K

\end{align*}

The Delta \( \frac{\partial V_t}{\partial S_t} \)is 1.

Let \( F(t,T) \) denote the futures price at time \( t \) for delivery of the asset at time \( T \). The futures contract is settled daily by an amount given by the change in futures price so the delta is \( \frac{\partial F(t,T)}{\partial S_t} \). In this simple situation \( F(t,T) = S_t e^{r (T-t)} \) so the delta is \( e^{r (T-t)} \)

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]]>The post What is the FRM Pass Rate? appeared first on Quantitative Finance Courses.

]]>The FRM pass rates are shown below. I have split the results in to two parts. The first table shows the pass rates when the exam was just one paper.

2009 | 2008 | 2007 | 2006 | 2005 | 2004 | 2003 | 2002 | 2001 | 2000 |
---|---|---|---|---|---|---|---|---|---|

44.1% | 42.4% | 44.4% | 44.8% | 44% | 52.5% | 53.8% | 58.9% | 63% | 53% |

The table below shows the results from 2010. The exam had become a two part exam.

Year | Nov 2014 | May 2014 | Nov 2013 | May 2013 | Nov 2012 | May 2012 | Nov 2011 | May 2011 | Nov 2010 | May 2010 |
---|---|---|---|---|---|---|---|---|---|---|

Part I | 48.4% | 42.5% | 50.9% | 45.9% | 46.7% | 47.3% | 46.6% | 53.1% | 39.3% | 52.5% |

Part II | 58.7% | 58.1% | 58.0% | 56.8% | 56.0% | 61.1% | 57.0% | 61.6% | 54.9% | 54.0% |

If you take parts I and II together on the same day you are only included in the stats for part II if you pass part I. You cannot pass part II without passing part I. Hence the part II pass rate in the table above is a conditional probability.

For fun let’s answer this FRM part 1 style exam question. Note it is not how the exam actually works.

**Q**. Assume all candidates who sat for the May 2013 exam sat for both FRM part 1 and FRM part 2 for the first time and that you can pass FRM part 1 or FRM part 2 without having to pass both. The probability of passing FRM part 1 is 45.9% and the probability of passing FRM part 2 is 56.8%. If we assume that the probability of passing FRM part 2 conditional on passing FRM part 1 is at least as high as the probability of passing FRM part 2 conditional on failing the FRM part 1 exam what are the upper and lower bounds for the probability of passing both FRM part 1 and FRM part 2 at the first attempt?

**A**. Let 2p denote the event of passing the FRM part 2 exam. Let 1p denote the event of passing the FRM part 1 exam. We have \(\mathbb{P}(1\text{p}) = x = 45.9\%\) and \(\mathbb{P}(2\text{p}) = y = 56.8\%\).

Now using \( \mathbb{P}(2\text{p} \;\&\; 1\text{p}) + \mathbb{P}(2\text{p} \;\&\; \bar{1\text{p}}) = y\) and combining with Bayes theorem we have

\begin{align*}

&\mathbb{P}(2\text{p}|1\text{p})*\mathbb{P}(1\text{p}) + \mathbb{P}(2\text{p}|\bar{1\text{p}})*(1 – \mathbb{P}(1\text{p})) = y \nonumber\\

&=\mathbb{P}(2\text{p}|1\text{p})*x + \mathbb{P}(2\text{p}|\bar{1\text{p}})*(1 – x) = y

\end{align*}

Clearly \(\mathbb{P}(2\text{p}\;\&\;1\text{p}) \leq \mathbb{P}(1\text{p}) \) and \(\mathbb{P}(2\text{p}\;\&\;1\text{p}) \leq \mathbb{P}(2\text{p}) \) i.e. we have \( \mathbb{P}(2\text{p} \;\&\; 1\text{p})\leq \min(x,y) = 45.9\%\).

We are given \(\mathbb{P}(2\text{p}|1\text{p})\geq \mathbb{P}(2\text{p}|\bar{1\text{p}})\) and the lower bound is achieved when \(\mathbb{P}(2\text{p}|1\text{p})= \mathbb{P}(2\text{p}|\bar{1\text{p}})\), which is the independence case so that \( \mathbb{P}(2\text{p} \; \&\; 1\text{p}) = xy = 26.07\%\)

So the upper bound is = 45.9% and the lower bound is 26.07%.

Check out my **FRM Part 1 practice questions **!!!

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]]>The post ABRACADABRA: Part 3 (Comment) appeared first on Quantitative Finance Courses.

]]>A colleague of mine asked me what if we wanted to know the expected time for the monkey to type the sequence of letters ABCDEFGHIJK and not ABRACADABRA what would the result be. Trudging through the mathematics gives \( \mathbb E[T] =26^{11} \). Why is the expected time for the monkey to type ABRACADABRA longer than to type ABCDEFGHIJK even though the probability to get either in the next 11 keystrokes is \(\frac{1}{26}^{11}\)? The simple answer is there are less ways to get ABRACADABRA for the first time in a sequence (if it is long enough). Consider a sequence of length 22. If we look at the number of ways of getting ABCDEFGHIJK from position 12.

\begin{align*}

???????????ABCDEFGHIJK

\end{align*}

Each question mark can be any letter of the alphabet except the sequence ABCDEFGHIJK hence there are \(26^{11} – 1\) ways to achieve this. Now lets consider the corresponding situation for ABRACADABRA.

\begin{align*}

???????????ABRACADABRA

\end{align*}

The letters immediately preceding ABRACADABRA can’t be ABRACAD otherwise we would not be getting ABRACADABRA for the first time so this rules out

\(26^4\) possible combinations. Also the immediately preceding letters can’t be ABRACADABR for the same reason so that rules out 26 more possible combinations.

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]]>The post ABRACADABRA: Part 2 appeared first on Quantitative Finance Courses.

]]>To answer the question what is the expected time for a monkey to type ABRACADABRA, assuming he types the letters of the alphabet randomly at times 1,2,3,… rigorously we will need to use a theorem from discrete martingale theory, Doob’s Optional Stopping Theorem.

This theorem states sufficient conditions for a stopped martingale to have the same expectation as its starting value.

Let \( X^j_i \) denote the wealth of the jth gambler at time i

\begin{align*}

X_i^j =1 \; \text{if $i < j$,}

\end{align*}

otherwise if \(i \geq j \),

\begin{align*}

X_i^j =

\begin{cases}

26^{i-j+1} \mathbb \;\text{with prob} \; p = {\frac{1}{26}}^{i-j+1} \;

\\

0 \mathbb \;\text{with prob} \; 1- p \;

\end{cases}

\end{align*}

Clearly \( X^j_i\) is a martingale with expectation 1. Lets define the martingale starting from 0 \( M_n = \sum_{j = 1}^{j = n} X_n^j – n\)

If we can show that \(\mathbb E[M_T] = \mathbb E[M_0]\) then the rest follows. This is where we invoke Doob’s Optional Stopping Theorem.

which gives us sufficient conditions. Showing that \(\mathbb E[T] < \infty\) and the increments \(M_n – M_{n-1}\) are bounded is sufficient.

1. \(\mathbb E[T] < \infty\)

To prove this we use a lemma from page 101 of Probability with Martingales.

Suppose that \(T\) is a stopping time such that for some \(N\) in \(\mathbb N\) and some \(\epsilon \geq 0 \), we have, for every n in \(\mathbb N\):

\begin{align*} \mathbb P(T\leq n+N | \mathcal F_n) > \epsilon, a.s \end{align*}

Then \(\mathbb E[T] \leq \infty\)

In our problem we take \(N = 11\) and the probability of the monkey typing ABRACADABRA in the next 11 keystrokes gives us a value for \(\epsilon\). We take \(\epsilon \; < \frac{1}{26}^{11}\)

2. \(|M_n(\omega) – M_{n-1}(\omega)| \leq K \; \forall (n,\omega)\)

The largest value attainable by \(M_n\) is \(26^{11} +26^4 + 26\) so the increments are bounded.

Hence at the stopping time \(T\) we know that \(\mathbb E[M_T] = 0\) from which follows \(\mathbb E[T] = 26^{11} +26^4 + 26\)

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]]>The post ABRACADABRA: Part 1 appeared first on Quantitative Finance Courses.

]]>Since I have just reviewed ‘Probability with Martingales’ I thought it would be nice to add one of the most famous puzzles in the book to the puzzles page.

If at each of times 1,2,3,.. a monkey types a capital letter at random. What is the expected time for the monkey to first produce the letters ABRACADABRA. On the face of it this is a hard question but martingale theory makes it easy. We can turn this question into a stopping time question about a fair game.

Let \(T\) be the stopping time when the monkey first produces the letters ABRACADABRA. Just before each time n = 1,2,3,… a new gambler arrives on the scene. He bets £1 on the nth letter being an ‘A’. If he loses he leaves, if he wins he receives £26 all of which he bets on the (n+1)th letter being a ‘B’. If he loses he leaves, If he wins he receives £26 * £26 and bets the entire amount on the (n+2)th letter being an ‘R’ and so on going through the letters of ABRACADABRA and then finally leaving if he gets that far. The key observation is that this is a fair game and that at each time, n, the total amount gambled (by all the gamblers) is n. The expected value of this game to the gamblers must be zero since it is a fair game.

At the stopping time \(T\)

the payout to the gambler who has guessed ABRACADABRA is \(26^{11}\)

the payout to the gambler who has guessed ABRA is \(26^4\)

and the payout to the gambler who has guessed A is \(26\)

The house has received \(T\) and since it is a fair game

\(\mathbb E [T] = 26^{11}+ 26^4 + 26\)

in ABRACADABRA: Part 2 we will re-do this question with a bit more rigour.

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