After completing this reading, you should be able to:
 Distinguish the key properties among the following distributions: uniform distribution, Bernoulli distribution, Binomial distribution, Poisson distribution, normal distribution, lognormal distribution, Chisquared distribution, Student’s t, and Fdistributions, and identify common occurrences of each distribution.
 Describe the central limit theorem and the implications it has when combining independent and identically distributed \(\left( i.i.d. \right) \) random variables.
 Describe \(i.i.d.\) random variables and the implications of the \(i.i.d.\) assumption when combining random variables.
 Describe a mixture distribution and explain the creation and characteristics of mixture distributions.
Parametric Distributions
There are two types of distributions namely parametric and nonparametric distributions. Parametric distributions are mathematically described by functions. On the other hand, one cannot use a mathematical function to describe a nonparametric distribution. Examples of parametric distributions are the uniform and the normal distributions.
Uniform Distribution
One of the most fundamental distributions in statistics is the uniform distribution.
Recall that, for a continuous distribution:
$$ P\left[ { b }_{ 1 }\le X\le { b }_{ 2 } \right] =\int _{ { b }_{ 1 } }^{ { b }_{ 2 } }{ f\left( x \right)dx } $$
\(f\left( x \right) \) is the PDF of \(X\). The following is the formula for a PDF of a given distribution:
$$ u\left( { b }_{ 1 },{ b }_{ 2 } \right) =\begin{cases} c\quad \forall { b }_{ 1 }\le x\le { b }_{ 2 } \\ 0\quad \forall { b }_{ 1 }>x>{ b }_{ 2 } \end{cases}:{ b }_{ 1 }>{ b }_{ 2 } $$
The likelihood of any outcome happening is always 1. The following is how to determine the value of \(c\):
$$ \int _{ \infty }^{ +\infty }{ u\left( { b }_{ 1 },{ b }_{ 2 } \right) } dx=1 $$
$$ \int _{ \infty }^{ +\infty }{ u\left( { b }_{ 1 },{ b }_{ 2 } \right) } dx=\int _{ \infty }^{ { b }_{ 1 } }{ 0dx } +\int _{ { b }_{ 2 } }^{ { b }_{ 1 } }{ cdx } +\int _{ { b }_{ 2 } }^{ +\infty }{ 0dx } =\int _{ { b }_{ 1 } }^{ { b }_{ 2 } }{ cdx } $$
$$ \int _{ { b }_{ 1 } }^{ { b }_{ 2 } }{ cdx } ={ \left[ cx \right] }_{ { b }_{ 1 } }^{ { b }_{ 2 } }=c\left( { b }_{ 2 }{ b }_{ 1 } \right) =1 $$
$$ c=\frac { 1 }{ { b }_{ 2 }{ b }_{ 1 } } $$
The mean is computed as follows:
$$ \mu =\int _{ { b }_{ 1 } }^{ { b }_{ 2 } }{ cxdx } =\frac { 1 }{ 2 } \left( { b }_{ 2 }+{ b }_{ 1 } \right) $$
On the other hand, we have the variance as:
$$ \sigma =\int _{ { b }_{ 1 } }^{ { b }_{ 2 } }{ c{ \left( x\mu \right) }^{ 2 }dx } =\frac { 1 }{ 12 } { \left( { b }_{ 2 }{ b }_{ 1 } \right) }^{ 2 } $$
The standard uniform distribution is a special case scenario where \(\left( { b }_{ 2 }={ b }_{ 1 }=0 \right) \).
The CDF of the uniform distribution is computed by integrating the PDF. We have:
$$ P\left[ X\le a \right] =\int _{ { b }_{ 1 } }^{ a }{ cdz } ={ \left[ cz \right] }_{ { b }_{ 1 } }^{ a }=\frac { a{ b }_{ 1 } }{ { b }_{ 2 }{ b }_{ 1 } } $$
If \(a={ b }_{ 1 }\) we are at the minimum, and if \(a={ b }_{ 2 }\) we are at the maximum and the \(CDF = 1\).
Bernoulli Distribution
A Bernoulli random variable is either equal to zero or one.
Assume \(p\) is the probability that \(X=1\), then:
$$ P\left[ X=1 \right] =p\quad and\quad P\left[ X=0 \right] =1p $$
Therefore, the mean and variance of the distribution are computed as:
$$ \mu =p\times 1+\left( 1p \right) \times 0=p $$
$$ { \sigma }^{ 2 }=p\times { \left( 1p \right) }^{ 2 }+{ \left( 1p \right) }\times { \left( 0p \right) }^{ 2 }=p{ \left( 1p \right) } $$
Binomial Distribution
A binomial distribution is a collection of Bernoulli random variables. Suppose we are given two bonds that are totally independent, with default likelihood of ten percent. Then we have the following possibilities:
 both of the bonds do not default
 both of them defaults, or
 only one of them defaults.
Let \(K\) represent the number of defaults:
$$ P\left[ K=0 \right] ={ \left( 110\% \right) }^{ 2 }=81\% $$
$$ P\left[ K=1 \right] =2\times 10\%\times \left( 110\% \right) =18\% $$
$$ P\left[ K=2 \right] ={ 10\% }^{ 2 }=1\% $$
If we possess three independent bonds having a 10% default probability then:
$$ P\left[ K=0 \right] ={ \left( 110\% \right) }^{ 3 }=72.9\% $$
$$ P\left[ K=1 \right] ={ 3\times 10\%\times \left( 110\% \right) }^{ 2 }=24.3\% $$
$$ P\left[ K=2 \right] ={ 3\times { 10\% }^{ 2 }\times \left( 110\% \right) }=2.7\% $$
$$ P\left[ K=3 \right] ={ { 10\% }^{ 3 } }=0.1\% $$
Suppose now that we have \(n\) bonds. The following combination represents the number of ways in which \(k\) of the \(n\) bonds can default:
$$ \left( \begin{matrix} n \\ k \end{matrix} \right) =\frac { n! }{ k!\left( nk \right) ! } \quad \quad \dots \dots \dots \dots equation\quad I $$
If \(p\) is the likelihood that one bond will default, then the chances that any particular \(k\) bonds will default is given by:
$$ { p }^{ k }{ \left( 1p \right) }^{ nk }\quad \quad \dots \dots \dots \dots \dots equation\quad II $$
Combining equation \(I\) and \(II\), we can determine the likelihood of \(k\) bonds defaulting as follows:
$$ P\left[ K=k \right] =\left( \begin{matrix} n \\ k \end{matrix} \right) { p }^{ k }{ \left( 1p \right) }^{ nk } $$
This is the PDF for the binomial distribution.
Poisson Distribution
Suppose that \(X\) is a Poisson random variable, then we have:
$$ P\left[ X=n \right] =\frac { { \lambda }^{ n } }{ n! } { e }^{ \lambda } $$
$$ \forall \lambda =constant,then,\quad \mu ={ \sigma }^{ 2 }=\lambda $$
Events are said to follow a Poisson process if they happen at a constant rate over time and the likelihood that one event will take place is independent of all the other events. Thus:
$$ P\left[ X=n \right] =\frac { { \left( \lambda t \right) }^{ n } }{ n! } { e }^{ \lambda t } $$
\(t\) is the amount of time that has elapsed.
Normal Distribution
Also called the Gaussian distribution, the normal distribution has a symmetrical PDF and the mean and median coincide with the highest point of the PDF. Furthermore, the normal distribution always has a skewness of 0 and a kurtosis of 3.
The following is the formula of a PDF that is normally distributed, for a given random variable \(X\):
$$ f\left( x \right) =\frac { 1 }{ \sigma \sqrt { 2\pi } } { e }^{ \cfrac { 1 }{ 2 } { \left( \cfrac { x\mu }{ \sigma } \right) }^{ 2 } } $$
When a variable is normally distributed, it is often written as follows, for convenience:
$$ X\sim N\left( \mu ,{ \sigma }^{ 2 } \right) $$
We read this as \(X\) is normally distributed, and has a mean, \(\mu\),and a variance of \({ \sigma }^{ 2 } \). Any linear combination of independent normal variables is also normal. To illustrate this, assume \(X\) and \(Y\) are two variables that are normally distributed. We also have constants \(a\) and \(b\). Then \(Z\) will be normally distributed such that:
$$ Z=aX+bY,\quad such\quad that\quad Z\sim N\left( a{ \mu }_{ X }+b{ \mu }_{ Y },{ a }^{ 2 }{ \sigma }_{ X }^{ 2 }+{ b }^{ 2 }{ \sigma }_{ Y }^{ 2 } \right) $$
A standard normal distribution is a normal distribution whose mean is 0 and standard deviation is 1.
$$ \emptyset =\frac { 1 }{ \sqrt { 2\pi } } { e }^{ \frac { 1 }{ 2 } { x }^{ 2 } } $$
To determine a normal variable whose standard deviation is \(\sigma\) and mean is \(\mu\), we compute the product of the standard normal variable with \(\sigma\) and then add the mean:
$$ X=\mu +\sigma \emptyset \Rightarrow X\sim N\left( \mu ,{ \sigma }^{ 2 } \right) $$
Three standard normal variables \({ X }_{ 1 }\), \({ X }_{ 2 }\), and \({ X }_{ 3 }\) are combined in the following way to construct two normal variables that are correlated:
$$ { X }_{ A }=\sqrt { \rho } { X }_{ 1 }+\sqrt { 1\rho } { X }_{ 2 } $$
$$ { X }_{ B }=\sqrt { \rho } { X }_{ 1 }+\sqrt { 1\rho } { X }_{ 3 } $$
Where \({ X }_{ A }\) and \({ X }_{ B }\) have a correlation of \(\rho\), and are standard normal variables.
The zvalue measures how many standard deviations the corresponding x value is above or below the mean.
xvalue 
zvalue 
μ 
0 
μ+1σ 
1 
μ+2σ 
2 
μ+nσ 
n 
Lognormal Distribution
The following density function represents the lognormal distribution:
$$ f\left( x \right) =\frac { 1 }{ x\sigma \sqrt { 2\pi } } { e }^{ \frac { 1 }{ 2 } { \left( \frac { lnx\mu }{ \sigma } \right) }^{ 2 } } $$
A variable is said to have a lognormal distribution if its natural logarithm has a normal distribution. The lognormal distribution is undefined, or zero, for negative values, as opposed to the normal distribution that has a range of values between negative infinity and positive infinity.
If the above equation of the density function of the lognormal distribution is rearranged, we obtain an equation that has a similar form to the normal distribution. That is:
$$ f\left( x \right) ={ e }^{ \frac { 1 }{ 2 } { \sigma }^{ 2 }\mu }\frac { 1 }{ \sigma \sqrt { 2\pi } } { e }^{ \frac { 1 }{ 2 } { \left( \frac { lnx\left( \mu { \sigma }^{ 2 } \right) }{ \sigma } \right) }^{ 2 } } $$
From the above, we notice that the lognormal distribution happens to be asymmetrical, and not symmetrical around \(\mu\), like it’s the case in the normal distribution, and peaks at \(exp\left( \mu { \sigma }^{ 2 } \right) \).
The following is the formula for the mean:
$$ E\left[ X \right] ={ e }^{ \mu +\frac { 1 }{ 2 } { \sigma }^{ 2 } } $$
This yields to an expression that closely resembles the Taylor expansion of the natural logarithm around 1. Recall that:
$$ r\approx R\frac { 1 }{ 2 } { R }^{ 2 } $$
where \(R\) is a standard return and \(r\) is the corresponding log return.
Following is the formula for the variance of the lognormal distribution:
$$ E\left[ \left( XE{ \left[ X \right] }^{ 2 } \right) \right] =\left( { e }^{ { \sigma }^{ 2 } }1 \right) { e }^{ 2\mu +{ \sigma }^{ 2 } } $$
Central Limit Theorem
Supposing that we are provided with an index that is composed of very may equities, or a bond comprising of very many bonds that are similar in nature. The constituent elements are often composed of random variables that are statistically identical but totally uncorrelated with each other. These variables are called independent and identically distributed \(\left( i.i.d \right) \) variables.
Assuming that we are given \(n\) \(i.i.d\) random variables, ranging from \({ X }_{ 1 }\) to \({ X }_{ n }\), and that the mean and standard deviation of each random variable is \(\mu\) and \(\sigma\) respectively. If \({ S }_{ n }\) is defined as the sum of those random variables, then we have that:
$$ \lim _{ n\rightarrow \infty }{ { S }_{ n }\sim N\left( n\mu ,n{ \sigma }^{ 2 } \right) } $$
This implies that the sum of \(n\) variables converges to a normal distribution as \(n\) approaches infinity. This is the central limit theorem. In finance, the central limit theorem is applicable when a normal distribution is used to justify the approximation of variables that have a financial aspect.
Monte Carlo Simulation
Most problems in risk management cannot be solved mathematically. To evaluate their solutions, we create a Monte Carlo simulation that is made up of several trials. Random inputs are then fed to a system of equations, for each trial. The statistical features of the output variables are then estimated by collecting the outputs from the system of equations for very many trials.
To illustrate an example of the application of Monte Carlo simulation, consider the following scenario: Suppose we are to compute the mean and standard deviation of profits from an option. The value, \(V\), of the option at expiry, is given as:
$$ V=max\left[ \frac { 1 }{ T } \sum _{ t=1 }^{ T }{ { S }_{ t }X,0 } \right] $$
Where \(X\) is the strike price, \({ S }_{ t }\) is the underlying asset’s closure price at time \(t\) and there are \(T\) periods throughout the option’s lifetime. If there are two hundred days until the option expires, and the returns of the underlying asset are lognormal, with a mean of 0.1 and a standard deviation of 0.2, then our Monte Carlo simulation will have inputs that are lognormal variables with the appropriate values of the standard deviation and the mean.
For each trial, two hundred random daily returns would be generated and used to compute a series of random prices. The average of the price series would be computed and applied in the computation of the option’s value.
As a different approach, we could apply the central limit theorem. If very many \(i.i.d\) uniform distributions are summed up, we can easily approximate any normal variable after the appropriate constants are summed up and multiplied accordingly. Under a classical approach, we simply sum up twelve standard uniform variables, and subtract six:
$$ X=\sum _{ j=1 }^{ 12 }{ { H }_{ j }6 } $$
This will result in a standard normal variable whose mean is 0 and standard deviation is 1. The approximation becomes more accurate as we increase the number of uniform variables.
ChiSquared Distribution
Assume we’ve got \(k\) standard normal variables that are independent ranging from \({ Z }_{ 1 }\) to \({ Z }_{ k }\). The sum of their squares will then have a ChiSquare distribution, written as follows:
$$ S=\sum _{ 1=1 }^{ k }{ { Z }_{ i }^{ 2 } } $$
$$ S\sim { X }_{ k }^{ 2 } $$
\(k\) is called the degree of freedom. It is important to note that two chisquared variables that are independent, with degrees of freedom as \({ k }_{ 1 } \) and \({ k }_{ 2 } \), respectively, have a sum that is chisquare distributed with \( \left( { k }_{ 1 } + { k }_{ 2 } \right) \) degrees of freedom.
The chisquared variable is usually asymmetrical and takes on nonnegative values only. The distribution has a mean of \(k\) and standard deviation of \(2k\).
The chisquared distribution takes the following PDF, for positive values of \(x\):
$$ f\left( x \right) =\frac { 1 }{ { 2 }^{ \frac { k }{ 2 } }\Gamma \left( \frac { k }{ 2 } \right) } { x }^{ \frac { k }{ 2 } 1 }{ e }^{ \frac { x }{ 2 } } $$
The gamma function, \(\Gamma\), is such that:
$$ \Gamma \left( n \right) =\int _{ 0 }^{ \infty }{ { x }^{ n1 }{ e }^{ x }dx } $$
This distribution is widely applicable in statistics and risk management when testing hypotheses.
Student’s t Distribution
This distribution is often called the \(t\) distribution. Let \(Z\) be the standard normal variable, \(U\) be the chisquare variable having \(k\) degrees of freedom, and is independent of \(Z\). Then, a random variable \(X\) that follows a \(t\) distribution having \(k\) degrees of freedom is such that:
$$ X=\frac { Z }{ \sqrt { \frac { U }{ k } } } $$
The following formula represents its PDF:
$$ f\left( x \right) =\frac { \Gamma \left( k+\frac { 1 }{ 2 } \right) }{ \sqrt { k\pi } \Gamma \left( \frac { k }{ 2 } \right) } { \left( 1+{ { x }^{ 2 } }/{ k } \right) }^{ \frac { k+1 }{ 2 } } $$
The mean of the \(t\) distribution is usually zero, and the distribution is symmetrical around it. The distribution converges to a standard normal distribution as \(k\) tends towards infinity. When \(k > 2\), the variance of the distribution becomes: \({ k }/{ \left( k2 \right) }\), and it converges to one as \(k\) increases. This distribution is widely applicable in hypotheses testing, and modeling the returns of financial assets due to the excess kurtosis it displays.
F – Distribution
The Fdistribution is often used in the analysis of variance (ANOVA). The F distribution is an asymmetric distribution that has a minimum value of 0, but no maximum value. Notably, the curve approaches but never quite touches the horizontal axis.
\(X\) is said to follow an \(F\)distribution with parameters \({ k }_{ 1 }\) and \({ k }_{ 2 }\) if:
$$ X=\frac { { { U }_{ 1 } }/{ { k }_{ 1 } } }{ { { U }_{ 2 } }/{ { k }_{ 2 } } } \sim F\left( { k }_{ 1 },{ k }_{ 2 } \right) $$
Provided that \({ U }_{ 1 }\) and \({ U }_{ 2 }\) are chisquared distributions that are independent having \({ k }_{ 1 }\) and \({ k }_{ 2 }\) as their degrees of freedom.
The \(F\)distribution has the following PDF:
$$ f\left( x \right) =\frac { \sqrt { \frac { { \left( { k }_{ 1 }X \right) }^{ { k }_{ 1 } }{ { k } }_{ 2 }^{ { k }_{ 2 } } }{ { \left( { k }_{ 1 }X+{ k }_{ 2 } \right) }^{ { k }_{ 1 }+{ k }_{ 2 } } } } }{ xB\left( \frac { { k }_{ 1 } }{ 2 } ,\frac { { k }_{ 2 } }{ 2 } \right) } $$
B(x,y) is a beta function such that:
$$ B\left( x,y \right) =\int _{ 0 }^{ 1 }{ { z }^{ x1 }{ \left( 1z \right) }^{ y1 }dz } $$
The distribution has the following mean and variance respectively:
$$ \mu =\frac { { k }_{ 2 } }{ { k }_{ 2 }2 } for\quad { k }_{ 2 }>2 $$
$$ { \sigma }^{ 2 }=\frac { 2{ k }_{ 2 }^{ 2 }\left( { k }_{ 1 }+{ k }_{ 2 }2 \right) }{ { k }_{ 1 }{ \left( { k }_{ 2 }2 \right) }^{ 2 }\left( { k }_{ 2 }4 \right) } for\quad { k }_{ 2 }>4 $$
Suppose that \(X\) is a random variable with a \(t\)distribution, and it has \(k\) degrees of freedom, then \({ X }^{ 2 }\) is said to have an \(F\)distribution with 1 and \(k\) degrees of freedom, i.e.,
$$ { X }^{ 2 }\sim F\left( 1,k \right) $$
The Triangular Distribution
This is a distribution whose PDF takes the form of a triangle. The following function describes the PDF of a triangular distribution whose minimum is \(a\) and maximum is \(b\):
$$ f\left( x \right) =\begin{cases} \frac { 2\left( xa \right) }{ \left( ba \right) \left( ca \right) } a\le x\le c \\ \frac { 2\left( bx \right) }{ \left( ba \right) \left( bc \right) } c\le x\le b \end{cases} $$
The following two equations represent the mean and variance of the triangular distribution:
$$ \mu =\frac { a+b+c }{ c } $$
$$ { \sigma }^{ 2 }=\frac { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }abacbc }{ 18 } $$
The Beta Distribution
This distribution is similar to the triangle distribution in the sense that they are both applicable in the modelling of default rates and recovery rates. Assuming that \(a\) and \(b\) are two positive constants, then the PDF of the beta distribution is written as:
$$ f\left( x \right) =\frac { 1 }{ B\left( a,b \right) } { x }^{ a1 }{ \left( 1x \right) }^{ b1 },\quad \quad \quad 0\le x\le 1 $$
The following two equations represent the mean and variance of the beta distribution:
$$ \mu =\frac { a }{ a+b } $$
$$ { \sigma }^{ 2 }=\frac { ab }{ { \left( a+b \right) }^{ 2 }\left( a+b+1 \right) } $$
The Mixture Distribution
This distribution comes from a weighted average distribution of density functions, and can be written as follows:
$$ f\left( x \right) =\sum _{ j=1 }^{ n }{ { w }_{ i }{ f }_{ i }\left( x \right) } \quad such\quad that:\sum _{ i=1 }^{ n }{ { w }_{ i }=1 } $$
\({ f }_{ i }\left( x \right) \)’s are the component distributions, with \({ w }_{ i }^{ \prime }\)s as the weights or the mixing proportions. The component weights must all sum up to one, for the resulting mixture to be legitimately distributed. These distributions are very flexible as they fall between parametric and nonparametric distributions.
Question 1
The number of new clients a wealth management company receives in a month is distributed as a Poisson random variable with mean 2. Calculate the probability that the company receives exactly 28 clients in a year.

 5.48%
 0.10%
 3.54%
 10.2%
The correct answer is A.
The number of clients in a year (2 × 12) has a Poi(24) distribution.
$$ P\left[ X=n \right] =\frac { { \lambda }^{ n } }{ n! } { e }^{ \lambda } $$
$$ P\left[ X=28 \right] =\frac { { 24 }^{ 28 } }{ 28! } { e }^{ 24 } $$