Friday, June 13, 2014

Optimal Overhedge

Consider a contract ( contingent claim ) with payoff that is dependent on the stock price at maturity \((P(S_T))\). The value of such a contract at time t with \(t\leq T\) will be: \[V_t= \mathbb{E}_t\left\{ P(S_T) \right\} \] ( For now we're assuming zero interest rates . )
If a firm sells such a contract and wishes to hedge the risk, one idea would be to buy \(\Delta\) units of the stock \(S_t\), thus the net protfolio value would be: \[Q_t= S_t \Delta - V_t \] To find out the exposure ( sensitivity ) of the portfolio to the stock would could take the partial derivative wrt \(S_t\): \[\frac{\partial Q_t}{ \partial S_t} = \Delta - \frac{\partial V_t}{\partial S_t} \] So if \[ \Delta = \frac{\partial V_t}{\partial S_t} \] then \[\frac{\partial Q_t}{ \partial S_t} = 0\] In which case, at that instance we're hedged against small moves in the stock price.

Leaving aside any formal mathematics for a moment...
Broadly speaking when the payoff \((P(S_T))\) has bumps and kinks we find that the contract valuation \(V_t(S_t)\) is a significantly more smooth function, since the value \(V_t\) is an expectation, so the kinks get averaged out. That said: \[\lim_{t\to T}V_t(S_t)=P(S_T)\] and also \[\lim_{t\to T}\Delta_t=\lim_{t\to T}\frac{\partial V_t}{\partial S_t}=\frac{\partial P}{\partial S_T}\] Thus when we have constraints on the delta ( first derivative ) and indeed gamma ( second derivative) of \(V_t\), we need to be sure that the constraints hold for the payoff \(P(S_T)\).

One practical problem with having high gamma is that it means the delta will change a lot for small changes in the underlying. So if a trader is attempting to keep a portfolio delta neutral, he'll need to rebalance frequently and the rebalances will be large. In such a scenario there is the potential to loose a significant amount of money on transaction costs. Also high gamma normally comes with high theta ( time sensitivity ) and vega ( sensitivity to volatility).

Suppose the contract that we wish to hedge is a barrier option with payout at time T: \[P(S_T)= \begin{cases} 0 \ \ \ \ \ when S_T < K \\ N \ \ \ when \ S_T \geq K \end{cases} \] This is a step function. It is not continuous and the partial derivative \(\frac{\partial P}{\partial S_T}\) does not exist at \(S_T=K\).
Thus we it seems we won't be able to hedge away the exposure to \(S_T\).
So what do we do?
Rather than hedge the actual contract, we could make a synthetic contract which does have nicely behaved partial derivatives. Consider a payoff which consists of the real payoff \(P(S_T)\) plus an "overhedge" \(H(S_T)\), such that \[H(S_T) \geq 0\ \ \forall S_T\] Note that the client does not receive \(H(S_T)\) but the contract we will hedge does include a contribution from it. When we work out a price that includes \(H(S_T)\) it will be higher than it would have been without the overhedge. But to be able to hedge the risks we need a "reasonable" contract, which is nicely behaved.
What we want is a minimal overhedge \(H(S_T)\) such that the delta and gamma are contained.
Let the synthetic contract with the overhedge have payoff: \[\Phi (S_T)= P(S_T) + H(S_T)\] (Why did we choose that Greek letter? Well adding P and H we get PHI )
We're going to insist that the first derivative ( \(\frac{\partial \Phi}{\partial S_T}\) )is continuous, because we want the second derivative ( gamma ) to exist and not be infinite.

Let's suppose we have the following caps and floors as constraints: \[-\Delta_F \leq \frac{\partial \Phi}{\partial S_T} \leq \Delta_C\] and \[-\Gamma_F \leq \frac{\partial ^2 \Phi}{\partial S_T^2} \leq \Gamma_C\]
With \(\Delta_F \geq 0 \), \(\Delta_C \geq 0 \)
and \(\Gamma_F \geq 0 \), \(\Gamma_C \geq 0 \)

So now the question is: what is the minimal overhedge which obeys those constraints?

Well for an analogy, suppose S was time t and H was position.
We are starting at rest at position N.
We wish to get home in our car as quickly as possible to position H=0
and we have a maximum acceleration, a maximum deceleration and a speed limit.
Then what should we do?
Clearly we should use maximum acceleration until we reach the speed limit.
We should then continue at the speed limit for as long as possible until near home
and then apply the breaks, using maximum deceleration until we neatly come to rest at home.

Similarly in this we case we break H up into 3 sections:
\(H_1(S_T)\) is the accelerating phase, ( quadratic in \(S_T\))
\(H_2(S_T)\) is the linear phase ( i.e. at the speed limit).
Then \(H_3(S_T)\) is the decelerating phase.
After (quite) a bit of algebra we find:
\(H_1(S_T)=N-(S_T-K)^2\ \Gamma_F\ \ \ \ \ \ \ \ \ \ \ \ \) in the domain: \(K \leq S_T < \Lambda_1\)
\(H_2(S_T)= H_1(\Lambda_1)-(S_T-\Lambda_1)\Delta_F\ \ \ \ \) in the domain: \(\Lambda_1 \leq S_T < \Lambda_2\)
\(H_3(S_T)=(S_T-\Lambda_3)^2\ \Gamma_C\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \) in the domain: \(\Lambda_2 \leq S_T \leq \Lambda_3\)

With:
\[\Lambda_1=K+\frac{\Delta_F}{2 \Gamma_F}\] \[\Lambda_2=\Lambda_1 + \frac{H_1(\Lambda_1)}{\Delta_F}-\frac{\Delta_F}{4\Gamma_C}\] \[\Lambda_3=\Lambda_1 + \frac{H_1(\Lambda_1)}{\Delta_F}+\frac{\Delta_F}{4\Gamma_C}\] We can see that: \[\Lambda_3=\Lambda_2+\frac{\Delta_F}{2\Gamma_C}\]
Note that \(H(S_T)\) is zero before K and after \(\Lambda_3\).

To derive the above equations we use the fact that, when we set: \[H_2(S_T)-H_3(S_T)= 0\] we have a quadratic in \(S_T\) and that quadratic should have a single root at \(S_T=\Lambda_2\).
I.e. the parabola \(H_3(S_T)\) is a tangent to the line \(H_2(S_T)\) at \(S_T=\Lambda_2\).

Returning to the analogy of using the car to get home quickly: Suppose you start off near home and the acceleration in the car is not very high. Then perhaps you'll need to start decelerating before you reach the speed limit.
In our case that happens when \[\Lambda_1 > \Lambda_2 \] So now \(H_2\) doesn't come into play and we need to find where is the transition from \(H_1\) to \(H_3\). I'll leave it as an exercise for the reader to work it out.

Question for the reader:
If you make some "reasonable" assumptions on the distribution of \(S_T\)
and given that \[\frac{\partial^2 P}{\partial S_T} \leq \Gamma_C \ \ \ \ \ \forall S_T\] then show that \[\frac{\partial^2 V_t}{\partial S_t} \leq \Gamma_C \ \ \ \ \ \forall S_t \ and \ t\] What assumptions did you need to make?
Suggestion: you might consider starting with the log-normal distribution case without drift or interest rates, then generalize.


Summary:
So we have derived the optimal overhedge for a barrier option subject to constraints on delta and gamma. And the synthetic contract which contains the overhedge is continuous and its first derivative is continuous.

Monday, April 28, 2014

Surname's on an island



On an isolated island each person has
1/5 probability of having 0 kids
1/5 probability of having 1 kid
1/5 probability of having 2 kids
1/5 probability of having 3 kids
1/5 probability of having 4 kids

Heterosexual partners are chosen at random from the population within the given generational cohort.

It is an old fashioned place where the men pass on their surname to their children,
the women don't.

Initially, the surname Klimpt is not very common. Just 1.00% of the total population.

After 100 generations what is the probability there there will be at least one person called Klimpt in the youngest generation if:
      (i) the initial generational cohort had 400 members
or   (ii) the initial generational cohort had 4,000 members

Express your answer as a percentage with two digits after the decimal point.

Assume that the starting population is half men, half women and that there is a 50-50 chance that each child born is a boy or girl.

Hint: you are probably going to need a computer for this one.

The Prince's Ancestors

Part A:
Once upon a time, there lived a fine young prince. He was next in line to the throne in a dynasty that stretched back for centuries. He had the official records that showed that his father's father's father's... 20 generations back had effectively created a country by unifying warring tribes.

Though it should be noted that his male line ancestors were not always very good husbands, and indeed their wives sometimes had lovers when the king was away fighting wars.

At each generation is was estimated that there was a 10% chance that the father's name on the birth certificate was not in fact the biological father.

What is probability that the man that the prince thinks is his father's father's father ... 20 generations back on the male line is indeed his true biological male line ancestor?

Background maths level required: high-school.
Difficulty level: fairly straight-forward.



Part B: 
Over the course of every generation, immigration caused a 10% population increase.

The true figures for the number of biological children each person had are as follows:
20% of people had no children that survived
20% of people had 1 child
20% of people had 2 children
20% of people had 3 children
20% of people had 4 children

Assume that when people paired off to reproduce, their partner was effectively just randomly chosen from their generation of population.

At the time of the founding king, his generation accounted for 1 million people, though the full population of the country was higher if we include the older people who were alive at the time.

What is the probability that the founding king was indeed an ancestor of the young prince?
Please give your answer as a percentage with two digits after the decimal point.

Level of maths required: some basic university maths is probably required.
Level of difficulty: for someone who can write a bit of code, fairly straight-forward


A witch with a compass

Wilma the Witch was given a new electronic compass that could tell her which way was north. It pointed to true north rather than magnetic north.

So she looked at her compass, found out which direction was north-east, put the compass away in her pocket and set out on her broom in a north-easterly direction. She continued as straight as she could over the surface of the earth for 1,000 km, without adjusting left or right.

She then took out her compass and found that even though she had started out by heading north-east and had continued going straight she was no longer exactly on a north-east bearing.

So she now found out which way was east and headed east. This time, for a change, she keep looking at her compass and diligently maintained a constant bearing, ( i.e. she kept going exactly east ) and continued for 6,000 km.

Then she turned to the south-east and headed on a constant south-east bearing. After a while she found herself back at the starting point. She noted that she had crossed every line of longitude exactly once.

She decided to continue on a constant south-east bearing, she crossed the equator and eventually found herself at the south pole.

She then stopped.

How far did she travel in total?

You can assume that the earth is a sphere and that the distance from the equator to either pole it is 10,000 km.
Please also ignore any distance covered due to altitude gain traversing mountains.

Level of background mathematics required: some undergraduate university maths
Question difficulty level: a tad tricky.

Sunday, April 27, 2014

Average number of matches won

In Wimbledon's women's singles tournament there are 128 players. It is a knock-out tournament. I.e. after each round the losers are excluded from the next round.
The first round losers will have a win percentage of 0%.
The overall winner will have a 100% win percentage.

Part A: What's the average win percentage, (assuming the weighting of each player is equal)?

Part B: What's the average win percentage if each player is weighted proportional to how many matches she played?

Level of background mathematics required: high-school.
Difficulty level: straight-forward.

Thursday, April 24, 2014

Maths puzzle: a bit of trig

Here's a little mathematics puzzle.



Suppose points A, B, C are all on a circle
and we have a point P outside the circle such that
the line that passes through A and P is a tangent to the
circle ( at A )
 the line that passes through B and P is a tangent to the
 circle ( at B)

also the line segments PA and BC are parallel.

 Suppose |AP| = 3
 and |BC| = 2

 What is |AC| ?

Background level of mathematics required: high-school
Difficulty level: by high-school standards it is a bit tricky.


In a while I'll post the answer ...