Transcript Common Probability Distributions in Finance - IAG PUC-Rio

Common Probability Distributions
in Finance
The Normal Distribution
• The normal distribution is a continuous, bell-shaped distribution
that is completely characterized by two parameters: its mean
and standard deviation
• If a random variable X follows a normal distribution with mean 
and variance 2, we write

X ~ N  , 2

Properties of Normal Distribution
• Since the normal distribution can be completely characterized
by its mean and variance, any probability question about a
normal random variable can be answered if these two
parameters are known
• The normal distribution is symmetric
– Skewness is zero
– Mean, median and mode are the same
Properties of Normal Distribution
• Due to the symmetric nature of the normal distribution, we can
derive the following statements
– Approximately 68% of the values of a normal variable fall within the
interval   
– Approximately 95% of the values of a normal variable fall within the
interval   2
– Approximately 99% of the values of a normal variable fall within the
interval   3
Properties of Normal Distribution
• To be more precise, the following intervals with their
corresponding cutoffs are frequently used in association with a
sample from a normal distribution
– 90% of the values of a normal variable lie within  1.65 sample
standard deviations from the sample mean
– 95% of the values of a normal variable lie within  1.96 sample
standard deviations from the sample mean
– 99% of the values of a normal variable lie within  2.58 sample
standard deviations from the sample mean
Properties of Normal Distribution
• Example: Suppose that the variable “approved mortgage
amount” follows a normal distribution
• Taking a sample of 200 loan approvals from a bank, it is found
that the sample mean is $150,000 and the sample standard
deviation is $55,000
• In this case, 95% of approved mortgages will be within$42,200
and $257,800
Normal Distribution and Portfolio Returns
• One potentially interesting application of the normal distribution
is in describing data on asset returns
• The normal distribution is a good fit for quarterly or annual
holding period returns on a diversified equity portfolio
• However, it does not fit equally well monthly, weekly or daily
period returns
• In general, the normal distribution tends to underestimate the
probability of extreme returns (the fat tails problem)
Normal Distribution and Portfolio Returns
• Relative to the normal distribution, the actual distribution of the
data may contain more observations in the center and in the
tails
• This implies that the actual distribution compared to the normal
distribution has
– More observations clustered near the mean
– A higher probability of observing extreme values on both tails of the
distribution (fat tails)
The Cumulative Distribution Function of a
Normal Distribution
• If a random variable X follows a normal distribution with mean 
and variance 2 , the cumulative distribution function is
Fx  x0   P X  x0 
• This probability is given by the area under the normal probability
function to the left of x0
The Cumulative Distribution Function of a
Normal Distribution
• Similarly, if a and b are two possible values of the normal
random variable X, with a < b, then the probability that X will
take values in between those two cutoffs is given by
Pa  X  b  Fx b  Fx a 
The Standard Normal Distribution
• The standard normal distribution is a normal distribution with
mean 0 and variance 1
• We denote a standard normal variable with Z and write
Z ~ N 0,1
• The cumulative distribution function of the standard normal
distribution is well documented and can be used to find
probabilities of normal random variables
Finding Areas Under the Normal Distribution
• We say that a normal random variable X is standardized if we
subtract from it its mean and divide by its standard deviation
• Thus, the new variable Z follows the standard normal
distribution
X     
Z
~ N 0,1

Finding Areas Under the Normal Distribution
• Using the above transformation of a normal into a standard
normal variable, we rewrite the result of the probability that a
normal variable takes values between two cutoffs as follows
a X  b
b   
a
Pa  X  b   P



F

F

 Z

Z











Finding Areas Under the Normal Distribution
• Example: Suppose that portfolio returns follow a normal
distribution, which we have estimated to have a mean return of
12% and standard deviation of return of 22% per year
• What is the probability that portfolio return will exceed 20%?
What is the probability that portfolio returns will be between 12%
and 20%?
• If X is portfolio return, the variable (X - .12)/.22 follows the
standard normal distribution
Finding Areas Under the Normal Distribution
• For X = .2, Z = (.2 - .12)/.22 = .363.
• We need to find P(Z > .363). But, P(Z > .363) = 1 – P(Z  .363)
= FZ(.363)
• From the table of the cumulative standard normal distribution,
we find that FZ(.363) is equal to .64 and, thus, the probability of
a return above 20% is 1 - .64 = .36.
Finding Areas Under the Normal Distribution
• For the second part, note that 12% is the mean of the
distribution, meaning that P(X < 12%) = .5 and the same will be
true for the corresponding value of the standard normal variable
• Thus, P(.12  X  .20) is the same as P (0  Z  .36), which is
equal to FZ(.36) - FZ(0) = .64 - .50 = .14
Finding Areas Under the Normal Distribution
• To expand upon the last question, what if we were interested in
the probability that portfolio returns are between 8% and 20%?
• Following the above steps and transforming the normal variable
into a standard normal, P(.08  X  .20) is equal to
 .08  .12 X   .20  .12 
P


  P .18  Z  .36

.22 
 .22
• To find the cumulative standard normal distribution for -.18,
which is FZ(Z  -.18), we subtract from 1 the cumulative normal
distribution for its symmetric value, i.e., 1 - FZ(Z  .18)
Finding Areas Under the Normal Distribution
• From the table of the standard normal distribution, FZ(Z  .18) =
.57
• Thus, FZ(Z  -.18) = 1 - FZ(Z  .18) = .43
• Finally, P(-.18  Z  .36) = FZ(Z  .36) - FZ(Z  -.18) = .64 - .43 =
.21