Common statistical distribution

Continuous Distributions

Normal distribution

The Central Limit Theorem says sample means and sample sums approach normal distributions as the sample size approaches infinity (normal, t, chisq, f distributions).

Probaility density function

\[ f(x)=\frac{e^{-(x-\mu)^2 /\left(2 \sigma^2\right)}}{\sigma \sqrt{2 \pi}} \]

where $\mu$ is the location (shift the graph)/center parameter and $\sigma$ is the scale (stretch out the graph)/spread parameter.

# numbers from  
x<-seq(-5,5,0.1)
 
# calculate the distribution function
# based on the parameters
pdf<- dnorm(x,(0),(2))
 
# Plotting the PDF 
plot(x,pdf,type = "l")

two different normal distribution multiple, approaching normal

# numbers from  
 
# calculate the distribution function
# based on the parameters
pdf1<- rnorm(1000,(9),(1))
pdf2<- rnorm(1000,(30),(10))
pdf=pdf1*pdf2
 
# Plotting the PDF 
hist(pdf,probability = T)
lines(density(pdf),col=2)

normal distribution *n, normal

# numbers from  
 
 
# calculate the distribution function
# based on the parameters
pdf1<- rnorm(1000,(1),(1))
pdf=pdf1*60
 
# Plotting the PDF 
hist(pdf,probability = T)
lines(density(pdf),col=2)

normal distribution *power, not normal

# numbers from  

# calculate the distribution function
# based on the parameters
pdf1<- rnorm(1000,(1),(1))
pdf=pdf1**4
 
# Plotting the PDF 
hist(pdf,probability = T)
lines(density(pdf),col=2)

normal distribution square root, approaching normal

# numbers from  

# calculate the distribution function
# based on the parameters
pdf1<- rnorm(1000,(10),(3))
pdf1=pdf1[pdf1>0]
pdf=sqrt(pdf1) 
 
# Plotting the PDF 
hist(pdf,probability = T)
lines(density(pdf),col=2)

two normal distribution plus, approaching normal

# numbers from  
 
# calculate the distribution function
# based on the parameters
pdf1<- rnorm(1000,(2),(1))
pdf2<- rnorm(1000,(8),(1))
pdf=pdf1+pdf2
 
# Plotting the PDF 
hist(pdf,probability = T)
lines(density(pdf),col=2)

normal distribution plus n, approaching normal

# numbers from  
 
 
# calculate the distribution function
# based on the parameters
pdf1<- rnorm(1000,(1),(1))
pdf=pdf1+100
 
# Plotting the PDF 
hist(pdf,probability = T)
lines(density(pdf),col=2)

two normal distribution, arithmetic operations (mixed distribution is approaching normal)

# numbers from  

# calculate the distribution function
# based on the parameters
pdf1<- rnorm(1000,(8),(1))
pdf2<- rnorm(1000,(2),(1))
pdf=pdf1**2+pdf2
 
# Plotting the PDF 
hist(pdf,probability = T)
lines(density(pdf),col=2)

normalization by scale (center and standard deviation) toward normal

# numbers from  

# calculate the distribution function
# based on the parameters
 
chi=rchisq(1000,4)
hist(chi,probability = T)
lines(density(chi),col=2)

pdf1<- scale(chi)
pdf=pdf1
# Plotting the PDF 
hist(pdf,probability = T)
lines(density(pdf),col=2)

Cummulative distribution function

\[ F(x)=\int_{-\infty}^x \frac{e^{-x^2 / 2}}{\sqrt{2 \pi}} \]

Note that this integral does not exist in a simple closed formula. It is computed numerically. it is not related to mu and sigma.

# numbers from  
x<-seq(-5,5,0.1)
 
# calculate the distribution function
# based on the parameters
cdf<- pnorm(x,(0),(2))
 
# Plotting the PDF 
plot(x,cdf,type = "l")

Properties

Mean The location parameter μ. Median The location parameter μ. Mode The location parameter μ. x Range −∞ to ∞. Standard Deviation The scale parameter σ. Coefficient of Variation σ/μ Skewness 0 (is systematic) Kurtosis 3 (>3 thin peak)

t distribution

PDF

\[ f(x)=\frac{\left(1+\frac{x^2}{\nu}\right)^{\frac{-(\nu+1)}{2}}}{B(0.5,0.5 \nu) \sqrt{\nu}} \]

where $\boldsymbol{B}$ is the beta function and $v$ is a positive integer shape parameter.

\[ B(0.5, 0.5\nu)=\int_0^1 t^{0.5-1}(1-t)^{0.5\nu-1} d t \]

In a testing context, the $t$ distribution is treated as a “standardized distribution” (i.e., no location or scale parameters).

# numbers from  
x<-seq(-5,5,0.1)
 
# calculate the distribution function
# based on the parameters
pdf1<- dt(x,1)
pdf2<- dt(x,30)
 
# Plotting the PDF 
plot(x,pdf1,type = "l")
lines(x,pdf2 ,col=2)

CDF

The formula for the cumulative distribution function of the t distribution is complicated.

# numbers from  
x<-seq(-5,5,0.1)
 
# calculate the distribution function
# based on the parameters
cdf1<- pt(x,1)
cdf2<- pt(x,30)
 
# Plotting the PDF 
plot(x,cdf1,type = "l")
lines(x,cdf2 ,col=2)

properties

$$ \[\begin{aligned} &\begin{array}{ll} \text { Mean } & 0 \text { (It is undefined for } v \text { equal to 1.) } \\ \text { Median } & 0 \\ \text { Mode } & 0 \\ \text { Range } & -\infty \text { to } \infty \\ \text { Standard } & \sqrt{\frac{\nu}{(\nu-2)}} \\ \text { Deviation } & \text { It is undefined for } v \text { equal to } 1 \text { or } 2 . \\ & \text { Undefined } \\ \text { Coefficient of } & \\ \text { Variation } & \begin{array}{l} \\ \text { } \end{array} \\ & \begin{array}{l} \text { Skewness: 3 } \\ \text { } \end{array} \\ \text { Kurtosis } & \frac{3(\nu-2)}{(\nu-4)} \end{array}\\ &\text { It is undefined for } v \text { less than or equal to } 4 \text {. } \end{aligned}\]

Chi-square distribution

$v$ independent variables with standard normal distributions are squared and summed.

\[ f(x)=\frac{e^{\frac{-x}{2}} x^{\frac{\nu}{2}-1}}{2^{\frac{\nu}{2}} \Gamma\left(\frac{\nu}{2}\right)} \quad \text { for } x \geq 0 \] where $v$ is the shape parameter and $\Gamma$ is the gamma function.

\[ \Gamma(v/2)=\int_0^{\infty} t^{v/2-1} e^{-t} d t \]

In a testing context, the chi-square distribution is treated as a “standardized distribution” (i.e., no location or scale parameters).

# numbers from  
x<-seq(0,5,0.1)
 
# calculate the distribution function
# based on the parameters
pdf1<- dchisq(x,1)
pdf2<- dchisq(x,5)
 
# Plotting the PDF 
plot(x,pdf1,type = "l")
lines(x,pdf2 ,col=2)

chisq= normal distribution**2 with freedom degree 1

# numbers from  

# calculate the distribution function
# based on the parameters
x<- rnorm(1000,(1),(1))
y=x**2
 
# Plotting the PDF 
hist(y,prob=T)
lines(density(y),col=2)

CDF

\[ F(x)=\frac{\gamma\left(\frac{\nu}{2}, \frac{x}{2}\right)}{\Gamma\left(\frac{\nu}{2}\right)} \quad \text { for } x \geq 0 \]

γ is the incomplete gamma function.

# numbers from  
x<-seq(0,5,0.1)
 
# calculate the distribution function
# based on the parameters
cdf1<- pchisq(x,1)
cdf2<- pchisq(x,5)
 
# Plotting the PDF 
plot(x,cdf1,type = "l")
lines(x,cdf2 ,col=2)

Properties

\[ \begin{array}{ll} \text { Mean } & v \\ \text { Median } & \text { approximately } v-2 / 3 \text { for large } v \\ \text { Mode } & \nu-2 \quad \text { for } \nu>2 \\ \text { Range } & 0 \text { to } \infty \\ \text { Standard } & \sqrt{2 \nu} \\ \text { Deviation } & \\ \text { Coefficient of Variation} & \sqrt{\frac{2}{\nu}} \\ \text { } & \frac{2^{1.5}}{\sqrt{\nu}} \\ \text { Skewness } & 3+\frac{12}{\nu} \\ \text { Kurtosis } & \end{array} \]

f distribution

PDF

The $\mathrm{F}$ distribution is the ratio of two chi-square distributions with degrees of freedom $v_1$ and $v_2$, respectively, where each chisquare has first been divided by its degrees of freedom.

\[ f(x)=\frac{\Gamma\left(\frac{\nu_1+\nu_2}{2}\right)\left(\frac{\nu_1}{\nu_2}\right)^{\frac{\nu_1}{2}} x^{\frac{\nu_1}{2}-1}}{\Gamma\left(\frac{\nu_1}{2}\right) \Gamma\left(\frac{\nu_2}{2}\right)\left(1+\frac{\nu_1 x}{\nu_2}\right)^{\frac{\nu_1+\nu_2}{2}}} \]

where $v_1$ and $v_2$ are the shape parameters and $\Gamma$ is the gamma function.

\[ \Gamma(y)=\int_0^{\infty} t^{y-1} e^{-t} d t \]

# numbers from  
x<-seq(0,5,0.1)
 
# calculate the distribution function
# based on the parameters
pdf1<- df(x,1,1)
pdf2<- df(x,10,10)
 
# Plotting the PDF 
plot(x,pdf1,type = "l")
lines(x,pdf2 ,col=2)

f= chisq1/chisq2

# numbers from  
 
# calculate the distribution function
# based on the parameters
cdf1<- rchisq(100,9)
cdf2<- rchisq(100,7)
f=cdf1/cdf2

# Plotting the PDF 
hist(f,probability = T)
lines(density(f),col=2)

CDF

$F(x)=1-I_k\left(\frac{\nu_2}{2}, \frac{\nu_1}{2}\right)$ where $k=\nu_2 /\left(\nu_2+\nu_1 \cdot x\right)$ and $\boldsymbol{I}_{\boldsymbol{k}}$ is the incomplete beta function.

\[ I_k(x, \alpha, \beta)=\frac{\int_0^x t^{\alpha-1}(1-t)^{\beta-1} d t}{B(\alpha, \beta)} \]

where $\boldsymbol{B}$ is the beta function

\[ B(\alpha, \beta)=\int_0^1 t^{\alpha-1}(1-t)^{\beta-1} d t \]

# numbers from  
x<-seq(0,5,0.1)
 
# calculate the distribution function
# based on the parameters
cdf1<- pf(x,1,1)
cdf2<- pf(x,10,10)
 
# Plotting the PDF 
plot(x,cdf1,type = "l")
lines(x,cdf2 ,col=2)

Properties

\[ \begin{array}{lll} \text { Mean } & \frac{\nu_2}{\left(\nu_2-2\right)} \quad \nu_2>2 & \\ \text { Mode } & \frac{\nu_2\left(\nu_1-2\right)}{\nu_1\left(\nu_2+2\right)} \quad \nu_1>2 & \\ \text { Range } & 0 \text { to } \infty & \\ \text { Standard Deviation} & \sqrt{\frac{2 \nu_2^2\left(\nu_1+\nu_2-2\right)}{\nu_1\left(\nu_2-2\right)^2\left(\nu_2-4\right)}} \quad \nu_2>4 \\ \text { } & \sqrt{\frac{2\left(\nu_1+\nu_2-2\right)}{\nu_1\left(\nu_2-4\right)}} \quad \nu_2>4 \\ \text { Coefficient of Variation} & \frac{\left(2 \nu_1+\nu_2-2\right) \sqrt{8\left(\nu_2-4\right)}}{\sqrt{\nu_1}\left(\nu_2-6\right) \sqrt{\left(\nu_1+\nu_2-2\right)}} \quad \nu_2>6 \\ \text { } & \text { Skewness } & \end{array} \]

Uniform distribution

Probaility density function

\[ f(x)=\frac{1}{B-A} \quad \text { for } A \leq x \leq B \] where $\mathrm{A}$ is the location parameter and $(\mathrm{B}-\mathrm{A})$ is the scale parameter.

# numbers from  
x<-seq(-5,5,0.1)
 
# calculate the distribution function
# based on the parameters
pdf<- dunif(x,(-5),(5))
 
# Plotting the PDF 
plot(x,pdf,type = "l")

Cummulative distribution function

\[ F(x)=x \quad \text { for } 0 \leq x \leq 1 \]

# numbers from  
x<-seq(-5,5,0.1)
 
# calculate the distribution function
# based on the parameters
cdf<- punif(x,(-5),(5))
 
# Plotting the PDF 
plot(x,cdf,type = "l")

Preporties

Mean (A + B)/2 Median (A + B)/2 Range B - A Standard Deviation √((B−A)*(B−A)/12) Coefficient of Variation (B−A)/√3(B+A) Skewness 0 Kurtosis 9/5

Discrete Distributions

If you ask how many successes there will be among $n$ Bernoulli trials, then the answer will have a binomial distribution, $\operatorname{Binomial}(n, p)$. The Bernoulli distribution, Bernoulli $(p)$, simply says whether one trial is a success. If you ask how many trials it will be to get the first success, then the answer will have a geometric distribution, Geometric $(p)$. If you ask how many trials there will be to get the $r^{\text {th }}$ success, then the answer will have a negative binomial distribution, NegativeBinomial $(p, r)$. Given that there are $M$ successes among $N$ trials, if you ask how many of the first $n$ trials are successes, then the answer will have a $\operatorname{HypergeOmetric~}(N, M, n)$ distribution. Sampling with replacement is described below under the binomial distribution, while sampling without replacement is described under the hypergeometric distribution.

Bernoulli Distribution

As in a Bernoulli process, you can ask various questions about a Poisson process, and the answers will have various distributions. If you ask how many events occur in an interval of length $t$, then the answer will have a Poisson distribution, Poisson $(\lambda t)$. If you ask how long until the first event occurs, then the answer will have an exponential distribution, $\operatorname{Exponential}(\lambda)$. If you ask how long until the $r^{\text {th }}$ event, then the answer will have a gamma distribution, $\operatorname{Gamma}(\lambda, r)$. If there are $\alpha+\beta$ events in a given time interval, if you ask what fraction of the interval it takes until the $\alpha^{\text {th }}$ event occurs, then the answer will have a $\operatorname{BETA}(\alpha, \beta)$ distribution.

PDF

The Bernoulli distribution is the probability distribution of Bernoulli trial, which only has 2 outcomes (eg: success or failure, win or lose).

\[ p(x)=p^x(1-p)^{1-x} \]

# numbers from  
x<-c(0,1)
 
# calculate the distribution function
# based on the parameters
pdf1<- dbinom(x, 1, 0.56)
pdf2<- dbinom(x, 1, 0.07)
 
# Plotting the PDF 
barplot(names.arg =x,height =pdf1)

barplot(names.arg =x,height =pdf2)

Binomial Distribution

PDF

the plot of the binomial probability density function is on p and n.

\[ P(x ; p, n)=\left(\begin{array}{l} n \\ x \end{array}\right)(p)^x(1-p)^{(n-x)} \quad \text { for } x=0,1,2, \cdots, n \]

approaching normal

# numbers from  
x<-seq(0,1000,1)
 
# calculate the distribution function
# based on the parameters
pdf1<- dbinom(x, 1000, 0.56)
pdf2<- dbinom(x, 1000, 0.07)
 
# Plotting the PDF 
plot(x,pdf1,type = "l")
lines(x,pdf2 ,col=2)

CDF

\[ F(x ; p, n)=\sum_{i=0}^x\left(\begin{array}{c} n \\ i \end{array}\right)(p)^i(1-p)^{(n-i)} \]

# numbers from  
x<-seq(0,1000,1)
 
# calculate the distribution function
# based on the parameters
cdf1<- pbinom(x, 1000, 0.56)
cdf2<- pbinom(x, 1000, 0.07)
 
# Plotting the PDF 
plot(x,cdf1,type = "l")
lines(x,cdf2 ,col=2)

Properties

\[ \begin{array}{ll} \text { Mean } & n p \\ \text { Mode } & p(n+1)-1 \leq x \leq p(n+1) \\ \text { Range } & 0 \text { to } n \\ \text { Standard } & \sqrt{n p(1-p)} \\ \text { Deviation } & \sqrt{\frac{(1-p)}{n p}} \\ \text { Coefficient of Variation} & \frac{(1-2 p)}{\sqrt{n p(1-p)}} \\ \text { } & 3-\frac{6}{n}+\frac{1}{n p(1-p)} \\ \text { Skewness } & \end{array} \]

Geometric distribution

with probability $p$ of success, the number of trials $X$ it takes to get the first success has a geometric distribution (failures prior to the first success).

\[ \begin{aligned} & f(x)=q^{x-1} p, \text { for } x=1,2, \ldots \\ & \mu=\frac{1}{p} \cdot \sigma^2=\frac{1-p}{p^2} . \\ & m(t)=\frac{p e^t}{1-q e^t} . \end{aligned} \]

the moment generating function m(t) = E(X^t).

# numbers from  
x<-seq(0,10,1)
 
# calculate the distribution function
# based on the parameters
pdf1<- dgeom(x, 0.23)
pdf2<- dgeom(x, 0.7)
 
# Plotting the PDF 
plot(x,pdf1,type = "l")
lines(x,pdf2 ,col=2)

Negative binomal distribution

with probability $p$ of success, and $X$ is the trial number when $r$ successes are first achieved (x failures before r successes).

$\operatorname{Geometric}(p)=\operatorname{NegativeBinomial}(p, 1)$.

\[ \begin{aligned} & f(x)=\left(\begin{array}{l} x-1 \\ r-1 \end{array}\right) p^r q^{x-r}, \text { for } x=r, r+1, \ldots \\ & \mu=\frac{r}{p} . \sigma^2=\frac{r q}{p^2} . \\ & m(t)=\left(\frac{p e^t}{1-q e^t}\right)^r . \end{aligned} \]

# numbers from  
x<-seq(0,10,1)
 
# calculate the distribution function
# based on the parameters
pdf1<- dnbinom(x, 2,0.23)
pdf2<- dnbinom(x, 2,0.7)
 
# Plotting the PDF 
plot(x,pdf1,type = "l")
lines(x,pdf2 ,col=2)

Hypergeometric distribution

there are $M$ successes among $N$ trials, the number $X$ of successes among the first $n$ trials has a HypergEOMETric $(N, M, n)$ distribution.

\[ \begin{aligned} & f(x)=\frac{\left(\begin{array}{c} M \\ x \end{array}\right)\left(\begin{array}{c} N-M \\ n-x \end{array}\right)}{\left(\begin{array}{l} N \\ n \end{array}\right)} \text {, for } x=0,1, \ldots, n . \\ & \mu=n p . \sigma^2=\left(\begin{array}{l} N-n \\ N-1 \end{array}\right) n p q . \end{aligned} \]

# numbers from  
x<-seq(0,5,1)
 
# calculate the distribution function
# based on the parameters
pdf1<- dhyper(x, 10,14,5)
pdf2<- dhyper(x, 14,10,5)
 
# Plotting the PDF 
plot(x,pdf1,type = "l")
lines(x,pdf2 ,col=2)

# m
# the number of white balls in the urn.
# n
# the number of black balls in the urn.
# k
# the number of balls drawn from the urn.
# x
# the number of white balls drawn without replacement

Poisson process

Poisson

PDF

\[ p(x ; \lambda)=\frac{e^{-\lambda} \lambda^x}{x !} \text { for } x=0,1,2, \cdots \]

$\lambda$ is the shape parameter which indicates the average number of events.

# numbers from  
x<-seq(0,30,1)
 
# calculate the distribution function
# based on the parameters
pdf1<- dpois(x, 1)
pdf2<- dpois(x, 7)
pdf3<- dpois(x, 12)
 
# Plotting the PDF 
plot(x,pdf1,type = "l")
lines(x,pdf2 ,col=2)
lines(x,pdf3 ,col=3)

CDF

\[ F(x ; \lambda)=\sum_{i=0}^x \frac{e^{-\lambda} \lambda^i}{i !} \]

# numbers from  
x<-seq(0,30,1)
 
# calculate the distribution function
# based on the parameters
cdf1<- ppois(x, 1)
cdf2<- ppois(x, 7)
cdf3<- ppois(x, 12)
 
# Plotting the PDF 
plot(x,cdf1,type = "l")
lines(x,cdf2 ,col=2)
lines(x,cdf3 ,col=3)

Properties

\[ \begin{array}{ll} \text { Mean } & \lambda \\ \text { Mode } & \text { For non-integer } \lambda, \text { it is the largest integer } \\ & \text { less than } \lambda . \text { For integer } \lambda, x=\lambda \text { and } x=\lambda \text { - } \\ & 1 \text { are both the mode. } \\ & 0 \text { to } \infty \\ \text { Range } & \sqrt{\lambda} \\ \text { Standard } & \\ \text { Deviation } & \\ \text { Coefficient of Variation} & \frac{1}{\sqrt{\lambda}} \\ \text { Skewness } & \frac{1}{\sqrt{\lambda}} \\ \text { Kurtosis } & 3+\frac{1}{\lambda} \end{array} \]

Exponential distribution

Probaility density function

let’s say mu=0 below.

\[ f(x)=\frac{1}{\beta} e^{-(x-\mu) / \beta} \quad x \geq \mu ; \beta>0 \]

where $\mu$=0 here is the location parameter and $\beta$ is the scale parameter (the scale parameter is often referred to as $\lambda$ which equals $1 / \beta$ ).

\[ f(x)=\lambda e^{-(x) \lambda} \quad x \geq 0 ; \lambda>0 \]

$\lambda=1 or 5$ is scale

# numbers from  
x<-seq(0,5,0.1)
 
# calculate the distribution function
# based on the parameters
pdf1<- dexp(x,1)
pdf2<- dexp(x,5)
 
# Plotting the PDF 
plot(x,pdf1,type = "l")
lines(x,pdf2,col=2)

Cummulative distribution function

\[ F(x)=1-e^{-\left(x^\gamma\right)} \quad x \geq 0 ; \gamma>0 \]

# numbers from  
x<-seq(0,5,0.1)
 
# calculate the distribution function
# based on the parameters
cdf<- pexp(x,2)
 
# Plotting the PDF 
plot(x,cdf,type = "l")

Properties

Mean β =1/lambda Median βln2 Mode μ Range μ to ∞ Standard Deviation β =1/lambda Coefficient of Variation 1 Skewness 2 Kurtosis 9

Weibull Distribution

let’s say mu=0 below.

Probaility density function

\[ f(x)=\frac{\gamma}{\alpha}\left(\frac{x-\mu}{\alpha}\right)^{(\gamma-1)} \exp \left(-((x-\mu) / \alpha)^\gamma\right) \quad x \geq \mu ; \gamma, \alpha>0 \]

where $\gamma$ is the shape parameter, $\mu$=0 here is the location parameter and $\alpha$ is the scale parameter.

The case where $\mu=0$ and $\alpha=1$ is called the standard Weibull distribution. \[ f(x)=\gamma x^{(\gamma-1)} \exp \left(-\left(x^\gamma\right)\right) \quad x \geq 0 ; \gamma>0 \]

shape/ gamma =0.5 or 3

# numbers from  
x<-seq(0,5,0.1)
 
# calculate the distribution function
# based on the parameters
pdf1<- dweibull(x,0.5,1)
pdf2<- dweibull(x,3,1)
 
# Plotting the PDF 
plot(x,pdf1,type = "l")
lines(x,pdf2 ,col=2)

same shape/ gamma =0.5 but different scales

# numbers from  
x<-seq(0,5,0.1)
 
# calculate the distribution function
# based on the parameters
pdf1<- dweibull(x,0.5,1)
pdf2<- dweibull(x,0.5,5)
 
# Plotting the PDF 
plot(x,pdf1,type = "l")
lines(x,pdf2 ,col=2)

Cummulative distribution function

\[ F(x)=1-e^{-\left(x^\gamma\right)} \quad x \geq 0 ; \gamma>0 \]

shape/ gamma=0.5 or 3

# numbers from  
x<-seq(0,5,0.1)
 
# calculate the distribution function
# based on the parameters
cdf1<- pweibull(x,0.5,1)
cdf2<- pweibull(x,3,1)
 
# Plotting the PDF 
plot(x,cdf1,type = "l")
lines(x,cdf2,col=2)

Properties

Mean $\quad \Gamma\left(\frac{\gamma+1}{\gamma}\right)$ where $\Gamma$ is the gamma function $\Gamma(a)=\int_0^{\infty} t^{a-1} e^{-t} d t$
Median & $\ln (2)^{1 / \gamma}$
Mode & $\left(1-\frac{1}{\gamma}\right)^{1 / \gamma}$ & $\gamma>1$ Range & 0 & $\gamma \leq 1$ Standard Deviation & $\sqrt{\Gamma\left(\frac{\gamma+2}{\gamma}\right)-\left(\Gamma\left(\frac{\gamma+1}{\gamma}\right)\right)^2}$ Coefficient of & $\sqrt{\frac{\Gamma\left(\frac{\gamma+2}{\gamma}\right)}{\left(\Gamma\left(\frac{\gamma+1}{\gamma}\right)\right)^2}-1}$

$\gamma$ if or shape and $\lambda$ is for scale.

Gamma Distribution

probability density function

\[ f(x)=\frac{\left(\frac{x-\mu}{\beta}\right)^{\gamma-1} \exp \left(-\frac{x-\mu}{\beta}\right)}{\beta \Gamma(\gamma)} \quad x \geq \mu ; \gamma, \beta>0 \]

where $\gamma$ is the shape parameter, $\mu$ is the location parameter, $\beta$ is the scale parameter, and $\Gamma$ is the gamma function which has the formula

\[ \Gamma(\gamma)=\int_0^{\infty} t^{\gamma-1} e^{-t} d t \]

The case where $\mu=0$ and $\beta=1$ is called the standard gamma distribution.

\[ f(x)=\frac{x^{\gamma-1} e^{-x}}{\Gamma(\gamma)} \quad x \geq 0 ; \gamma>0 \]

# numbers from  
x<-seq(0,5,0.1)
 
# calculate the distribution function
# based on the parameters
pdf1<- dgamma(x,0.5,1)
pdf2<- dgamma(x,2,1)
 
# Plotting the PDF 
plot(x,pdf1,type = "l")
lines(x,pdf2 ,col=2)

Cummulative distribution function

\[ F(x)=\frac{\Gamma_x(\gamma)}{\Gamma(\gamma)} \quad x \geq 0 ; \gamma>0 \]

where $\Gamma$ is the gamma function defined

\[ \Gamma_x(\gamma)=\int_0^x t^{\gamma-1} e^{-t} d t \]

# numbers from  
x<-seq(0,5,0.1)
 
# calculate the distribution function
# based on the parameters
cdf1<- pgamma(x,0.5,1)
cdf2<- pgamma(x,2,1)
 
# Plotting the PDF 
plot(x,cdf1,type = "l")
lines(x,cdf2 ,col=2)

Properties

\[ \begin{array}{ll} \text { Mean } & \gamma \\ \text { Mode } & \gamma-1 \quad \gamma \geq 1 \\ \text { Range } & 0 \text { to } \infty . \\ \text { Standard } & \sqrt{\gamma} \\ \text { Deviation } & \\ \text { Skewness } & \frac{2}{\sqrt{\gamma}} \\ \text { Kurtosis } & 3+\frac{6}{\gamma} \\ \text { Coefficient of } & \frac{1}{\sqrt{\gamma}} \\ \text { Variation } & \end{array} \]

Beta distribution

PDF

\[ f(x)=\frac{(x-a)^{p-1}(b-x)^{q-1}}{B(p, q)(b-a)^{p+q-1}} \quad a \leq x \leq b ; p, q>0 \] where $p$ and $q$ are the shape parameters, $a$ and $b$ are the lower and upper bounds, respectively

\[ \begin{aligned} & \text { location }=a \\ & \text { scale }=b-a \end{aligned} \]

$B(p, q)$ is the beta function. The beta function has the formula

\[ B(p, q)=\int_0^1 t^{p-1}(1-t)^{q-1} d t \]

The case where $a=0$ and $b=1$ is called the standard beta distribution.

\[ f(x)=\frac{x^{p-1}(1-x)^{q-1}}{B(p, q)} \quad 0 \leq x \leq 1 ; p, q>0 \]

# numbers from  
x<-seq(0,2,0.01)
 
# calculate the distribution function
# based on the parameters
pdf1<- dbeta(x,0.5,0.5)
pdf2<- dbeta(x,0.5,1)
pdf3<- dbeta(x,1,0.5)
pdf4<- dbeta(x,2,2)
 
# Plotting the PDF 
plot(x,pdf1,type = "l")
lines(x,pdf2 ,col=2) #red
lines(x,pdf3 ,col=3) #green
lines(x,pdf4 ,col=4) #blue

Properties

\[ \begin{array}{ll} \text { Mean } & \frac{p}{p+q} \\ \text { Mode } & \frac{p-1}{p+q-2} \quad p, q>1 \\ \text { Range } & 0 \text { to } 1 \\ \text { Standard } & \sqrt{\frac{p q}{(p+q)^2(p+q+1)}} \\ \text { Deviation } & \sqrt{\frac{q}{p(p+q+1)}} \\ \text { Coefficient of variance} & \frac{2(q-p) \sqrt{p+q+1}}{(p+q+2) \sqrt{p q}} \end{array} \]

Common statistical distribution

2024-01-29

Continuous Distributions

Normal distribution

Probaility density function

Cummulative distribution function

Properties

t distribution

PDF

CDF

properties

Chi-square distribution

CDF

Properties

f distribution

PDF

CDF

Properties

Uniform distribution

Probaility density function

Cummulative distribution function

Preporties

Discrete Distributions

Bernoulli Distribution

PDF

Binomial Distribution

PDF

CDF

Properties

Geometric distribution

Negative binomal distribution

Hypergeometric distribution

Poisson process

Poisson

PDF

CDF

Properties

Exponential distribution

Probaility density function

Cummulative distribution function

Properties

Weibull Distribution

Probaility density function

Cummulative distribution function

Properties

Gamma Distribution

probability density function

Cummulative distribution function

Properties

Beta distribution

PDF

Properties