Volatility and Greeks

8/13/2019 Volatility and Greeks

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Volatility and Greeks

By A V Vedpuriswar

February 8, 2009


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Basics of continuous compounding


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Time value of money for frequent compounding

Principal = Rs. 100. Interest rate = 10%

Annual compounding means

amount = 100 (1.1) = 110

Semi Annual compounding means

amount = 100 (1.05)2 = 110.3

Quarterly compounding means

amount = 100 (1.025)4 = 110.38


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Daily compounding means

amount = 100 (1.000274)365= 110.52

Continuous compounding means

amount = 100 (1+.10/m)m

= [100( 1 +.10/m)m/.10].10

= 100e.10

= 110.52


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Another example

Interest period

k

Rate of interest

(R/k)

Future value after one year (start with CHF 1)

VT= V0 (1 + R/k)T kV1= 1 (1 + R/k)

1 k

Year 1 0.06/1= 0.06

1.061

= 1.06

Six months 2 0.06/2= 0.03

1.032

=1.0609

Quarter 4 0.06/4

= 0.015

1.0154

= 1.061363550625

Month 12 0.06/12= 0.005

1.00512

= 1.061677811865

Week 52 0.06/52= 0.00115385

1.0011538552

= 1.061799819549

Day 365 0.06/365= 0.00016438

1.00016438365= 1.061831310678

Hour 8760(365 24)

0.06/8760= 0.0000068493

1.00000684938760

= 1.061836328360

Minute 525 600(8760 60)

0.00000011416 1.00000011416525600

= 1.061836542996

Second 31 536 000(525 600 60)

0.0000000019026 1.000000001902631536000

= 1.061836542758


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Recap of Continuous compounding

If interest rate payments are made continuously over the term at interest

rate r per year, then the value of the sum invested increases to:

with r = Continuous interest rate per year in decimal form

T = Term of investment in years

V0 = Sum invested at time t = 0 (today)

VT = Future value after T years

e = Euler's number e (e = 2.7183)

TreVV

0T


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Relationship between simple and continuouscompounding

How high must the continuous interest rate r be for the same final amount of

capital to accumulate as with a simple interest payment at interest rate R?

The following applies for starting capital V0:

Investment with Investment with

simple interest payments continuous interest payments

V0(1 + R)T V0e

rT

As a result:

(1 + R)T= erT

r = ln(1 + R)


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Basics of volatility

Volatility is a huge issue in risk management.

The science of volatility measurement has advanced a lot inrecent years.

Here we look at some basic concepts and tools.


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8

Estimating Volatility

Calculate daily return u1= ln Si/ Si-1

Variance rate per day

We can simplify this formula by making the following

simplifications.

ui= (SiSi-1) / Si-1

= 0 m-1 = m

If we want to weight

2

1

1

)(1

1

uum

n

m

i

12

1

2 1 nm

i

um

n

122

niun )1(

i


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9

Estimating Volatility

Exponentially weighted moving average model means

weights decrease exponentially as we go back in time.

n2= 2n-1+ (1 - ) u

2n-1

= [n-12+ un-2

2] + 2n-22

= (1-) [un-12+ u2n-2+ 2un-32] + 32n-3

If we apply GARCH model,

n2= Y VL+ un-1

2 + 2n-1

VL= Long run average variance rate

Y + + = 1. If Y = 0, = 1-, = , it becomesexponentially weighted model.

GARCH incorporates the property of mean reversion.


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Risk Management and Financial Institutions, Chapter 15, Copyright John C. Hull 2006 15.10

Volatility Smiles

A volatility smile shows the variation of the implied volatilitywith the strike price.


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The Volatility Smile for Foreign Currency Options(Figure 15.1, page 347)

Implied

Volatility

Strike

Price


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The Volatility Smile for Equity Options (Figure 15.2, page 348)

Implied

Volatility

Strike

Price


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Volatility Term Structure

In addition to calculating a volatility smile, traders alsocalculate a volatility term structure.

This shows the variation of implied volatility with the time tomaturity of the option.


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Example of a Volatility Surface(Table 15.1, page 350)

Strike Price

0.90 0.95 1.00 1.05 1.10

1 mnth 14.2 13.0 12.0 13.1 14.5

3 mnth 14.0 13.0 12.0 13.1 14.2

6 mnth 14.1 13.3 12.5 13.4 14.3

1 year 14.7 14.0 13.5 14.0 14.8

2 year 15.0 14.4 14.0 14.5 15.1

5 year 14.8 14.6 14.4 14.7 15.0


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Problem

Consider a stock whose current price is 20,

expected return is 20% per annum. What is theexpected stock price, E (ST) in 1 year ?


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Solution

E (ST) = S0eT

= (20)e(.20)(1) = 24.43


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17

Problem

Consider a stock with an initial price of $100. Its price one

year from now is given by S = 100 erwhere the rate of return, ris normally distributed with a mean of 0.1 and a standarddevn of 0.2. What will the 95% confidence interval for S be?


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Solution

S 100er

ln S 100 r

100e.10+1.96x0.2 = 100e.492 = 163.56

100e.10-1.96x0.2 = 100e-.292 = 74.68

So S will lie between 74.68 and 163.56


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Problem

The current estimate of daily volatility is 1.5%. The closing

price of an asset yesterday was $30. The closing price ofthe asset today is $30.50. Using the EWMA model, with =0.94, calculate the updated estimate of volatility .


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Solution

ht = 2t-1+ ( 1 ) rt-12 = .94

rt-1 = ln[(30.50 )/ 30]

= .0165

ht = (.94) (.015)2+ (1-.94) (.0165)2

Volatility = .01509 = 1.509 %


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Problem

Suppose a variable follows a Wiener process. The initial

value is 25. What will be the probability distribution at the endof

A) 1 year ?

B) 5 years ?


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Solution

z = zt

At the end of one year, the value of the variablewill be normally distributed with mean of 25 andstd devn of 1.

At the end of 5 years, the value of the variable

will be normally distributed with mean of 25 andstd devn of (1) T = (1)5 = 2.236


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26

Problem

Suppose the cash position of a company follows a

generalised Wiener process with a drift of 20 per year and avariance rate of 900 per year. Initial cash position is 50. Whatwill be the mean and std devn at the end of

A) 1 year

B) 6 months


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Solution

dx = adt + bdz

At the end of 1 year, it will be a normal distributionwith mean = 50+20 = 70 and standard deviation of900 or 30.

At the end of 6 months, it will be a normal

distribution with mean = 50+10 = 60 and standarddeviation of 30.5 = 21.21


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Problem

Consider a stock that pays no dividends, has a volatility of

30% per annum and provides an expected return of 15%per annum with continuous compounding. What is the processfor the stock price for a time of 1 week ? Assume the initialstock price is 100.


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Solution

= .15 = .30

ds/s = .15dt + .30dz

s/s = .15 t+ .30 t

or s = 100 (.15t+ .30 t)

t = 1/52 = .0192

or s = 100 (.15 x .0192 + .30 (.0192)

= 100 (.00288 + .0416)

or s = .288 + 4.16


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Problem

Suppose an existing short option position is delta

neutral and has a gamma of

6000. Here, gammais negative because we have sold options. Assumethere exists a traded option with a delta of 0.6 andgamma of 1.25. Create a gamma neutral position.


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Solution

To gamma hedge, we must buy 6000/1.25 = 4800 options.

Then we must sell (4800) (.6) = 2880 shares to maintaina gamma neutral and original delta neutral position.


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Problem

A delta neutral position has a gamma of 3200.

There is an option trading with a delta of 0.5 andgamma of 1.5. How can we generate a gammaneutral position for existing portfolio whilemaintaining delta neutral hedge?


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35

Solution

Buy 3200/1.5 = 2133 options

Sell (2133) (.5) = 1067 shares

Volatility and Greeks

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