Question

Plot the probability mass function (PMF) and the cumulative distribution function (CDF) of 3 random variables following (1) binomial distribution [p,n], (2) a geometric distribution [p], and (3) Poisson distribution [?]. You have to consider two sets of parameters per distribution which can be chosen arbitrarily. The following steps can be followed: Setp1: Establish two sets of parameters of the distribution: For Geometric and Poisson distributions take two values of p (p1 and p2) and take two values of [?], (?1 and ?2) respectively. For the binomial you should take two values of p and two values of n, first keep p fixed and change n, in the second keep n fixed and change p. Step 2: Generate the random variables (at least 10 values) [column 1] Step 3: Calculate the PMF [column 2] and CDF [column 3] Step 4: Plot the PMF and CDF in different graphics (It is recommended to combine the graphics while changing the parameters). Comment on how the parameters values affect the distribution.

Answer 1

ANSWER::

I am using R, I am providing the R code here, one can replicate and generate new graphs from this code by varying the repeatations and value of the parameters,

Here is the code for geomteric

##geometric
p1=0.5
p2=0.25

x1=rgeom(50,p1)
x2=rgeom(50,p2)

hist(x1,breaks=30,col="black", main="PMF of Geo(0.5)")
hist(x2,breaks=50,col="black", main="PMF of Geo(0.25)")

plot(ecdf(x1),main="CDF of Geo(0.5)")
plot(ecdf(x2),main="CDF of Geo(0.25)")

and here are the plots

Here as we see when lowered the parameter, we get much more values in the distribution as expected, the structire of the distribution however remains same

Now for Poisson

##Poisson
p1=3
p2=8

x1=rpois(50,p1)
x2=rpois(50,p2)
dev.new()
par(mfrow=c(1,2))

hist(x1,breaks=30,col="black", main="PMF of Pois(3)")
hist(x2,breaks=50,col="black", main="PMF of Pois(8)")

dev.new()
par(mfrow=c(1,2))

plot(ecdf(x1),main="CDF of Pois(3)")
plot(ecdf(x2),main="CDF of Pois(8)")

Now for the plots

Here also we see even though the bell shaped structure of the distribution remains same in both the cases. Increasing the parameter increases the spread of the distribution

Finally for Binomial

First we keep n constant and vary p

##binomial
p1=0.5
p2=0.25

n1=10
n2=20

x11=rbinom(100,n1,p1)
x12=rbinom(100,n1,p2)
dev.new()
par(mfrow=c(1,2))

hist(x11,breaks=40,col="black", main="PMF of Binom(10,0.5)")
hist(x12,breaks=40,col="black", main="PMF of Binom(10,0.25)")

dev.new()
par(mfrow=c(1,2))

plot(ecdf(x11),main="CDF of Binom(10,0.5)")
plot(ecdf(x12),main="CDF of Binom(10,0.25)")

The plots are

Interesting to see the change of structure in the two cases, although the domain remains same p1=0. gives symmetric distribution whereas p1=0.25 gives a skewed one

Now keep p=0.5 and vary n

p1=0.5
p2=0.25

n1=10
n2=50

x11=rbinom(100,n1,p1)
x12=rbinom(100,n2,p1)
dev.new()
par(mfrow=c(1,2))

hist(x11,breaks=40,col="black", main="PMF of Binom(10,0.5)")
hist(x12,breaks=40,col="black", main="PMF of Binom(50,0.5)")

dev.new()
par(mfrow=c(1,2))

plot(ecdf(x11),main="CDF of Binom(10,0.5)")
plot(ecdf(x12),main="CDF of Binom(50,0.5)")

The plots here are

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