Chapter 18
The Lognormal Distribution
1
The normal distribution
 Normal distribution (or density):
1
( x; , ) 
e
 2
1  x  2
 

2  
2
The normal distribution
 Normal density is symmetric: ( x; , )  ( x; , )
 If a random variable x is normally distributed
with mean  and standard deviation , x ~ N ( ,  2 )
 z is a random variable distributed standard
normal: z ~ N (0,1)
 The value of the cumulative normal [P(z<a)]
distribution function N(a) or NormSDist(a) in
Excel equals to the probability P of a number
z drawn from the normal distribution to be
1
 x2
a
1
less than a. N (a ) 
  2 e 2 dx
3
The normal distribution
The Normal Distribution allows us to make
statements and inferences regarding future
stock price levels.
We are able to estimate probabilities of the
terminal stock price being between two
values, above a value, or below a value.
Note: the probability of reaching an EXACT value is zero.
4
The normal distribution (cont.)
5
The normal distribution (cont.)
 The probability of a number drawn from the standard
normal distribution of being between a and –a is:
Prob (z < –a) = N(–a)
Prob (z < a) = N(a)
therefore
Prob (–a < z < a) =
N(a) – N(–a) = N(a) – [1 – N(a)] = 2·N(a) – 1
 Example 18.1: Prob (–0.3 < z < 0.3) = 2·0.6179 – 1 =
0.2358
6
The normal distribution (cont.)
 Converting a normal random variable to
standard normal:

If
x ~ N ( ,  2 )
, then
z ~ N (0,1)
 And vice versa:
 If z ~ N (0,1)
, then x ~ N ( ,  2 )
x

if
z
if
x    z
 Example 18.2: Suppose x ~ N (3,5)
and z ~ N (0,1) , then
x3
~ N (0,1)
5
and
3  5  z ~ N (3,25)
7
The normal distribution (cont.)
 Example:
The number 7 is drawn from a Normal
distribution of mean 4 and variance 9.
What is the equivalent draw from a standard
normal distribution?
What is the probability of drawing a number
larger than 7 ? Lower than 7 ? Exactly the
number 7 ?
8
Cumulative Normal Table
d
-3
-2.95
-2.9
-2.85
-2.8
-2.75
-2.7
-2.65
-2.6
-2.55
-2.5
-2.45
-2.4
-2.35
-2.3
-2.25
-2.2
-2.15
-2.1
-2.05
-2
-1.95
-1.9
-1.85
-1.8
-1.75
-1.7
-1.65
-1.6
-1.55
-1.5
N(d)
0.00135
0.00159
0.00187
0.00219
0.00256
0.00298
0.00347
0.00402
0.00466
0.00539
0.00621
0.00714
0.00820
0.00939
0.01072
0.01222
0.01390
0.01578
0.01786
0.02018
0.02275
0.02559
0.02872
0.03216
0.03593
0.04006
0.04457
0.04947
0.05480
0.06057
0.06681
d
-1.45
-1.4
-1.35
-1.3
-1.25
-1.2
-1.15
-1.1
-1.05
-1
-0.95
-0.9
-0.85
-0.8
-0.75
-0.7
-0.65
-0.6
-0.55
-0.5
-0.45
-0.4
-0.35
-0.3
-0.25
-0.2
-0.15
-0.1
-0.05
0
N(d)
0.07353
0.08076
0.08851
0.09680
0.10565
0.11507
0.12507
0.13567
0.14686
0.15866
0.17106
0.18406
0.19766
0.21186
0.22663
0.24196
0.25785
0.27425
0.29116
0.30854
0.32636
0.34458
0.36317
0.38209
0.40129
0.42074
0.44038
0.46017
0.48006
0.50000
d
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
0.55
0.6
0.65
0.7
0.75
0.8
0.85
0.9
0.95
1
1.05
1.1
1.15
1.2
1.25
1.3
1.35
1.4
1.45
1.5
N(d)
0.51994
0.53983
0.55962
0.57926
0.59871
0.61791
0.63683
0.65542
0.67364
0.69146
0.70884
0.72575
0.74215
0.75804
0.77337
0.78814
0.80234
0.81594
0.82894
0.84134
0.85314
0.86433
0.87493
0.88493
0.89435
0.90320
0.91149
0.91924
0.92647
0.93319
d
1.55
1.6
1.65
1.7
1.75
1.8
1.85
1.9
1.95
2
2.05
2.1
2.15
2.2
2.25
2.3
2.35
2.4
2.45
2.5
2.55
2.6
2.65
2.7
2.75
2.8
2.85
2.9
2.95
3
N(d)
0.93943
0.94520
0.95053
0.95543
0.95994
0.96407
0.96784
0.97128
0.97441
0.97725
0.97982
0.98214
0.98422
0.98610
0.98778
0.98928
0.99061
0.99180
0.99286
0.99379
0.99461
0.99534
0.99598
0.99653
0.99702
0.99744
0.99781
0.99813
0.99841
0.99865
9
The normal distribution (cont.)
 Answer:
Equivalent standard draw is: z* = (7- 4)/9 = 1.
The probability of drawing a number larger than 7 is
Prob(x>7) or Prob(z>z*). This is equal to 1-N(z*) or
1-N(1) = 1 - 0.841345 = 15.87%.
The probability of drawing a number lower than 7 is
Prob(x<7) or Prob(z<z*). This is equal to N(z*) or
N(1) = 0.841345 = 84.13%.
Obtaining exactly the number 7 has a zero probability.
10
The normal distribution (cont.)
 The sum of normal random variables is also normal:
n n
 n

  i xi ~ N    i  i ,    i  j ij 
 i 1

i 1
i 1 j 1
n
where xi, i = 1,…,n, are n random variables, with
mean E(xi) = i, variance Var(xi) =i2, covariance
Cov(xi,xj) = ij = rijij
 Ex: ax1+bx2 ~ N(a1+b2, a221+b222+2abr12)
Question: how is this variance obtained ?
11
The normal distribution (cont.)
 Example:
The variance of the sum of the two variables is:
Var(ax1+bx2) = Cov(ax1+bx2 , ax1+bx2)
= a2Cov(x1,x1) + 2abCov(x1,x2) + b2Cov(x2,x2)
= a221 + 2ab12 + b222
= a221 + b222 + 2abr12
12
The normal distribution (cont.)
 Example: take the case of two identically and
independently distributed returns x1 and x2
(assume we are talking about continuously
compounded returns, so you can add them).
Using the last two slides, what would be the
resulting means and variances?
13
The normal distribution (cont.)
 If x1 and x2 are iid and are summed, we have:
 1=2=
 1=2=
 r =0
(since they are independent)
 a=1 and b=1
 Hence we get:
x1+x2 ~ N(2, 22)
14
The lognormal distribution
 A random variable x is lognormally distributed if ln(x)
is normally distributed


If x is normal, and ln(y) = x (or y = ex), then y is
lognormal
If continuously compounded stock returns are normal
then the stock price is lognormally distributed
 Product of lognormal variables is lognormal
 If x1 and x2 are normal, then y1=ex and y2=ex are
lognormal.
 The product of y1 and y2: y1 x y2 = ex x ex = ex +x
 Since x1+x2 is normal, ex +x is lognormal
1
2
1
1
2
1
2
2
15
The lognormal distribution (cont.)
 The lognormal density function:
1
g ( S ; m, v, S0 ) 
e
Sv 2
1  ln( S / S0 )  m 
 

2
v

2
 where S0 is initial stock price, and
ln(S/S0)~N(m,v2), S is the future stock price,
m is the mean of the continuously
compounded return and v is standard
deviation of the continuously compounded
return
1
m v
x
 If x ~ N(m,v2), then E (e )  e 2
2
16
The lognormal distribution (cont.)
17
A lognormal model of stock prices
 If the stock price St is lognormal, St / S0 = ex, where x,
the continuously compounded return from 0 to t is
normal
 If R(t, s) is the continuously compounded return from
t to s, and, t0 < t1 < t2, then
R(t0, t2) = R(t0, t1) + R(t1, t2)
 From 0 to T, E[R(0,T)] = nah , and Var[R(0,T)] = nh2
(we showed it in the two-period cases earlier)
 If returns are iid, the mean and variance of the
continuously compounded returns are proportional to
time
18
A lognormal model of stock prices (cont.)
 If we assume that
ln(St / S0) ~ N [(a – d – 0.52)t, 2t]
then ln(St / S0) = (a – d – 0.52)t + tz
and therefore St = S0
2)t + tz
(a
–
d
–
0.5
e
 Exercise: compute the value of E(St) by using
the fact that if x ~ N(m,v2), then
E(ex) = em+(1/2)v2 . (Note that S0 is constant)
19
A lognormal model of stock prices (cont.)
 Let x = ln(St / S0) ~ N [(a – d – 0.52)t, 2t]
Thus m = (a – d – 0.52)t
and v2 = 2t
Then E(ex) = E(St / S0) = em+(1/2)v2
=
2)t +(1/2)2t
(a
–
d
–
0.5
e
=
e(a – d)t
and therefore E(St ) = S0e(a – d)t
20
A lognormal model of stock prices (cont.)
 If the stock price is lognormally distributed,
we can use the fact that the distribution is
known to compute a number of probabilities
and expectations:
 Probability that the option will expire in-the-
money.
 Expected stock price, given that the option
expires in the money.
21
A lognormal model of stock prices (cont.)
 If current stock price is S0, the probability that
the option will expire in the money, i.e.,
Prob( St  K )  N (d2 )
where the expression for d2 contains a, the
true expected return on the stock instead of
the risk-free rate r.
(if you used r, you would obtain the riskneutral probability of expiring in the money)
22
A lognormal model of stock prices (cont.)
Using the lognormal distribution, confidence
intervals can be derived.
 One might be interested in computing the
prices StL and StU such that:
Prob (St<StL) = p/2 and Prob (StU < St ) = p/2
 The probability of being in the tails of the
distribution is split in two, half for each tail.
 We will then be (1-p)% confident that the final
stock price will be between SL and SU.
23
Lognormal probability calculations
 Prices StL and StU such that Prob (St<StL) =
p/2 and Prob (StU < St ) = p/2 are:
StL  S0e
1
(a d   2 ) t  t N 1 ( p / 2)
2
StU  S0e
1
(a d   2 ) t  t N 1 ( p / 2)
2
 Example: use the cumulative normal
distribution table seen earlier to derive the
95% confidence interval if S0=100, t=2,
a=0.10, d=0, =0.30
24
Lognormal probability calculations
 Answer:
S  S0 e
L
t
1 2
(0.10 0  0.3 )(2) (0.3) 2 (1.96)
2
S  S0 e
U
t
1
(0.10  0  0.32 )(2)  (0.3) 2 (1.96)
2
 48.6
 256.4
 There is a 95% probability that in two years the stock
will be between $48.60 and $256.40
25
Lognormal probability calculations (cont.)
 Given the option expires in the money, what
is the expected stock price? The conditional
expected price
E ( St | St  K )  Se

(a  d)t
N (d1 )
N (d2 )
where the expressions for d1 and d2 contain a,
the true expected return on the stock in place
of r, the risk-free rate
26
How are asset prices distributed?
27
Is volatility constant?
28