Quantitative Interview Questions

Question

A lady walks into a shop to buy some stuff worth Rs. 200. She hands a Rs. 1000 note to the shopkeeper. The shopkeeper doesn't have change for Rs. 1000, so he walks to the shop next door, hands the Rs. 1000 note to the next door shopkeeper and gets the change. He completes the transaction with the lady. After an hour, the next door shopkeeper comes to him and says the Rs. 1000 note is fake, returns it and takes the money from him. How much did the shopkeeper lose?

Solution

It's a simple problem - I am posting a solution to highlight a step-by-step process in reaching an answer

There are two ways we can look at this problem.

Balance Sheet or the End state method

In this method, we take a look at the final state of each party

Lady : Gave a fake Rs. 1000 note, received Rs. 800 in change and stuff worth Rs. 200.

Total P/L: 1000

Next-door shopkeeper: Gave change for fake Rs. 1000, and received his money back an hour later.

Total P/L: 0

Shopkeeper: Looking at the above two, we can say the lady gained Rs. 1000 and that came from the shopkeeper.

Total P/L: -1000

Final loss to the shopkeeper is Rs. 1000

Cash flows or intermediate state method

In this method, we provide the state of each party at the end of each transaction event.

Lady Paid fake Rs. 1000 - Lady: 0, Shopkeeper: 0, Next-door: 0

Shopkeeper got change for Rs. 1000 from next-door - Lady: 0, Shopkeeper: 1000, Next-door: -1000

Lady bought stuff and left - Lady: 1000, Shopkeeper: 0, Next-door: -1000

Next-door returns the fake Rs. 1000 and gets his money - Lady: 1000, Shopkeeper: -1000, Next-door: 0

Question

You and your friend decide to meet between 4:00 and 5:00 PM tomorrow and head for a movie. Both of you will arrive between 4:00 and 5:00, and have an equal chance to show up at any time during the one hour. You decide that whoever comes first waits for 20 minutes for the other person and leaves if the other person doesn't show up. What is the probability that you meet your friend and watch the movie tomorrow?

Solution

Let us first try and convert this problem into a mathematical abstraction we'd be most comfortable with ;).

To start with, we can map the one hour interval from 4:00 PM to 5:00 PM to an interval \( (0, 1) \).

With this mapping, 20 minutes will map to \( \frac{1}{3} \)

\( T_{1} \) - Number representing the time you reach the place

\( T_{2} \) - Number representing the time your friend reaches the place

Now we can break this problem into two scenarios when you reach before your friend. The case of your friend reaching before you should lead to the same probability and we can just double this probability for our answer.

Scenario 1: You reach at a time \( T_{1} \in (0, \frac{2}{3}) \) and your friend reaches at a time \( T_{2} \in (T_{1}, T_{1}+\frac{1}{3}) \)

\[ P[T_{1} \in (0, \frac{2}{3})  \&  T_{2} \in (T_{1}, T_{1}+\frac{1}{3})] = \int_{0}^{\frac{2}{3}}\int_{x_{1}}^{x_{1}+\frac{1}{3}}dx_{2}dx_{1} = \int_{0}^{\frac{2}{3}}\frac{1}{3}dx_{1} = \frac{1}{3}[x_{1}]_{0}^{\frac{2}{3}} = \frac{2}{9} \]

Scenario 2: You reach at a time \( T_{1} \in (\frac{2}{3}, 1) \) and your friend reaches at a time \( T_{2} \in (T_{1}, 1) \) 

\[ P[T_{1} \in (\frac{2}{3}, 1)  \&  T_{2} \in (T_{1}, 1)] = \int_{\frac{2}{3}}^{1}\int_{x_{1}}^{1}dx_{2}dx_{1} = \int_{\frac{2}{3}}^{1}(1-x_{1})dx_{1} = [x_{1} - \frac{x_{1}^{2}}{2}]_{\frac{2}{3}}^{1} = \frac{1}{18} \]

Probability that you will meet your friend given you get there first

\[ P[T_{1} \in (0, \frac{2}{3})  \&  T_{2} \in (T_{1}, T_{1}+\frac{1}{3})] + P[T_{1} \in (\frac{2}{3}, 1)  \&  T_{2} \in (T_{1}, 1)] = \frac{2}{9} + \frac{1}{18} = \frac{5}{18} \]

Similarly, probability that you will meet your friend given he gets there first

\[ P[T_{2} \in (0, \frac{2}{3})  \&  T_{1} \in (T_{2}, T_{2}+\frac{1}{3})] + P[T_{2} \in (\frac{2}{3}, 1)  \&  T_{1} \in (T_{2}, 1)] = \frac{5}{18} \]

Finally, probability that you will meet your friend = \( 2 * \frac{5}{18} = \frac{5}{9} \)

Question

"N" people of different heights are standing in a line. What is the expected position of the tallest guy?

Solution

We can go about calculating the probability of tallest guy being at position \( k \) using two methods

Method 1

There is a \( \frac{1}{N} \) chance that the tallest guy is present at any of the \( N \) available positions. So, probability that the tallest guy is at position k can be written as

\[ P[T = k] = \frac{1}{N} \]

Method 2

# ways in which the \( N \) people with different heights can stand in a line = \( N! \)

# ways in which the tallest person is at position \( k \) = \( (N-1)! \)

\[ P[T=k] = \frac{(N-1)!}{N!} = \frac{1}{N} \]

\[ E[T] = \sum_{1}^{N}k\frac{1}{N} = \frac{N(N+1)}{2N} = \frac{N+1}{2} \]

Question

Given \( x_{1}, x_{2}, ... \) are iids from a continuous uniform distribution with a range \( [0, 1] \), and define \( S_{n} \) as

\[ S_{n} = \sum_{i = 1}^{n}x_{i} \]

Let \( N \) be the minimal \( n \) such that \( S_{n} > 1 \). What is E[N]?

Solution

Let us start with the probability that the sum \( S_{n} \) is less than or equal to 1.

\( P[S_{1} \le 1] = 1 \)

\( P[S_{2} \le 1] = \int_{0}^{1}\int_{0}^{1-x_{1}}dx_{2}dx_{1} = \int_{0}^{1}(1-x_{1})dx_{1} = [-\frac{(1-x_{1})^2}{2}]_{0}^{1} = \frac{1}{2} \)

For a general \( S_{n} \)

\( P[S_{n} \le 1] = \int_{0}^{1}\int_{0}^{1-x_{1}}\int_{0}^{1-x_{1}-x_{2}}...\int_{0}^{1-x_{1}-...-x_{n-1}}dx_{n}dx_{n-1}...dx_{1} \)

\( = \int_{0}^{1}\int_{0}^{1-x_{1}}...\int_{0}^{1-x_{1}-...-x_{n-2}}(1-x_{1}-...x_{n-1})dx_{n-1}...dx_{1} \)

\( = \int_{0}^{1}\int_{0}^{1-x_{1}}...\int_{0}^{1-x_{1}-...-x_{n-3}}[-\frac{(1-x_{1}-...-x_{n-1})^2}{2}]_{0}^{1-x_{1}-...-x_{n-2}}dx_{n-2}...dx_{1} \)

\( = \int_{0}^{1}\int_{0}^{1-x_{1}}...\int_{0}^{1-x_{1}-...-x_{n-3}}\frac{(1-x_{1}-...-x_{n-2})^2}{2}dx_{n-2}...dx_{1} \)

\( = \int_{0}^{1}\int_{0}^{1-x_{1}}...\int_{0}^{1-x_{1}-...-x_{n-4}}[-\frac{(1-x_{1}-...-x_{n-2})^3}{3!}]_{0}^{1-x_{1}-...-x_{n-3}}dx_{n-3}...dx_{1} \)

\( = \int_{0}^{1}\int_{0}^{1-x_{1}}...\int_{0}^{1-x_{1}-...-x_{n-3}}\frac{(1-x_{1}-...-x_{n-3})^3}{3!}dx_{n-3}...dx_{1} \)

\( ... \)

\( ... \)

\( ... \)

\( = \int_{0}^{1}\frac{(1-x_{1})^{n-1}}{(n-1)!}dx_{1} \)

\( = [-\frac{(1-x_{1})^{n}}{n!}]_{0}^{1} \)

\( = \frac{1}{n!} \)

We now calculate the conditional probability that the sum \( S_{n} \) is greater than 1 given that the sum till then \( S_{n-1} \) is less than or equal to 1. Since the possibilities leading to \( S_{n} \le 1 \) is a subset of \( S_{n-1} \le 1 \), we can calculate our conditional probability as

\[ P[S_{n} > 1 | S_{n-1} \le 1] = \frac{\frac{1}{(n-1)!} - \frac{1}{n!}}{\frac{1}{(n-1)!}} = \frac{n-1}{n} \]

Now we calculate the probability of n being the minimal iid where the sum exceeds 1.

\[ P[S_{n} > 1  \&  S_{n-1} \le 1] = P[S_{n-1} < 1]P[S_{n} > 1 | S_{n-1} < 1] = \frac{1}{(n-1)!}*\frac{n-1}{n} = \frac{n-1}{n!} \]

Finally the expectation

\[ E[N] = \sum_{n=2}^{\infty}n*P[S_{n} > 1 \& S_{n-1} \le 1] = \sum_{n=2}^{\infty}n*\frac{n-1}{n!} = \sum_{n=2}^{\infty}\frac{1}{(n-1)!} = \sum_{n=1}^{\infty}\frac{1}{n!} = e \]

Question

Given

\[ \frac{1}{a} + \frac{1}{b} = \frac{1}{c} \]

and

\[ gcd(a, b, c) = 1 \]

is \( a + b \) a perfect square?

Solution

Let \( gcd(a, b) = k \)

We can write \( a = ka_{1}  \&  b = kb_{1} \), where \( gcd(a_{1}, b_{1}) = 1 \) and that implies

\[ gcd(a_{1}, a_{1}+b_{1}) = 1  \&  gcd(b_{1}, a_{1}+b_{1}) = 1 \]

\[ \frac{1}{a} + \frac{1}{b} = \frac{1}{c} \]

\[ \implies (a + b)c = ab \]

\[ \implies k(a_{1} + b_{1})c = k^2a_{1}b_{1} \]

\[ \implies (a_{1} + b_{1})c = ka_{1}b_{1} \]

\[ gcd(a_{1}+b_{1}, a_{1}) = 1  \&  gcd(a_{1}+b_{1}, b_{1}) = 1 \implies (a_{1}+b_{1}) | k \]

\[ gcd(a_{1}, a_{1}+b_{1}) = 1, gcd(b_{1}, a_{1}+b_{1}) = 1  \&  gcd(c, k) = 1 \implies k | (a_{1}+b_{1}) \]

\[ (a_{1}+b_{1}) | k  \&  k | (a_{1}+b_{1}) \implies a_{1}+b_{1} = k \]

\[ a + b = k(a_{1} + b_{1}) = k^{2} \]

Question:

You are about to get on a plane to Seattle. You want to know if you should bring an umbrella. You call 3 random friends of yours who live in Seattle and ask them independently whether its raining there. Each of your friends has a \( \frac{2}{3} \) chance of telling you the truth and \( \frac{1}{3} \) chance of messing with you. All 3 friends tell you that "Yes" it's raining. What is the probability that it's actually raining in Seattle.

Solution:

We'll start by considering the various events

1. Probability of rain in Seattle - \( P[R] \)

2. Probability of not raining in Seattle - \( P[!R] = 1 - P[R] \)

3. Probability of all three saying yes - \( P[Y, Y, Y] \)

4. Probability of all three saying yes given it is actually raining - \( P[Y, Y, Y | R] \)

5. Probability of all three saying yes given it is actually not raining - \( P[Y, Y, Y | !R] \)

6. Probability of it actually raining given all three said yes - \( P[R | Y, Y, Y] \). This is the value we need to evaluate in our question.

The first observation is that when all three friends say it is raining, it is either that all three are speaking truth or all three are lying. If it is actually raining, all three are speaking truth, else all three are lying.

Probability of all three saying Yes

\[ P[Y, Y, Y] = P[R] * P[Y, Y, Y | R] + P[!R] * P[Y, Y, Y | !R] = (\frac{2}{3})^3P[R] + (\frac{1}{3})^{3}(1 - P[R]) = \frac{1 + 7P[R]}{27} \]

Using Baye's Theorem

\[ P[R | Y, Y, Y] = \frac{P[Y, Y, Y | R] * P[R]}{P[Y, Y, Y]} = \frac{\frac{8}{27}P[R]}{\frac{1 + 7P[R]}{27}} = \frac{8P[R]}{1 + 7P[R]} \]

Specific Scenarios

If \( P[R] = 0.5 \),

\[ P[R | Y, Y, Y] = \frac{8P[R]}{1 + 7P[R]} = \frac{8*0.5}{1 + 7*0.5} = \frac{8}{9} \]

if \( P[R] = 0.25 \),

Question:

Two friends A and B who have a long time to pass, decide to play a game with a fair coin where A flips the coin 1000 times and since B is younger he gets to flip it 1001 times. The person who gets more heads than the other wins. But since B gets to flip one more than A, A gets to win if they both get the same number of heads. Whats the probability of B winning the game?

Solution:

General Problem:

Lets take a quick look at a more general problem.

\( n \) - Number of times A flips the coin

\( m \) - number of times B flips the coin

I'll also assume for the sake of simplicity that \( m > n \)

\( k \) - number of heads A gets

\( r \) - number of heads B gets

Probability that B gets more heads than A

\[ P[r > k] = \sum_{k = 0}^{n} {n \choose k} \frac{1}{2^k} \frac{1}{2^{n-k}} \left( \sum_{r = k+1}^{m} {m \choose r} \frac{1}{2^{r}} \frac{1}{2^{m-r}} \right) = \frac{1}{2^{n+m}} \sum_{k = 0}^{n} {n \choose k} \left( \sum_{r = k+1}^{m} {m \choose r} \right) \]

\( \)

Special Case:

The above summation assumes some special properties when n = m+1

\[ \frac{1}{2^{2n+1}} \sum_{k = 0}^{n} {n \choose k} \left( \sum_{r = k+1}^{n+1} {n+1 \choose r} \right) = \frac{2^{2n}}{2^{2n+1}} = \frac{1}{2} \]

The above formula as can be seen evaluates to \( \frac{1}{2} \) owing to the symmetry of pascal's triangle and binomial coefficients as shown below.

\[ \sum_{k = 0}^{n} {n \choose k} \left( \sum_{r = k+1}^{n+1} {n+1 \choose r} \right) = \sum_{k = 0}^{n} {n \choose k} \left( \sum_{r = 0}^{k} {n+1 \choose r} \right) \]

\( \)

Current Game:

Question:

An employee at a call center works from 8:00 am until 5:00 pm with breaks between 10:30 - 10:45, 12:30 - 13:30, and 14:45 - 15:00. Assume that calls come in according to a poisson process with expected number of calls per hour equal to 6.

1. What is the probability that there are at-most 10 calls during the breaks?

2. What is the probability that the first call of the day is after 8:10 am?

3. What is the probability that the employee can do something for 45 minutes without being disturbed by a call?

Solution:

Poisson process:

From the information we have, the probability of getting \( k \) calls in \( t \) hours at the call center can be represented using poisson process as

\[ P[N(t) = k] = \frac{(\lambda t)^{k}}{k!}e^{-\lambda t} \]

where

\[ \lambda = 6/hour \]

The quantity \( \lambda t \) here represents the average number of calls received in a given time interval \( t \) starting at any point in time. This property makes it very useful in solving the questions.

\( \)

At-most 10 calls during the breaks:

Total time during the breaks = 1.5 hours

Though the breaks are spread out in time, we can club them together into one contiguous time frame owing to the property of poisson process mentioned above.

\[ P[N(1.5) \le 10] = \sum_{n = 0}^{10} \frac{(6*1.5)^{n}}{n!}e^{-6*1.5} = \sum_{n = 0}^{10}\frac{9^n}{n!}e^{-9} \]

\( \)

First call of the day is after 8:10

We can rephrase this question as getting no calls in the first 10 minutes and then we can also say that this probability will be same as not getting call in any given time frame of 10 minutes. So the probability of getting the first call of the day after 8:10 will be

\[ P[N(\frac{1}{6}) = 0] = \frac{(6*\frac{1}{6})^0}{0!}e^{-6*\frac{1}{6}} = e^{-1} = 0.368 \]

\( \)

No calls for 45 minutes

This probability follows straight from the definition of the random variable for number of calls in a given time interval

Question:

You were to meet a girl on a date at 7:00 pm. You reach the place on time to get a message from her informing she is stuck elsewhere, her phone ran out of charge, and a hint that chances of her making it are high currently but they drop exponentially with passing time. Anyway she says there is a 90% chance she would make it before 7:30. She also mentions that she has a constant rate \( h \) of showing up at her dates. Since you have some time to pass, you try to make some calculations about your impending future.

1. What would be her rate \( h \) of showing up at dates?

2. What is the chance that she will be there by 7:10?

3. What would be your expected wait time?

Solution:

Defining a distribution:

\( T \) - Time in minutes from 7:00 pm when the girl would show up

Since the chance of her making it decreases exponentially with time, we can model it with an exponential distribution using her rate of showing up for dates as

\[ P[T > t] = e^{-ht} \]

This distribution incorporates all her proclivities when it comes to making it to a date. The chances of her showing up reduce to zero with passing time.

\[ \lim_{t \to \infty} P[T > t] = \lim_{t \to \infty} e^{-ht} = 0 \]

The probability of her arriving before a given time t would be 

\[ P[T < t] = 1 - P[T > t] = 1 - e^{-ht} \]

The probability of her arriving at some time in the eternity is

\[ P[T < \infty] = 1 - e^{-h\infty} = 1 \]

\( \)

The girl's showing up rate for dates:

\[ P[T < 30] = 0.9  \implies 1 - e^{-30h} = 0.9 \implies e^{-30h} = 0.1 \implies h = \frac{1}{30}\ln{10} = 7.68\% \]

\( \)

Chances of her arriving before 7:10:

\[ P[T < 10] = 1 - e^{-10h} = 1 - \sqrt[3]{0.1} = 0.536 \]

\( \)

Your expected wait time:

We need to first evaluate the probability density function for our distribution to answer this question.

\[ F(t) = P[T < t] = \int_{0}^{t}f(s)ds \implies f(t) = \frac{dF(t)}{dt} = \frac{d}{dt}(1 - e^{-ht}) = he^{-ht} \]

Once we have the pdf, the expected wait time would be

\[ E[T] = \int_{0}^{\infty}Tf(T)dT = \int_{0}^{\infty}The^{-hT}dT = [-Te^{-hT}]_{0}^{\infty} + \int_{0}^{\infty}e^{-hT}dT = [-\frac{1}{h}e^{-hT}]_{0}^{\infty} = \frac{1}{h} \approx 13  mins \]

\( \)

Question:

What is Value at Risk? What are different ways of measuring Var? What are the advantages and disadvantages of VaR?

Solution:

Value at Risk is a measure of expected maximum/minimum loss on an asset or a portfolio of assets. Given a portfolio of assets and a time frame we are interested in, VaR defines a loss threshold at a particular probability known as level of significance. It is different from the traditional measure of risk - volatility in that VaR measures the downside threshold whereas volatility measures a magnitude of movement without reference to direction of movement.

\( \)

Examples:

5% Daily VaR on a Portfolio is $200,000 means that there is a 5% probability (once in 20 days) that the portfolio may lose more than $200,000 in a day.

1% Weekly VaR on a portfolio is $400,000 means that there is a 1% probability (i.e once in 100 weeks) that the portfolio may lose more than $400,000 in a week.

\( \)

VaR Terms:

Significance Level:

The number 5% in the VaR measure above indicating the probability of the loss crossing the threshold is called significance level.

Level of Confidence:

\[ Level  of  Confidence = 1 - Significance  Level \]

A $200,000 Daily-VaR at a 95% level of confidence means 95% of the times, the portfolio may not lose more than $200,000 on a given day. This is the same as 5% Daily VaR of $200,000 said in terms of level of confidence.

Time Frame:

VaR calculations can be done for different time-frames. We can look at daily VaR to see how much the portfolio might lose in a day and weekly VaR to check the possible loss in a week and so on.

Threshold:

The threshold indicates the minimum loss at a given significance level or a maximum loss at a given confidence level.

\( \)

Measuring VaR:

Historical method

Historical method, as the name suggest uses a history of asset price/returns to calculate Value at Risk at a particular significance level.

Estimation:

Let us say we are interested in calculating s% daily VaR on a portfolio of assets.

Input:

A history of daily returns on the portfolio over a period of time.

Process:

1. Sort the given set in the ascending order of daily returns.

2. Find the boundary in this sorted daily returns that separates the least s% of daily returns from the rest of the daily returns and you have your s% daily VaR.

We don't have to actually sort these returns and there is a cheaper way to do this using a heap structure.

Advantages:

1. The key advantage in using historical VaR is that it doesn't assume any parametric model for the asset values allowing the possibility of the risk factors following any distribution.

2. The independence from parametric models for risk factors removes errors related to estimating model parameters.

Disadvantages:

1. The key disadvantage in using historical method is its assumption that asset returns in future will completely mimic their returns from specified historical window. This is valid only when all underlying risk-factor dynamics remain the same in the future to what they were in the reference window.

2. In the purest form, validating the above assumption requires collecting all possible risk factors and over a reasonably long historical window. Gathering all possible risk-factors will be quite a task.

\( \)

Variance-Covariance/Delta-normal method:

We use the same input here as in the Historical method, but we don't assume that the history is going to repeat itself. Instead we assume that the asset prices follow a particular distribution and calculate the parameters for the distribution based on the historical data.

Estimation:

Let us say we are interested in calculating s% daily VaR on a portfolio of assets assuming the portfolio returns follow a normal distribution.

Input:

A history of daily returns on the portfolio over a period of time.

Process:

1. Find the mean \( \mu \) and standard deviation \( \sigma \) for portfolio daily returns. These are the parameters for normal distribution.

\[ \mu = \frac{1}{N}\sum_{i = 1}^{N}R_{i} \]

\[ \sigma = \sqrt{\frac{1}{N}(R_{i} - \mu)^2} \]

2. Once you have the parameters for the distribution, you can calculate VaR using the corresponding one-sided table.

Advantages:

1.The main advantage for this method is that it is very simple to calculated VaR. As the historical data changes, the parameters used for estimating VaR change. and you calculate VaR using corresponding formulas.

Disadvantages:

1.The main disadvantage is that the assumption of normal distribution might not be true for many asset classes. A lot of asset class returns exhibit positive kurtosis and so a larger tail. This means that variance/covariance method would underestimate VaR for these asset classes.

2.This also doesn't scale well with non-linear combinations of assets. I will talk about this in a post on calculating portfolio VaR.

\( \)

Monte-Carlo simulation method:

This method doesn't consider historical portfolio values to predict the future. Instead, it looks to a probabilistic model for future asset values based on certain risk-factors with provision for non-linearity, various distributional properties like skewness and kurtosis, and for varying parameters. In a nut-shell it offers more flexibility in modeling the future prices.

Estimation:

Let us say we are interested in calculating s% VaR for a portfolio of assets.

Input:

A model for future asset prices and corresponding parameters and a number N of simulations.

Process:

1. Run the simulation for asset prices based on the model for the given time period and calculate the return in each simulation.

2. Run N such simulations and make a list of the returns.

3. Take the smallest s% of the returns from all the simulations. That highest number there will be our s% VaR.

Again as in the historical method, we can just use a heap structure to calculate s% VaR.

Advantages:

1. Monte-Carlo simulations are the most flexible of the methods used for calculating VaR and thereby minimize your assumptions.

Disadvantages:

Question:

Suppose \( S_{1}(t) \) and \( S_{2}(t) \) follow Geometric Brownian Motions which are instantaneously correlated with positive correlation \( \rho \). Price an option with pay-off \( max(S_{1}(T) - S_{2}(T), 0) \). If you are the owner of such an option, are you long or short correlation?

Solution:

The option in question is an Exchange Option. From the link above, we can see that the option satisfies the following equation

\[ V(S_{1}(t), S_{2}(t), t) = S_{2}(t)H(\xi_{t}, t)  where  \xi_{t} = \frac{S_{1}(t)}{S_{2}(t)} \]

\[ \& \]

\[ H = \frac{1}{D_{2}}(\frac{\partial H}{\partial t} + (D_{2} - D_{1})\xi_{t}\frac{\partial H}{\partial \xi_{2}} + \frac{1}{2}(\sigma_{1}^2 + \sigma_{2}^2 - 2\rho\sigma_{1}\sigma_{2})\xi_{2}\frac{\partial^2H}{\partial \xi_{t}^2}) \]

From the equation, we can see that

\[ \frac{\partial^2H}{\partial \xi_{t}^2} > 0 \implies short-correlation \]

\[ \& \]

\[ \frac{\partial^2H}{\partial \xi_{t}^2} < 0 \implies long-correlation \]

The equation for \( H \) with the boundary conditions resembles a European call option. As discussed in Option Greeks - Gamma, the gamma for a long position \( H \) should be positive. This means that the exchange option holder is short correlation.

Arguing in another way, the option holder expects to make money by an increase in the difference in the asset values when they move away from each other in a particular direction.

\[ S_{1}(t)\uparrow  \&  S_{2}(t)\downarrow \]

Question:

What is an exchange option? How do you price it?

Solution:

Exchange Option:

An exchange option is an option written on two assets \( S_{1}  \&  S_{2}\) with their difference \( S_{1} - S_{2} \) as payoff when \( S_{1} > S_{2} \). Put another way, the asset price \( S_{2} \) acts as the strike price for the option, and the option allows you to exchange \( S_{2} \) for \( S_{1} \) when it expires in the money, hence the name exchange option. Mathematically, the payoff function for exchange option can be written as

\[ V(S_{1,T}, S_{2,T}, T) = max(S_{1} - S_{2}, 0) \]

A more general form for the exchange option payoff where you exchange \( n_{2} \) units of \( S_{2} \) for \( n_{1} \) units of \( S_{1} \) would be

\[ V(S_{1,T}, S_{2,T}, T) = max(n_{1}S_{1,T} - n_{2}S_{2,T}, 0) \]

The corresponding PDE as can be derived using the equation for Multi-Asset options is

\[ \frac{\partial V}{\partial t} + (r - D_{1})S_{1,t}\frac{\partial V}{\partial S_{1,t}} + (r - D_{2})S_{2,t}\frac{\partial V}{\partial S_{2,t}} + \frac{1}{2}\sigma_{1}^2S_{1}^2\frac{\partial^2V}{\partial S_{1,t}^2} + \frac{1}{2}\sigma_{2}^2S_{2,t}^2\frac{\partial^2V}{\partial S_{2,t}^2} + \rho\sigma_{1}\sigma_{2}S_{1,t}S_{2,t}\frac{\partial^2V}{\partial S_{1,t}\partial S_{2,t}} - rV= 0 \]

\( \)

Similarity Reduction:

This equation can be simplified by considering a similarity reduction for the option price \( V(S_{1,t}, S_{2,t}, t) \) as

\[ V(S_{1,t}, S_{2,t}, t) = n_{1}S_{2}H(\xi_{t}, t) \]

where

\[ \xi_{t} = \frac{S_{1,t}}{S_{2,t}} \]

The boundary condition for \( H(\xi, t) \) becomes

\[ H(\xi_{T}, T) = max(\xi_{T} - \frac{n_{2}}{n_{2}}, 0) \]

We can now substitute for \( V(S_{1,t}, S_{2,t}, t) \) in the equation after evaluating each of the partial derivatives

\[ \frac{\partial V}{\partial t} = \frac{\partial(n_{1}S_{2,t}H)}{\partial t} = n_{1}S_{2,t}\frac{\partial H}{\partial t} \]

\[ \frac{\partial V}{\partial S_{1,t}} = \frac{\partial(n_{1}S_{2,t}H)}{\partial S_{1,t}} = n_{1}S_{2,t}\frac{\partial H}{\partial S_{1,t}} \]

\[ \frac{\partial V}{\partial S_{2,t}} = \frac{\partial(n_{1}S_{2,t}H)}{\partial S_{2,t}} = n_{1}S_{2,t}\frac{\partial H}{\partial S_{2,t}} + n_{1}H \]

\[ \frac{\partial^2V}{\partial S_{1,t}^2} = \frac{\partial}{\partial S_{1,t}}(\frac{\partial V}{\partial S_{1,t}}) = \frac{\partial }{\partial S_{1,t}}(n_{1}S_{2,t}\frac{\partial H}{\partial S_{1,t}}) = n_{1}S_{2,t}\frac{\partial^2H}{\partial S_{1,t}^2} \]

\[ \frac{\partial^2V}{\partial S_{2,t}^2} = \frac{\partial}{\partial S_{2,t}}(\frac{\partial V}{\partial S_{2,t}}) = \frac{\partial }{\partial S_{2,t}}(n_{1}S_{2,t}\frac{\partial H}{\partial S_{2,t}} + n_{1}H) = n_{1}S_{2,t}\frac{\partial^2H}{\partial S_{2,t}^2} + 2n_{1}\frac{\partial H}{\partial S_{2,t}} \]

\[ \frac{\partial^2V}{\partial S_{1,t}\partial S_{2,t}} = \frac{\partial}{\partial S_{1,t}}(\frac{\partial V}{\partial S_{2,t}}) = \frac{\partial }{\partial S_{1,t}}(n_{1}S_{2,t}\frac{\partial H}{\partial S_{2,t}} + n_{1}H) = n_{1}S_{2,t}\frac{\partial^2H}{\partial S_{1,t}\partial S_{2,t}} + n_{1}\frac{\partial H}{\partial S_{1,t}} \]

Substituting for the derivatives in the equation and removing the common factor \( n_{1}S_{2,t} \), we get

\[ \frac{\partial H}{\partial t} + (r - D_{1})S_{1,t}\frac{\partial H}{\partial S_{1,t}} + (r - D_{2})S_{2,t}\frac{\partial H}{\partial S_{2,t}} + (r - D_{2})H + \frac{1}{2}\sigma_{1}^2S_{1,t}^2\frac{\partial^2H}{\partial S_{1,t}^2} + \frac{1}{2}\sigma_{2}^2S_{2,t}^2\frac{\partial^2H}{\partial S_{2,t}^2} + \sigma_{2}^2S_{2,t}\frac{\partial H}{\partial S_{2,t}} + \rho\sigma_{1}\sigma_{2}S_{1,t}S_{2,t}\frac{\partial^2H}{\partial S_{1,t}\partial S_{2,t}} + \rho\sigma_{1}\sigma_{2}S_{1,t}\frac{\partial H}{\partial S_{1,t}} - rH= 0 \]

\[ \frac{\partial H}{\partial t} + (r - D_{1} + \rho\sigma_{1}\sigma_{2})S_{1,t}\frac{\partial H}{\partial S_{1,t}} + (r - D_{2} + \sigma_{2}^2)S_{2,t}\frac{\partial H}{\partial S_{2,t}} + \frac{1}{2}\sigma_{1}^2S_{1,t}^2\frac{\partial^2H}{\partial S_{1,t}^2} + \frac{1}{2}\sigma_{2}^2S_{2,t}^2\frac{\partial^2H}{\partial S_{2,t}^2} + \rho\sigma_{1}\sigma_{2}S_{1,t}S_{2,t}\frac{\partial^2H}{\partial S_{1,t}\partial S_{2,t}} - D_{2}H= 0 \]

Moving from \( S_{1,t}  \&  S_{2,t} \) to \( \xi_{t} \)

\[ \frac{\partial H}{\partial S_{1,t}} = \frac{\partial \xi_{t}}{\partial S_{1,t}}\frac{\partial H}{\partial \xi_{t}} = \frac{1}{S_{2,t}}\frac{\partial H}{\partial \xi_{t}} \]

\[ \frac{\partial H}{\partial S_{2,t}} = \frac{\partial \xi_{t}}{\partial S_{2,t}}\frac{\partial H}{\partial \xi_{t}} = -\frac{\xi_{t}}{S_{2,t}}\frac{\partial H}{\partial \xi_{t}} \]

\[ \frac{\partial^2H}{\partial S_{1,t}^2} = \frac{\partial}{\partial S_{1,t}}(\frac{1}{S_{2,t}}\frac{\partial H}{\partial \xi_{t}}) = \frac{1}{S_{2,t}^2}\frac{\partial^2H}{\partial \xi_{t}^2} \]

\[ \frac{\partial^2H}{\partial S_{2,t}^2} = \frac{\partial}{\partial S_{2,t}}(-\frac{S_{1,t}}{S_{2,t}^2}\frac{\partial H}{\partial \xi_{t}}) = \frac{\xi_{t}^2}{S_{2,t}^2}\frac{\partial^2H}{\partial \xi_{t}^2} + \frac{2\xi_{t}}{S_{2,t}^2}\frac{\partial H}{\partial \xi_{t}} \]

\[ \frac{\partial^2H}{\partial S_{1,t}\partial S_{2,t}} = \frac{\partial}{\partial S_{2,t}}(\frac{1}{S_{2,t}}\frac{\partial H}{\partial \xi_{t}}) = -\frac{\xi_{t}}{S_{2,t}^2}\frac{\partial^2H}{\partial \xi_{t}^2} - \frac{1}{S_{2,t}^2}\frac{\partial H}{\partial \xi_{t}} \]

Substituting in the equation, we have

\[ \frac{\partial H}{\partial t} + (r - D_{1} + \rho\sigma_{1}\sigma_{2})\xi_{t}\frac{\partial H}{\partial \xi_{t}} - (r - D_{2} + \sigma_{2}^2)\xi_{t}\frac{\partial H}{\partial \xi_{t}} + \frac{1}{2}\sigma_{1}^2\xi_{t}^2\frac{\partial^2H}{\partial \xi_{t}^2} + \frac{1}{2}\sigma_{2}^2\xi_{t}^2\frac{\partial^2H}{\partial \xi_{t}^2} + \sigma_{2}^{2}\xi_{t}\frac{\partial H}{\partial \xi_{t}} - \rho\sigma_{1}\sigma_{2}\xi_{t}^2\frac{\partial^2H}{\partial \xi_{t}^2} -\rho\sigma_{2}\sigma_{2}\xi_{t}\frac{\partial H}{\partial \xi_{t}} - D_{2}H= 0 \]

\[ \implies \frac{\partial H}{\partial t} + (D_{2} - D_{1})\xi_{t}\frac{\partial H}{\partial \xi_{t}} + \frac{1}{2}(\sigma_{1}^2 + \sigma_{2}^2 - 2\rho\sigma_{1}\sigma_{2})\xi_{t}^2\frac{\partial^2H}{\partial \xi_{t}^2} - D_{2}H= 0 \]

This equation now has the same structure as the Black-Scholes PDE and the boundary conditions similar to a vanilla call option. So the price of the Exchange option would be

\[ V(S_{1,t}, S_{2,t}, t) = n_{1}S_{1,t}e^{-D_{1}(T-t)}\Phi(d_{1}) - n_{2}S_{2,t}e^{-D_{2}(T-t)}\Phi(d_{2}) \]

where

\[ d_{1} = \frac{ln\frac{n_{1}S_{1}}{n_{2}S_{2}} + (D_{2} - D_{1} + \frac{\sigma^2}{2})(T-t)}{\sigma\sqrt{T-t}}  \&  d_{2} = d_{1} - \sigma\sqrt{T-t} \]

where

Question:

Derive the Option Pricing Partial Differential Equation for Multi-Asset options.

Solution:

Given a universe of \( N \) assets \( S_{1}, \cdots, S_{N} \) that follow a log-normal random walk based on \( N \) brownian motions \( W_{1}, \cdots, W_{N}\)

\[ dS_{i,t} = \mu_{i}S_{i,t}dt + \sigma_{i}S_{i,t}dW_{i} \]

with correlation \( \rho_{ij} \) between brownian two motions \( W_{i}  \& W_{j} \), i.e

\[ dW_{i}dW_{j} = \rho_{ij}dt \]

For an option \( V(S_{1,t}, \cdots, S_{N,t}, t) \) on the \( N \) assets, we can use the multi-dimensional Ito to get

\[ dV(S_{1,t}, \cdots,S_{N,t}) = (\frac{\partial V}{\partial t} + \frac{1}{2}\sum_{i=1}^{N}\sum_{j=1}^{N}\rho_{ij}\sigma_{i}\sigma_{j}S_{i}S_{j}\frac{\partial^2V}{\partial S_{i}\partial S_{j}})dt + \sum_{i=1}^{N}\frac{\partial V}{\partial S_{i,t}}dS_{i,t} \]

\( \)

Delta Hedging:

The delta hedging for a multi-asset option follows the same dynamics as the concet of delta hedging used in deriving Black Scholes PDE for single asset options

Let us consider a portfolio \( \Pi \) consisting long an option and short \( \Delta_{i,t} \) of asset \( S_{i,t} \) for all the \( N \) assets.

\[ \Pi = V(S_{1,t}, \cdots, S_{N,t}) - \sum_{i=1}^{N}\Delta_{i,t}S_{i,t} \]

A change in this portfolio would be

\[ d\Pi = dV(S_{1,t, \cdots, S_{N,t}}) - \sum_{i=1}^{N}\Delta_{i,t}dS_{i,t} \]

\[ \implies d\Pi = (\frac{\partial V}{\partial t} + \frac{1}{2}\sum_{i=1}^{N}\sum_{j=1}^{N}\rho_{ij}\sigma_{i}\sigma_{j}S_{i}S_{j}\frac{\partial^2V}{\partial S_{i}\partial S_{j}})dt + \sum_{i=1}^{N}(\frac{\partial V}{\partial S_{i,t}} - \Delta_{i,t})dS_{i,t} \]

By choosing

\[ \Delta_{i,t} = \frac{\partial V}{\partial S_{i,t}}, 1 \le i \le N \]

we can eliminate the random/diffusion term from the equation and be left with only the deterministic/drift term making

\[ d\Pi = (\frac{\partial V}{\partial t} + \frac{1}{2}\sum_{i=1}^{N}\sum_{j=1}^{N}\rho_{ij}\sigma_{i}\sigma_{j}S_{i}S_{j}\frac{\partial^2V}{\partial S_{i}\partial S_{j}})dt \]

Now since the randomness in the portfolio is removed, the portfolio should grow at a risk-free rate. i.e

\[ d\Pi = r\Pi dt = r(V - \sum_{i=1}^{N}\Delta_{i,t}S_{i,t})dt \]

From the two equations for \( d\Pi \) above, we have

\[ r(V - \sum_{i=1}^{N}\Delta_{i,t}S_{i,t}) = \frac{\partial V}{\partial t} + \frac{1}{2}\sum_{i=1}^{N}\sum_{j=1}^{N}\rho_{ij}\sigma_{i}\sigma_{j}S_{i}S_{j}\frac{\partial^2V}{\partial S_{i}\partial S_{j}} \]

\[ \frac{\partial V}{\partial t} + r\sum_{i=1}^{N}S_{i,t}\frac{\partial V}{\partial S_{i,t}} + \frac{1}{2}\sum_{i=1}^{N}\sum_{j=1}^{N}\rho_{ij}\sigma_{i}\sigma_{j}S_{i}S_{j}\frac{\partial^2V}{\partial S_{i}\partial S_{j}} - rV = 0 \]

Each asset with a continuous dividend yield \( D_{i} \), the corresponding equation becomes

Question:

Write an implementation for pricing options using explicit finite difference method.

Solution:

This is an implementation for Explicit Finite Difference method for pricing options.

Question:

What are Finite Difference methods and how are they used in option pricing?

Solution:

Finite Difference Methods:

Finite Difference methods are numerical methods used to find approximate solutions to partial differential equations/Boundary Value problems. This is achieved by writing PDEs as difference equations and derivatives approximated with difference quotients.

It is not always possible/simple to find a continuous closed form solution to partial differential equations. Instead, we divide the continuous space into a discrete n-dimensional grid (n is the number of independent variables in the PDE) of closely spaced points. We then use the boundary conditions and the difference quotients to numerically evaluate the values at these discrete points.

\( \)

Option Pricing:

In the option pricing world, we can take a look at Black Scholes equation and see how it can be solved numerically for evaluating the option price \( V(S_{t}, t) \) using finite difference methods. The option value is a function of 2 independent variables \( S_{t}  \&  t \), which means option value space will be represented by a two-dimensional finite difference grid as shown below.

\[ \begin{array}\mbox{V(i\delta S, k\delta t)_{\searrow}} & M\delta t & \cdots & \cdots & k\delta t & \cdots & 0 \end{array} \]

\[ \begin{array}\mbox{N\delta S} & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet\end{array} \]

\[ \begin{array}\mbox{\cdots} & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet\end{array} \]

\[ \begin{array}\mbox{\cdots} & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet\end{array} \]

\[ \begin{array}\mbox{\cdots} & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet\end{array} \]

\[ \begin{array}\mbox{i\delta S} & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet\end{array} \]

\[ \begin{array}\mbox{\cdots} & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet\end{array} \]

\[ \begin{array}\mbox{\cdots} & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet\end{array} \]

\[ \begin{array}\mbox{0} & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet\end{array} \]

\[ t \longrightarrow \]

\[ \longleftarrow \tau = T-t \]

where

Time axis - represents time till expiry and divides the horizontal axis into discrete time points separated by \( \delta t \).

Asset axis - represents asset price and divides the vertical axis into discrete Asset price points separated by \( \delta S \).

Grid point \( (i, k) \) - represents Asset prcie \( i\delta S \) and time \( T - k\delta t \)

N - Number of points along Asset axis implying maximum asset price \( N\delta S \)

M - Number of points along Time axis

\( V_{i}^{k} = V(i\delta S, T - k\delta t) \) - The value of the option at a given point on the grid.

\( \)

Boundary Conditions:

Once we have our grid space, we need some boundary conditions in the form of option values at certain points on the grid. The processing starts with the values defined by these conditions and move along the grid by evaluating the rest of the values using what we already know and the PDE that binds them. For option pricing we get one set of boundary conditions through option payoff. We start with option values at option expiry in the form of payoff for each asset price - represented in blue below. The other boundary conditions will be with respect to \( S_{t} = 0 \)-yellow and \( S_{t} = \infty or S_{Max} \)-red which are specific to the option at hand.

\[ \begin{array}\mbox{\color{red}\bullet} & \color{red}\bullet & \color{red}\bullet & \color{red}\bullet & \color{red}\bullet & \color{red}\bullet & \color{red}\bullet & \color{red}\bullet & \color{red}\bullet & \color{red}\bullet & \color{red}\bullet & {\color{Blue}\bullet}\end{array} \]

\[ \begin{array}\mbox{\bullet} & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & {\bullet} & {\color{Blue}\bullet}\end{array} \]

\[ \begin{array}\mbox{\bullet} & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & {\color{Blue}\bullet}\end{array} \]

\[ \begin{array}\mbox{\bullet} & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & {\color{Blue}\bullet}\end{array} \]

\[ \begin{array}\mbox{\bullet} & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & {\color{Blue}\bullet}\end{array} \]

\[ \begin{array}\mbox{\bullet} & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & {\color{Blue}\bullet}\end{array} \]

\[ \begin{array}\mbox{\bullet} & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & {\color{Blue}\bullet}\end{array} \]

\[ \begin{array}\mbox{\color{yellow}\bullet} & \color{yellow}\bullet & \color{yellow}\bullet & \color{yellow}\bullet & \color{yellow}\bullet & \color{yellow}\bullet & \color{yellow}\bullet & \color{yellow}\bullet & \color{yellow}\bullet & \color{yellow}\bullet & \color{yellow}\bullet & {\color{Blue}\bullet}\end{array} \]

Evaluating Grid Values:

The process of finite difference methods for option pricing involves using the option values at the boundary to go backwards and evaluate the values at every point on the grid. We make these evaluations with the option value at a given point expressed as a function of derivatives.

\[ V(S_{t}, t) = \frac{1}{r}(\frac{\partial V(S_{t}, t)}{\partial t} + (r - D)S_{t}\frac{\partial V(S_{t}, t)}{\partial S_{t}} + \frac{1}{2}\sigma^{2}S_{t}^{2}\frac{\partial^2V(S_{t}, t)}{\partial S_{t}^2}) \]

or

\[ V(S_{t}, t) = \frac{1}{r}(\Theta(S_{t}, t) + (r - D)S_{t}\Delta(S_{t}, t) + \frac{1}{2}\sigma^{2}S_{t}^2\Gamma(S_{t}, t)) \]

We need to evaluate each of these partial derivatives by expressing them as finite differences and plug them into the above equation to get option value.There are different ways to evaluate these values. I will discuss the explicit method here and will talk about others in a different post.

\( \)

Explicit Finite Difference:

With the explicit finite difference method, three values on the grid are used to calculate another value on the grid as shown below.

\[ \hspace{4 cm}_\swarrow\color{blue}\bullet V_{i+1}^{k} \]

\[ V_{i}^{k-1}\color{red}\bullet\leftarrow\color{blue}\bullet V_{i}^{k} \]

\[ \hspace{4 cm}^\nwarrow\color{blue}\bullet V_{i-1}^{k} \]

Here are the expressions for the derivatives

\[ \Theta = \frac{\partial V}{\partial t} = \frac{V_{i}^{k} - V_{i}^{k-1}}{\delta t} \]

\[ \Delta = \frac{\partial V}{\partial S_{t}} = \frac{V_{i+1}^{k} - V_{i-1}^{k}}{2\delta S} \]

\[ \Gamma = \frac{\partial^2V}{\partial S_{t}^2} = \frac{V_{i+1}^{k} -2V_{i}^{k} + V_{i-1}^{k}}{\delta S^2} \]

Substituting in the Black Scholes equation, we have

\[ rV_{i}^{k} = \frac{V_{i}^{k} - V_{i}^{k-1}}{\delta t} + (r - D)i\delta S \frac{V_{i+1}^{k} - V_{i-1}^{k}}{2\delta S} + \frac{1}{2}\sigma^2i^2\delta S^2\frac{V_{i+1}^{k} - 2V_{i}^{k} + V_{i-1}^{k}}{\delta S^2} \]

\[\implies rV_{i}^{k} = \frac{V_{i}^{k} - V_{i}^{k-1}}{\delta t} + (r - D)i \frac{V_{i+1}^{k} - V_{i-1}^{k}}{2} + \frac{1}{2}\sigma^2i^2(V_{i+1}^{k} - 2V_{i}^{k} + V_{i-1}^{k}) \]

\[ \implies 2r\delta tV_{i}^{k} = 2(V_{i}^{k} - V_{i}^{k-1}) + (r - D)i\delta t (V_{i+1}^{k} - V_{i-1}^{k}) + \sigma^2i^2\delta t(V_{i+1}^{k} - 2V_{i}^{k} + V_{i-1}^{k}) \]

\[ \implies 2V_{i}^{k-1} = \delta t(\sigma^2i^2 - (r - D)i)V_{i-1}^{k} + 2(1 - \delta t(r + \sigma^2i^2))V_{i}^{k} + \delta t(\sigma^2i^2 + (r - D)i)V_{i+1}^{k} \]

\[ \implies V_{i}^{k-1} = \frac{1}{2}\delta t(\sigma^2i^2 - (r - D)i)V_{i-1}^{k} + (1 - \delta t(r + \sigma^2i^2))V_{i}^{k} + \frac{1}{2}\delta t(\sigma^2i^2 + (r - D)i)V_{i+1}^{k} \]

Question:

Write an implementation for evaluating the Minimum variance portfolios for a given asset universe - i.e asset returns, standard deviations and correlations

Solution:

Here is an implementation using c++ boost libraries for evaluating Minimum Variance Portfolio and Global Minimum Variance Portfolio.

Question:

What is implied volatility? Write an implementation for calculating the implied volatility of an option.

Solution:

Implied Volatility:

Implied Volatility is obtained by plugging in quoted option price into Black-Scholes formula or your option pricing formula to solve for volatility. We can look at implied volatility as the volatility in the underlying as implied by the option prices - a market view or perceived future volatility of the underlying.

More theory on the relation between Implied Volatility and Option Prices

Implementation:

This implementation is part of the COptionPricingBlackScholes class, the rest of which can be found at Option Pricing Black Scoles