mfe lesson 7 slides (1)

Lesson 7: Modeling stock prices with thelognormal distribution.

ACTS 4302

Natalia A. Humphreys

September 22, 2011

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Acknowledgement

This work is based on the material in ASM MFE Study Manual forExam MFE/Exam 3F. Financial Economics (7th Edition), 2009, by

Abraham Weishaus.

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Lognormal distribution of the stock price.

Calculating the price of a derivative instrument requires a modelfor the underlying asset. Let’s assume it’s a stock.

We’ve discussed modeling stock prices using binomial trees.

We now look at an alternative model using the lognormaldistribution.

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Normal distribution of the growth rate of a stock’s price.

Let St be the price of the stock at time t. Then the growth ratefrom time 0 to time t is the A such that

St/S0 = eAt ⇔ A =1

tln (St/S0)

The growth rate A is a random variable and it’s plausible to assumethat it’s normally distributed: A ∼ N(µ, σ2) for some σ and µ.

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Normal distribution. Properties.

By definition, the normal distribution has the following properties:

1. The distribution has two parameters: µ and σ. µ is the meanand σ is the standard deviation. If a random variable X has anormal distribution with parameters µ and σ, we say thatX ∼ N(µ, σ2).

2. A standard normal distribution has µ = 0 and σ = 1. It’sprobability density function is

φ(x) =1√2π

e−x2

2

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Normal distribution. Properties (cont.)

3. The normal distribution is symmetric around its mean, e.g. forthe standard normal distribution, where µ = 0 the graph ofthe probability density function is symmetric around the 0.

4. A variable Y ∼ N(µ, σ2) can be expressed in terms of astandard normal random variable X ∼ N(0, 1) as Y = µ+σX .

5. The probability density function for Y is

φ(y , µ, σ) =1

σ√

2πe−

(y−µ)2

2σ2

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Normal distribution. Properties (cont.)

6. The normal distribution is a stable distribution. This meansthat if X ∼ N(µX , σ

2X ) and Y ∼ N(µY , σ

2Y ), then

Z = aX + bY ∼ N(µZ , σ2Z ) - a linear combination of normal

random variables is normal, even if the variables are notindependent.

7. If X1,X2, · · · ,Xn are identically distributed random variables,then

limn→∞

n∑i=1

Xi = Y ∼ N(µ, σ2)

the limit of the sum of independent identically distributed(i.d.d.) random variables, regardless of their distribution, isnormal. This is a version of the Central Limit Theorem.

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Lognormal distribution. Discussion.

In general, if X ∼ N(µ, σ2), then eX is a lognormal randomvariable with parameters µ and σ.

For stocks, if the growth rate A ∼ N(µA, σ2A), then St

S0= eAt has a

lognormal distribution. Note that this is the ratio of the stockprice at time t to the original stock price.

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Lognormal distribution. Discussion (cont.)

Let m and v be parameters of a lognormal distribution. Then for alognormal random variable Y , m = µt, v = σ

√t and the mean,

variance and k-th moment are:

E [Y ] = em+0.5v2

Var [Y ] = e2m+v2(ev

2 − 1)

E [Y k ] = ekm+0.5k2v2

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Jensen’s inequality.

Recall that a convex function g is defined as having the property

g(ax + (1− a)y) ≤ ag(x) + (1− a)g(y) for 0 ≤ a ≤ 1

Jensen’s inequality states that if X is a random variable and g is aconvex function, then

E [g(x)] ≥ g (E [X ])

If g(x) = ex , then the Jensen’s inequality could be read as

E [ex ] ≥ eE [X ]

The inequalities are reversed if function g is concave (g′′ ≤ 0).

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The lognormal distribution as a model for stock prices.

Let σ be the volatility of the stock price. If

StS0

= eXt and X ∼ N(µ, σ2),

Then

StS0∼ LND(m, v2), with parameters m = µt, v = σ

√t

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The lognormal distribution as a model for stock prices.Expected growth rate.

If α is the continuously compounded expected rate of return on astock; andδ is the continuously compounded annual dividend return. Thenα− δ is the continuously compounded expected growth rate of thestock price St .

In a lognormal model with parameters µ and σ, the continuouslycompounded expected growth rate is µ+ 0.5σ2.Therefore, α− δ = µ+ 0.5σ2 and µ = α− δ − 0.5σ2. Thus,

StS0∼ LND(m, v), where

m = µt = (α− δ − 0.5σ2)t,

v = σ√t

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Example 7.1.

St is the price of a non-dividend paying stock at time t. St followsa lognormal model (St ∼ LND). You are given:

1. S0 = 40.

2. The stock’s continuously compounded expected growth rate isα = 0.15.

3. The stock’s volatility σ = 0.3.

Answer the following questions.

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Example 7.1. Question 1.

Determine the average price of the stock in t years.

Solution. The average price of the stock in t years is

E [St ] = S0E

[StS0

]= S0e

m+0.5v2=

= S0e(α−δ−0.5σ2)t+0.5σ2t = S0e

(α−δ)t

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Determine the average price of the stock after one and four years.

Solution. Since by above for any t,

E [St ] = S0e(α−δ)t ,

we obtain:

E [S1] = 40e0.15·1 = 46.4734

E [S4] = 40e0.15·4 = 72.8848

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Example 7.1. Question 3.Determine the median price of the stock after t years.

Solution. Note that the median is the 50-th percentile of adistribution. By definition, Zα is the αth percentile of adistribution, if Pr (Z > Zα) = α. Hence, denoting S0.5 the medianof St ,

Pr (St > S0.5) = 0.5⇔ Pr

(StS0

>S0.5S0

)= 0.5⇔

Pr

(ln

(StS0

)> ln

(S0.5S0

))= 0.5⇔

Pr

(Xt > ln

(S0.5S0

))= 0.5

But Xt ∼ N(m, v) and for a normal distribution the median isequal to the mean. Hence,

ln

(S0.5S0

)= m = µt ⇔ S0.5 = S0e

µt

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Determine the median price of the stock after one and four years.


S0.5(t) = S0eµt , µ = α− δ − 0.5σ2 = 0.15− 0.5 · 0.32 = 0.105

we obtain:

S0.5(1) = 40e0.105·1 = 44.4284

S0.5(4) = 40e0.105·4 = 60.8785

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Determine the average t-th year growth rate of the stock.

Solution. Recall that the growth rate of a stock is a randomvariable A such that

StS0

= eAt , At ∼ N(m, v2)

Hence, the average t-th year growth rate of the stock is

E

[ln

(StS0

)]= m = µt

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Determine the average first and fourth year growth rate of thestock.


E

[ln

(StS0

)]= m = µt, µ = 0.105

we obtain:

E

[ln

(S1S0

)]= 0.105 · 1 = 0.105

E

[ln

(S4S0

)]= 0.105 · 4 = 0.42

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What lognormal model helps us determine about a stock.

Using lognormal model, we can answer questions on thedistribution of the price of the stock.

We can calculate the probability that the price of a stock is in acertain range.

We can also calculate ”confidence intervals” for the price of thestock

By definition, the p-confidence interval is an interval centered atthe mean of a random variable for which p is the probability thatthe random variable is in the interval.

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Example 7.2.

A stock’s prices follow a lognormal distribution. You are given:

1. α = 0.14

2. δ = 0.02

3. σ = 0.3

Answer the following questions.

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Example 7.2. Question 1.Determine the probability that the stock’s price at the end of onemonth will be greater than its current price.

Solution. For any t,

Pr (St > S0) = Pr

(StS0

> 1

)=

Pr

(ln

(StS0

)> ln(1) = 0

)= Pr (Xt > 0)

Recall that Xt ∼ N(m, v2). Hence,

Pr (Xt > 0) = 1− Pr (Xt < 0) = 1− N

(0−m

v

)= N

(mv

)Since m = µt and v = σ

√t

N(mv

)= N

(µ√t

σ

)22 / 41

Example 7.2. Question 1 (cont.)

Solution (cont.) Using

t =1

12= 0.0833, µ = α−δ−0.5σ2 = 0.14−0.02−0.5·0.32 = 0.075

,and the normal distribution table, we obtain:

N

(µ√t

σ

)= N

(0.075

√0.0833

0.3

)= N (0.0722) ≈

≈ N (0.07) = 0.5279

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Determine the probability that the stock’s price at the end of onemonth will be greater than the expected price.

Solution. For any t, we need to find Pr (St > E [St ]). Recall fromQuestion 1 of the Example 7.1,

E [St ] = S0e(α−δ)t

Hence, repeating the argument of Question 1 of the currentExample,

Pr (St > E [St ]) = Pr(St > S0e

(α−δ)t)

= Pr

(StS0

> e(α−δ)t)

=

= Pr

(ln

(StS0

)> ln(e(α−δ)t) = (α− δ)t

)=

= Pr (Xt > (α− δ)t)

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Solution (cont.) Recall that Xt ∼ N(m, v2). Hence,

Pr (Xt > (α− δ)t) = 1− Pr (Xt < (α− δ)t) =

= 1− N

((α− δ)t −m

v

)= 1− N

((α− δ)t − µt

v

)=

= 1− N

(0.5σ2t

σ√t

)= 1− N

(σ√t

2

)Using t = 1

12 = 0.0833 and the normal distribution table, weobtain:

1− N

(σ√t

2

)= 1− N

(0.3√

0.0833

2

)= 1− N (0.0433) ≈

≈ 1− N (0.04) = 1− 0.516 = 0.484

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Example 7.2. Question 3.Construct a 95% symmetric confidence interval for the ratio of thestock’s price at the end of one month to the current price.

Solution. If Z ∈ N(m, v2), then the 100(1− α)% confidenceinterval is defined to be(

m − z1−α/2v ,m + z1−α/2v), where

z1−α/2 is the number such that

P[Z < z1−α/2

]= 1− α

2

Since StS0

= eXt , the confidence interval for St will be(S0e

m−z1−α/2v ,S0em+z1−α/2v

), where

z1−α/2 is the number such that

P[Z < z1−α/2

]= 1− α

2, Z ∼ N(0.1)

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Solution (cont.) If α = 1− 0.95 = 0.05, then

1− α

2= 0.975, z0.975 = 1.96

Hence,

m − z1−α/2v = 0.075 · 0.0833− 1.96 · 0.3√

0.0833 = −0.1635

m + z1−α/2v = 0.075 · 0.0833 + 1.96 · 0.3√

0.0833 = 0.176

Thus, the confidence interval for StS0

will be

em−z1−α/2v = e−0.1635 = 0.8492

em+z1−α/2v = e0.176 = 1.1924

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Pricing European options using the lognormal parameters.

In this section we’ll make some progress towards pricing Europeanoptions. Let us calculate the probability that an option will pay off.Suppose

1. S0 is the price of a stock

2. α is the continuously compounded expected rate of return ona stock

3. σ is the volatility of the stock price

4. δ is the continuously compounded annual dividend return

5. K is the strike price

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Probability that an option will pay off. Put.

A put will pay off if St < K :

Pr (St < K ) = Pr

(StS0

<K

S0

)=

= Pr

(ln

(StS0

)< ln

(K

S0

))=

= Pr

(Xt < ln

(K

S0

))= N

ln(

KS0

)−m

v

= N

ln(

KS0

)− (α− δ − 0.5σ2)t

σ√t

=

= N

− ln(S0K

)+ (α− δ − 0.5σ2)t

σ√t

= N(−d̂2

)

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Probability that a put option will pay off. Final formula.

Pr (St < K ) = N(−d̂2

), where

d̂2 =ln(S0K

)+ (α− δ − 0.5σ2)t

σ√t

N(x) is the standard normal cumulative distribution function at x -probability that a standard normal random variable X is less thanor equal to x :

N(x) = Pr (X < x) , X ∼ N(0, 1)

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Probability that an option will pay off. Call.

Similarly, for a call option, a call will pay off if St > K :

Pr (St > K ) = 1− Pr (St < K ) = 1− N(−d̂2

)= N

(d̂2)

d̂2 =ln(S0K

)+ (α− δ − 0.5σ2)t

σ√t

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Example 7.3

A stock’s price follows a lognormal model. You are given:

1. S0 = 60

2. α = 0.15

3. σ = 0.2

4. δ = 0.05

A European call option on the stock with strike price 70 expires in3 months.Calculate the probability that the option pays off.

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Example 7.3. Solution.

Solution.

Pr (St > K ) = N(d̂2)

d̂2 =ln(S0K

)+ (α− δ − 0.5σ2)t

σ√t

=

=ln(6070

)+ (0.15− 0.05− 0.5 · 0.22)0.25

0.2√

0.25= −1.3415

Pr (S0.25 > 70) = N (−1.3415) = 1− N (1.3415) = 0.0901

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Conditional payoff of the option, given that it pays off.

Defn. The partial expectation of a continuous random variable Xhaving probability density function f (x), given that it is in theinterval [a, b] is the contribution to the expectation from values inthe interval [a, b]:

E [X |X ∈ [a, b]] =

∫ b

axf (x) dx

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Conditional expectation. Lognormal r.v.

For a lognormal random variable Y ∼ LND(m, v2),

PE (X |X < K ) = E (X )N

(lnK −m − v2

v

)Applying this for stocks,

PE[St |St < K ] = E

(StS0

)N

(ln K

S0−m − v2

v

)=

= S0em+0.5v2

N

(ln K

S0−m − v2

v

)= S0e

(α−δ)tN(−d̂1)

d̂1 =ln(S0K

)+ (α− δ + 0.5σ2)t

σ√t

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Expectation. Lognormal r.v.

SincePr(St < K ) = N(−d̂2),

it follows that

E[St |St < K ] =S0e

(α−δ)tN(−d̂1)

N(−d̂2)

Note thatd̂2 = d̂1 − σ

√t

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Expected payoff of a put option

One who owns a European put option has the following cash flowsat expiry:

1. Receipt of K if the stock price is below K :

E [K |St < K ] = K · Pr(St < K ) = K · N(−d̂2)

2. Payment of stock, if the stock price is below K:

E[−St |St < K ] = −S0e(α−δ)tN(−d̂1)

N(−d̂2)

Pr(St < K ) = N(−d̂2)⇒E [−St ] = E[−St |St < K ] · Pr(St < K ) = −S0e(α−δ)tN(−d̂1)

Therefore, the expected put option payoff:

E[max(0,K − St)] = KN(−d̂2)− S0e(α−δ)tN(−d̂1)

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Expected payoff of a call option

One who owns a European call option has the following cash flowsat expiry:

1. Payment of K if the stock price is above K :

E [−K |St > K ] = −K · Pr(St > K ) = −K · N(d̂2)

2. Receipt of stock, if the stock price is above K:

E[St |St > K ] =S0e

(α−δ)tN(d̂1)

N(d̂2)

Pr(St > K ) = N(d̂2)⇒E [St ] = E[St |St > K ] · Pr(St > K ) = S0e

(α−δ)tN(d̂1)

Therefore, the expected call option payoff:

E[max(0, St − K )] = S0e(α−δ)tN(d̂1)− KN(d̂2)

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Example 7.4

A stock price follows a lognormal model. You are given:

1. S0 = 50

2. α = 0.15

3. σ = 0.3

4. δ = 0

Determine the conditional expected value of the stock’s price after3 months, given that it is higher than 75.

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Example 7.4. Solution.Solution. By the above,

E[St |St > K ] =S0e

(α−δ)tN(d̂1)

N(d̂2)= ∗

Let’s calculate d̂1 and d̂2.

d̂1 =ln(S0K

)+ (α− δ + 0.5σ2)t

σ√t

=

=ln(5075

)+ (0.15 + 0.5 · 0.32)0.25

0.3√

0.25= −2.3781

d̂2 = d̂1 − σ√t = −2.3781− 0.3

√0.25 = −2.5281

Hence,

∗ = 50e0.15·0.25N(−2.3781)

N(−2.528)≈ 50e0.0375

N(−2.38)

N(−2.53)=

= 51.9106 · 0.0087

0.0057= 79.23

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mfe lesson 7 slides (1)

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