Discrete Probability Distribution

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Introduction

As we know
\(f( x,y )=\frac{xy}{36}\)for \(x=1,2,3,y=1,2,3\)
is a probability function of discrete random variable
\( f( x,y )=\frac{3}{5}x( x+y )\) for \( 0 \le x \le 1,0 \le y \le 2\)
is probability function for continuous random variable
Among such various types of probability functions (discrete or continuous), some particular types of discrete probability functions are

1. binomial probability function
2. hypergeometric probability function
3. poisson probability function

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Binomial probability function

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Binomial probability function is a special kind of discrete probability function.
It satisfy following assumptions.

1. The experiment consist finite trail. The number of trail is denoted by n. For example, if we toss a coin 5 times, then n=5.
2. The experiment consists only two mutually exclusive outcomes. One is called success and other is called failure. The probability of success is denoted by \(p\) and probability of failure is denoted by \(q=(1-p)\).
3. The probability of success (or failure) in each trail remain constant. i.e. the probability of outcome must be independent from trail to trail.

Below is a simulation for different \(p \in \mathbb{N}\) taking n=20.

Definition

A random variable X is said to have binomial probability distribution if its probability function is
\( f( x)=b( x;n,p)= \begin{pmatrix} n \\ x \end{pmatrix} p^xq^{n-x}\) for \( x=0,1,2,..,n\)
where \(n \in \mathbb{N}\) and \(p \in [0,1]\) are parameters and \( q=1-p\)

Example

A fair coin is tossed eight times, find the probability of getting exactly six heads.
Solution
Since a fair coin is tossed eight times,
Total number of trails \( n=8\)
Probability of success (Head) in each trial \(p=0.5\)
Probability of failure (Tail) in each trial \(q=1-p=0.5\)
Concern of success \( x=6\)
Thus
\( f(x)= \begin{pmatrix} n \\ x \end{pmatrix} \times p^x\times q^{n-x}\)
or \( f(x)= \begin{pmatrix} 8 \\ 6 \end{pmatrix}  \times {{0.5}^{6}}\times {{0.5}^2}\)
or \( f(x)=\frac{28}{256}=\frac{7}{64}=0.11\)

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pdf Theorem

Show that \( \displaystyle \sum_X f(x) =1 \) where \(f( x )\) is binomial probability function
Solution
Let X be a random variable having binomial probability distribution with parameter n and p, then total probability of X is
\( \displaystyle \sum_X f(x)=\sum_{x=0}^nf(x) \)
or \( \displaystyle \sum_X f(x)=\sum_{x=0}^n \begin{pmatrix} n \\ x \end{pmatrix} \times p^x\times q^{n-x}\)
According to Binomial Theorem, for some values a and b and a positive integer n:
\( \displaystyle \sum_{x=0}^n \begin{pmatrix} n \\ x \end{pmatrix} \times a^x\times b^{n-x}=( a+b )^n\)
Thus we have
\( \displaystyle \sum_X f(x)= \sum_{x=0}^n \begin{pmatrix} n \\ x \end{pmatrix} \times p^x\times q^{n-x}\)
or \( \displaystyle \sum_X f(x)=( p+q )^n\)
or \( \displaystyle \sum_X f(x)=1\)

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Mean of Binomial Distribution: \( E( X )=np\)

Let X be a random variable having binomial distribution with parameter n and p, then
\( E( X )=\displaystyle \sum_X x f( x )\)
or \( E( X )= \displaystyle \sum_{x=0}^n x. \begin{pmatrix} n \\ x \end{pmatrix} p^xq^{n-x}\)
or \( E( X )=\displaystyle \sum_{x=1}^n x\times \frac{n}{x} \times \begin{pmatrix} n-1 \\ x-1 \end{pmatrix} p^xq^{n-x}\) [binomial theorem अनुसार \( \begin{pmatrix} n \\ x \end{pmatrix} = \frac{n}{x} \times \begin{pmatrix} n-1 \\ x-1 \end{pmatrix} \) हुने भएकोले]
or \( E( X )=np \displaystyle \sum_{x=1}^n \begin{pmatrix} n-1 \\ x-1 \end{pmatrix} \times p^{x-1}\times q^{( n-1 )-( x-1 )}\)
Let \( y=x-1\), then
when \(x=1\), then \(y=0\)
when \(x=n\), then \(y=n-1\), say \(n-1=m \)
Now, mean of binomial distribution is
\( E( X )=np \displaystyle \sum_{y=0}^m \begin{pmatrix} m \\ y \end{pmatrix} p^yq^{m-y}\)
or \( E( X )=np \times 1 \) [pdf theorem अनुसार \( \displaystyle \sum_{y=0}^m \begin{pmatrix} m \\ y \end{pmatrix} p^yq^{m-y}=1\) हुने भएकोले]
or \( E( X )=np\)

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Variance of Binomial Distribution: \( V( X )=npq\)

Let X be a random variable having binomial distribution with parameter n and p, then
Variance of X is
\( V(X)=E(X^2)-E(X)^2\)
We know that
\(E(X^2)=E[X(X-1)+X]\)
or \(E(X^2)=E[X(X-1)]+E[X]\)
Thus, Variance of X is
\( V(X)=E(X^2)-E(X)^2\)
or \( V(X)=E[X(X-1)]+E[X]-E(X)^2\)
Now, we compute \( E[ X(X-1) ]\), so
\( E[ X(X-1) ]=\displaystyle \sum_X x(x-1) f( x )\)
or \( E[ X(X-1)= \displaystyle \sum_{x=0}^n x(x-1). \begin{pmatrix} n \\ x \end{pmatrix} p^xq^{n-x}\)
or \( E[ X(X-1)]=\displaystyle \sum_{x=2}^n x(x-1)\times \frac{n}{x}; \times\frac{n-1}{x-1} \times \begin{pmatrix} n-2 \\ x-2 \end{pmatrix} p^xq^{n-x}\) [binomial theorem अनुसार \( \begin{pmatrix} n \\ x \end{pmatrix} = \frac{n}{x} \times \frac{n-1}{x-1} \times \begin{pmatrix} n-2 \\ x-2 \end{pmatrix} \) हुने भएकोले]
or \(E[ X(X-1)]=n(n-1)p^2 \displaystyle \sum_{x=2}^n \begin{pmatrix} n-2 \\ x-2 \end{pmatrix} \times p^{x-2}\times q^{( n-2 )-( x-2 )}\)
Let \( y=x-2\), then
when \(x=2\), then \(y=0\)
when \(x=n\), then \(y=n-2\), say \(n-2=m \)
Now, \( E[ X(X-1)] \) of binomial distribution is
\( E[ X(X-1)]=n(n-1)p^2 \displaystyle \sum_{y=0}^m \begin{pmatrix} m \\ y \end{pmatrix} p^yq^{m-y}\)
or \( E[ X(X-1)]=n(n-1)p^2 \times 1 \) [pdf theorem अनुसार \( \displaystyle \sum_{y=0}^m \begin{pmatrix} m \\ y \end{pmatrix} p^yq^{m-y}=1\) हुने भएकोले]
or \(E[ X(X-1)]=n(n-1)p^2\)
Finally, the Variance of X is
\( V(X)=E[X(X-1)]+E[X]-E(X)^2\)
or \( V(X)=n(n-1)p^2+np-(np)^2\)
or \( V(X)=n^2p^2-np^2+np-(np)^2\)
or \( V(X)=-np^2+np\)
or \( V(X)=np(1-p)\)
or \( V(X)=npq\) [definition अनुसार \( 1-p=q\) हुने भएकोले]

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Moment Generating Function of Binomial Distribution: \( M_X(t) ={[ 1+p( e^t-1 ) ]^n}\)

Let X be a random variable having binomial distribution with parameter n and p, then
\( M_X(t) =E[e^{tx}] \)
or \( M_X(t) =\displaystyle \sum_Xe^{tx}.f( x )\)
or \( M_X(t) =\displaystyle \sum_{x=0}^ne^{tx}. \begin{pmatrix} n \\ x \end{pmatrix} p^xq^{n-x}\)
or \( M_X(t) = \displaystyle \sum_{x=0}^n \begin{pmatrix} n \\ x \end{pmatrix}( pe^{t} )^xq^{n-x}\)
or \( M_X(t) =[pe^{t} +q ]^n\) binomial theorem अनुसार \(\displaystyle \sum_{r=0}^n\begin{pmatrix} n \\ r \end{pmatrix}a^rb^{n-r} = (a+b)^n\) हुने भएकोले
or \( M_X(t) =[pe^{t} +1-p]^n\)
or \( M_X(t) =[ 1+p( e^t-1 ) ]^n\)

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Exercise

1. Roll 12 dice simultaneously, and let X denote the number of 6’s that appear. Find the probability of getting seven, eight, or nine 6’s.
2. A four-child family. Each child may be either a boy (B) or a girl (G). For simplicity we suppose that P(B) = IP(G) = 1/2 and that the genders of the children are determined independently. If we let X count the number of B’s, then find E(X) and V(X).
3. Toss a coin three times and let X be the number of Heads observed. Find E(X) and V(X).
4. A recent national study showed that approximately 44.7% of college students have used Wikipedia as a source in at least one of their term papers. Let X equal the number of students in a random sample of size n = 31 who have used Wikipedia as a source. Find the probability that X is equal to 17
5. Suppose there are 10 multiple choice questions in a class quiz. Each question has four possible answers, and only one of them is correct. Find the probability of having five or less correct answers if a student attempts to answer every question at random.
6. Toss a fair coin 100 times, and count the number of heads that appear. Find the mean, variance, and standard deviation of this experiment

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Relationships : Binomial and Normal

Let X is a random variable giving the number of successes in n trials and p is the probability of success. The normal approximation becomes better with increasing n and is exact in the limiting case. In practice, the approximation is very good if both np and nq are greater than 5. The fact that the binomial distribution approaches the normal distribution can be described by writing
\( z=\frac{x-np}{npq}\)
In words, we say that the standardized random variable \(z\) is asymptotically normal.

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Hypergeometric probability function

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Hypergeometric probability distribution is particular case of discrete probability function. It satisfy following assumptions.

1. Experiment must consist with finite population object, denoted by N. For example, 20 students in a class.
2. The \( N \) objects in the population is bi-categorical. The objects in one category is \( M \) and the remaining objects in another category is \( N-M \) .
3. The experiment consist finite trail denoted by n. For example, if we select 5 students from a class of 20 students, then n=5.
4. The probability in each trail remain dependent to previous trail.

A random variable X is said to have hypergeometric distribution if its probability function is
\( f(x)=\frac{\left( \begin{matrix} M \\ x \\ \end{matrix} \right)\left( \begin{matrix} N-M \\ n-x \\ \end{matrix} \right)}{\left( \begin{matrix} N \\ n \\ \end{matrix} \right)} \) for x=0,1,2,...n
where n,N,M are parameters, and x is variable.

Here
Total object: N, on which M are of first kind and N-M are of second kind
Selection : n objects is selected without replacement
Probability : getting x objects out of M objects

Below is a graph of x=0,1,2,...,5 taking N=20, M=15, n=5.

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pdf Theorem : Show that\( \sum f(x)=1\) where \( f(x) \) is hypergeometric probability function

Let X be a random variable having hypergeometric distribution with parameter M, N and n, then total probability is
\( \displaystyle \sum_{x=0}^nf(x)= \displaystyle \sum_{x=0}^n\frac{\left( \begin{matrix} M \\ x \\ \end{matrix} \right)\left( \begin{matrix} N-M \\ n-x \\ \end{matrix} \right)}{\left( \begin{matrix} N \\ n \\ \end{matrix} \right)}\)
According to Binomial theorem,
\( \displaystyle \sum_{r=0}^k\left( \begin{matrix} m \\ r \\ \end{matrix} \right)\left( \begin{matrix} n \\ k-r \\ \end{matrix} \right)=\left( \begin{matrix} m+n \\ k \\ \end{matrix} \right)\)
So we have
\( \displaystyle \sum_{x=0}^nf(x)=\frac{ \displaystyle \sum_{x=0}^n\left( \begin{matrix} M \\ x \\ \end{matrix} \right)\left( \begin{matrix} N-M \\ n-x \\ \end{matrix} \right)}{\left( \begin{matrix} N \\ n \\ \end{matrix} \right)} \)
\( \displaystyle \sum_{x=0}^nf(x)=\frac{\left( \begin{matrix} N \\ n \\ \end{matrix} \right)}{\left( \begin{matrix} N \\ n \\ \end{matrix} \right)}=1 \)

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Mean of Hyper Geometric distribution

Let X be a random variable having hyper geometrical distribution with parameter n, N and M, then
\( E(X)=\displaystyle \sum_X x. f(x)\)
or\( E(X)= \displaystyle \sum_{x=0}^nx.\frac{\left( \begin{matrix} M \\ x \\ \end{matrix} \right)\left( \begin{matrix} N-M \\ n-x \\ \end{matrix} \right)}{\left( \begin{matrix} N \\ n \\ \end{matrix} \right)}\)
or\( E(X)= \displaystyle \sum_{x=1}^n x.\frac{M}{x}\frac{n}{N}\frac{\left( \begin{matrix} M-1 \\ x-1 \\ \end{matrix} \right)\left( \begin{matrix} N-M \\ n-x \\ \end{matrix} \right)}{\left( \begin{matrix} N-1 \\ n-1 \\ \end{matrix} \right)}\)
or\( E(X)=\frac{Mn}{N} \displaystyle \sum_{x=1}^n \frac{\left( \begin{matrix} M-1 \\ x-1 \\ \end{matrix} \right)\left( \begin{matrix} \left( N-1 \right)-\left( M-1 \right) \\ \left( n-1 \right)-\left( x-1 \right) \\ \end{matrix} \right)}{\left( \begin{matrix} N-1 \\ n-1 \\ \end{matrix} \right)}\)
Let \( y=x-1\) Then \( y=0\) for x = 1 and \( y=n-1=m\) (say) for x = n
Now, mean of hyper geometric distribution is
or \( E(X)=\frac{Mn}{N} \displaystyle \sum_{y=0}^m \frac{\left( \begin{matrix} M-1 \\ y \\ \end{matrix} \right)\left( \begin{matrix} \left( N-1 \right)-\left( M-1 \right) \\ m-y \\ \end{matrix} \right)}{\left( \begin{matrix} N-1 \\ m \\ \end{matrix} \right)}\)
or \( E(X)=\frac{Mn}{N} \frac{\left( \begin{matrix} N-1 \\ m \\ \end{matrix} \right)}{\left( \begin{matrix} N-1 \\ m \\ \end{matrix} \right)}\)
or\( E(X)=\frac{Mn}{N}\)

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Variance of Hyper Geometric distribution

Let X be a random variable having hyper geometric distribution with parameter n, N and M, then variance of X is
\( V(X)=E\left[ X\left( X-1 \right) \right]+E(X)-E(X)^2\)
Now,
\( E\left[ X\left( X-1 \right) \right]=\displaystyle \sum_X X\left( X-1 \right). f(x)\)
or\( E\left[ X\left( X-1 \right) \right]= \displaystyle \sum_{x=0}^nx\left( x-1 \right).\frac{\left( \begin{matrix} M \\ x \\ \end{matrix} \right)\left( \begin{matrix} N-M \\ n-x \\ \end{matrix} \right)}{\left( \begin{matrix} N \\ n \\ \end{matrix} \right)}\)
or\( E\left[ X\left( X-1 \right) \right]= \displaystyle \sum_{x=2}^n x\left( x-1 \right).\frac{M}{x}\frac{M-1}{x-1}\frac{n}{N}\frac{n-1}{N-1}\frac{\left( \begin{matrix} M-2 \\ x-2 \\ \end{matrix} \right)\left( \begin{matrix} N-M \\ n-x \\ \end{matrix} \right)}{\left( \begin{matrix} N-2 \\ n-2 \\ \end{matrix} \right)}\)
or\( E\left[ X\left( X-1 \right) \right]=\frac{Mn\left( M-1 \right)\left( n-1 \right)}{N\left( N-1 \right)} \displaystyle \sum_{x=2}^n \frac{\left( \begin{matrix} M-2 \\ x-2 \\ \end{matrix} \right)\left( \begin{matrix} \left( N-2 \right)-\left( M-2 \right) \\ \left( n-2 \right)-\left( x-2 \right) \\ \end{matrix} \right)}{\left( \begin{matrix} N-2 \\ n-2 \\ \end{matrix} \right)}\)
Let \( y=x-2\)
Then \( y=0\) for \( x=2\) and
\( y=n-2=m\) (say) for \( x=n\)
Then
\( E\left[ X\left( X-1 \right) \right]=\frac{Mn\left( M-1 \right)\left( n-1 \right)}{N\left( N-1 \right)} \displaystyle \sum_{y=0}^m \frac{\left( \begin{matrix} M-2 \\ y \\ \end{matrix} \right)\left( \begin{matrix} \left( N-2 \right)-\left( M-2 \right) \\ m-y \\ \end{matrix} \right)}{\left( \begin{matrix} N-2 \\ m \\ \end{matrix} \right)}\)
or \( E\left[ X\left( X-1 \right) \right]=\frac{Mn\left( M-1 \right)\left( n-1 \right)}{N\left( N-1 \right)} \frac{\left( \begin{matrix} N-2 \\ m \\ \end{matrix} \right)}{\left( \begin{matrix} N-2 \\ m \\ \end{matrix} \right)}\)
or \( E\left[ X\left( X-1 \right) \right]=\frac{Mn\left( M-1 \right)\left( n-1 \right)}{N\left( N-1 \right)}\)
Therefore, variance of the distribution is
\( V(X)=E\left[ X\left( X-1 \right) \right]+E(X)-E(X)^2\)
or \( V(X)=\frac{Mn\left( M-1 \right)\left( n-1 \right)}{N\left( N-1 \right)}+\frac{Mn}{N}-{{\left( \frac{Mn}{N} \right)}^{2}}\)
or \( V(X)=\frac{Mn\left( N-M \right)\left( N-n \right)}{N^2\left( N-1 \right)}\)

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Exercise

1. Suppose in a certain production of 250 objects, there are 17 defective production. A quality control consultant randomly collects 5 production for inspection to determine whether or not they are defective. Let X denote the number of defectives in the sample. Find the probability of exactly 3 defectives in the sample, that is, find P(X = 3)

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Poisson probability function

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Poisson probability is particular case of discrete probability distribution. It is also called special case of binomial probability distribution. In binomial probability distribution, if \(p\) is very small, and \(n\) is relatively large, then, Poisson probability distribution can be used. The Poisson probability satisfy following assumptions.

1. The number of experiment \( (n) \) is large
2. The experiment results in outcomes that can be classified as successes or failures.
3. The number of successes in two disjoint time intervals is independent.
4. The probability of success (\(p\) ) that will occur in an extremely small region is virtually zero.
5. The average number of success \( \lambda = n p) \) in a specified region is constant

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Some example where Poisson probability can occur

• number of deaths by TB
• number of bike accidents
• number of suicide
• number of telephone calls
• number of defective production
• number of birth defects
• number of typing errors on a page
• customers arriving in a bank

A random variable X is said to have Poisson probability distribution if its probability function is given by
\( f(x)=\frac{ \lambda^x e^{-\lambda }}{x!}\)
where \(x=0,1,2,...\) number of success, \(e=2.72\) constant and \(\lambda =np\) average successes in an interval or region or space
\( \lambda \) is the shape parameter which indicates the average number of events in the given time interval.

Examples

1. The average number of accidents is 2 per day. What is the probability that exactly 3 accidents will be happen tomorrow?
   Solution
   This is a Poisson experiment in which
   \(\lambda = 2\)
   \(x = 3\); since we want to find the likelihood that 3 accidents will be happen
   e = 2.71828; since e is a constant equal to approximately 2.71828.
   Hence
   \(P(x; \lambda) = P(3; 2) = 0.180\)
   Thus, the probability of happening 3 accidents is 0.180 .
2. Suppose the average number of lions seen on a 1-day Chitwan safari is 5. What is the probability that tourists will see fewer than four lions on the next 1-day Chitwan safari?
   Solution
   This is a Poisson experiment in which
   \( \lambda = 5\); since 5 lions are seen per safari, on average.
   \(x = 0, 1, 2, 3\) since we want to find the likelihood that tourists will see fewer than 4 lions
   e = 2.71828; since e is a constant equal to approximately 2.71828.
   Thus, we need to calculate the sum of four probabilities
   \(P(0; 5) + P(1; 5) + P(2; 5) + P(3; 5)\).
   To compute this sum, we use the Poisson formula:
   \( P(x \le 3, 5) = 0.2650\)
   Thus, the probability of seeing at no more than 3 lions is 0.2650.

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pdf Theorem

Show that \( \sum f(x)=1 \) where f(x) is Poisson probability function
Let X be a random variable having Poisson distribution with parameter \( \lambda \) , then total probability within the range of X is
\( \displaystyle \sum_{x=0}^{\infty} p(x; \lambda )= \sum_{x=0}^{\infty} \frac{{ \lambda^x}{{e}^{-\lambda }}}{x!}\)
or \( \displaystyle \sum_{x=0}^{\infty} p (x; \lambda )={{e}^{-\lambda }} \displaystyle \sum_{x=0}^{\infty} \frac{{ \lambda^x}}{x!}\)
or \( \displaystyle \sum_{x=0}^{\infty} p(x; \lambda )={{e}^{-\lambda }}\left( 1+\frac \lambda{1!}+\frac{{ \lambda^{2}}}{2!}+\frac{{ \lambda^{3}}}{3!}+\ldots +\ldots \right)\)
or \( \displaystyle \sum_{x=0}^{\infty} p(x; \lambda )={{e}^{-\lambda }}{{e}^ \lambda}\)
or \( \displaystyle \sum_{x=0}^{\infty} p(x; \lambda )=1\)

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Mean of Poisson Distribution: \( E( X )=\lambda\)

Let X be a random variable having poisson distribution with parameter \(\lambda \), then
\( E( X )=\displaystyle \sum_X x f( x )\)
or \( E( X )= \displaystyle \sum_{x=0}^\infty x. \frac{ \lambda^x e^{-\lambda }}{x!}\)
or \( E( X )=\displaystyle \sum_{x=1}^\infty x\times\frac{ \lambda \lambda^{x-1} e^{-\lambda }}{ x (x-1)!}\)
or \( E( X )=\lambda \displaystyle \sum_{x=1}^\infty \frac{ \lambda^{x-1} e^{-\lambda }}{ (x-1)!}\)
Let \( y=x-1\), then
\( y=0\) for \( x=1\) and
\( y=n-1=m\) (say) for \( x=n\)
Now, mean of Poisson distribution is
\( E( X )=\lambda \displaystyle \sum_{y=0}^\infty \frac{ \lambda^y e^{-\lambda }}{y!}\)
or \( E( X )=\lambda \times 1 \) [pdf theorem अनुसार \( \displaystyle \sum_{y=0}^\infty \frac{ \lambda^y e^{-\lambda }}{y!}=1\) हुने भएकोले]
or \( E( X )=\lambda\)

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Variance of Poisson Distribution: \( V( X )=\lambda\)

Let X be a random variable having Poisson distribution with parameter \(\lambda\), then
Variance of X is
\( V(X)=E(X^2)-E(X)^2\)
We know that
\(E(X^2)=E[X(X-1)+X]\)
or \(E(X^2)=E[X(X-1)]+E[X]\)
Thus, Variance of X is
\( V(X)=E(X^2)-E(X)^2\)
or \( V(X)=E[X(X-1)]+E[X]-E(X)^2\)
Now, we compute \( E[ X(X-1) ]\), so
\( E[ X(X-1)]= \displaystyle \sum_{x=0}^\infty x(x-1). \frac{ \lambda^x e^{-\lambda }}{x!}\)
or \( E[ X(X-1)]=\displaystyle \sum_{x=2}^\infty x(x-1) \times\frac{ \lambda^2 \lambda^{x-2} e^{-\lambda }}{ x(x-1) (x-2)!}\)
or \( E[ X(X-1)]=\lambda^2 \displaystyle \sum_{x=2}^\infty \frac{ \lambda^{x-2} e^{-\lambda }}{ (x-2)!}\)
Let \( y=x-2\), then
\( y=0\) for \( x=2\) and
\( y=n-2=m\) (say) for \( x=n\)
Now, \( E[ X(X-1)]\) of Poisson distribution is
\( E[ X(X-1)]=\lambda^2 \displaystyle \sum_{y=0}^\infty \frac{ \lambda^y e^{-\lambda }}{y!}\)
or \( E[ X(X-1)]=\lambda^2 \times 1 \) [pdf theorem अनुसार \( \displaystyle \sum_{y=0}^\infty \frac{ \lambda^y e^{-\lambda }}{y!}=1\) हुने भएकोले]
or \( E[ X(X-1)]=\lambda^2\)
Finally, the Variance of X is
\( V(X)=E[X(X-1)]+E[X]-E(X)^2\)
or \( V(X)=\lambda^2+\lambda-\lambda^2\)
or \( V(X)=\lambda\)

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Moment Generating Function of Poisson Distribution: \( M_X(t) =e^{\lambda (e^t-1) }\)

Let X be a random variable having Poisson distribution with parameter \( \lambda \), then
\( M_X(t) =E[e^{tx}] \)
or \( M_X(t) =\displaystyle \sum_X e^{tx}.f( x )\)
or \( M_X(t) =\displaystyle \sum_{x=0}^\infty e^{tx}.\frac{ \lambda^x e^{-\lambda }}{x!} \)
or \( M_X(t) = \displaystyle \sum_{x=0}^\infty \frac{ (\lambda e^t)^x e^{-\lambda }}{x!} \)
or \( M_X(t) = e^{-\lambda } \displaystyle \sum_{x=0}^\infty \frac{ (\lambda e^t)^x }{x!} \)
or \( M_X(t) = e^{-\lambda } e^{\lambda e^t } \)
or \( M_X(t) = e^{\lambda (e^t-1) } \)

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Exercise

1. On the average, five cars arrive at a particular car wash every hour. Let X count the number of cars that arrive from 10AM to 11AM. What is the probability that no car arrives during this period?
2. If there are twelve bike crossing a bridge per minute on average, find the probability of having seventeen or more bikes crossing the bridge in a particular minute.
3. If the probability that an individual will suffer a bad reaction from injection of a given serum is 0.001, determine the probability that out of 2000 individuals, (a) exactly 3, (b) more than 2, individuals will suffer a bad reaction
4. The average number of accidents in a city every year is 18. Calculate the probability that there are exactly 2 accidents this year.

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Relationships : Binomial and Poisson

In the binomial distribution, if n is large while the probability p of occurrence of an event is close to zero, so that q = 1 – p is close to one, the event is called a rare event. In practice, we shall consider an event as rare if the number of trials is at least 50 (n ≥ 50) while np is less than 5. For such cases, the binomial distribution is very closely approximated by the Poisson distribution with λ = np.

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