91Ó°ÊÓ

A discrete random variable is one that can only take on a finite number of different values.

\({\rm{b(x;n,p) = }}\left\{ {\begin{array}{*{20}{l}}{\left( {\begin{array}{*{20}{l}}{\rm{n}}\\{\rm{x}}\end{array}} \right){{\rm{p}}^{\rm{x}}}{{{\rm{(1 - p)}}}^{{\rm{n - x}}}}}&{{\rm{,x = 0,1,2, \ldots ,n}}}\\{\rm{0}}&{{\rm{, otherwise }}}\end{array}} \right.\)

Intuitively, the probabilities provided should be the same because you can select S and F at random (you can choose what a success is and what a failure is). The events that occurred

\({\rm{\{ x successes when P(S) = 1 - p\} }}\)

Which appears on the left side, is the same as the event

\({\rm{\{ n - x failures when P(F) = p\} }}\)

On the right hand side However, for \({\rm{x}} \in {\rm{R}}\), the formal evidence is as follows:

\right){{\rm{p}}^{{\rm{n - x}}}}{{\rm{(p - 1)}}^{\rm{x}}}\\\mathop {\rm{ = }}\limits^{{\rm{(3)}}} {\rm{b(n - x;n,p)}}\end{array}\)

(1): From the above-mentioned theorem;

(2): It is then proven that:

\(\left( {\begin{array}{*{20}{l}}{\rm{n}}\\{\rm{x}}\end{array}} \right){\rm{ = }}\left( {\begin{array}{*{20}{c}}{\rm{n}}\\{{\rm{n - x}}}\end{array}} \right)\)

(3): from the above-mentioned theorem.

Therefore, it is proven:: \(\left( {\begin{array}{*{20}{l}}{\rm{n}}\\{\rm{x}}\end{array}} \right){\rm{ = }}\left( {\begin{array}{*{20}{c}}{\rm{n}}\\{{\rm{n - x}}}\end{array}} \right)\).

The cdf of a binomial random variable X with parameters n and p is the Cumulative Density Function.

\(\begin{array}{c}{\rm{B(x;n,p) = P(X}} \le {\rm{x)}}\\{\rm{ = }}\sum\limits_{{\rm{y = 0}}}^{\rm{x}} {\rm{b}} {\rm{(y;n,p),}}\quad \\{\rm{x = 0,1, \ldots ,n}}\end{array}\)

Where \({\rm{P(X}} \le {\rm{x)}}\) denotes a maximum of x successes. The following is also true:

\({\rm{B(x;n,1 - p) = }}\sum\limits_{{\rm{y = 0}}}^{\rm{x}} {\rm{b}} {\rm{(y;n,p - 1)}}\)

Since the complement of event \({\rm{\{ X}} \le {\rm{x\} }}\) is {X > x}

\(\sum\limits_{{\rm{y = 0}}}^{\rm{n}} {\rm{b}} {\rm{(y;n,p) = 1}}\)

Then, we have that:

\(\begin{array}{c}{\rm{P(X > x) = 1 - }}\sum\limits_{{\rm{y = 0}}}^{\rm{x}} {\rm{b}} {\rm{(y;n,p - 1)}}\\{\rm{ = }}\sum\limits_{{\rm{y = x + 1}}}^{\rm{n}} {\rm{b}} {\rm{(y;n,p - 1)}}\end{array}\)

Therefore, it is proven: \(\begin{array}{c}{\rm{P(X > x) = 1 - }}\sum\limits_{{\rm{y = 0}}}^{\rm{x}} {\rm{b}} {\rm{(y;n,p - 1)}}\\{\rm{ = }}\sum\limits_{{\rm{y = x + 1}}}^{\rm{n}} {\rm{b}} {\rm{(y;n,p - 1)}}\end{array}\).

If we utilise (a), we get the following for \({\rm{y}} \in {\rm{R}}\):

\({\rm{b(y;n,p - 1) = b(n - y;n,p)}}\)

So there you have it.

\({\rm{P(X > x) = }}\sum\limits_{{\rm{y = x + 1}}}^{\rm{n}} {\rm{b}} {\rm{(n - y;n,p)}}\)

Alternatively, if we write sum as:

\(\begin{array}{c}{\rm{P(X > x) = b(n - x - 1;n,p) + b(n - x - 2;n,p) + b(n - x - 3;n,p) + \ldots + b(1;n,p) + b(0;n,p)}}\\\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{b(0;n,p) + b(1;n,p) + \ldots + b(n - x - 1;n,p)}}\\{\rm{ = }}\sum\limits_{{\rm{y - x - 1}}}^{\rm{n}} {\rm{b}} {\rm{(y;n,p)}}\\\mathop {\rm{ = }}\limits^{{\rm{(2)}}} {\rm{B(n - x - 1;n,p}}\end{array}\)

(1) Rearrange the numbers in the total;

(2): see the cdf provided before.

Finally, the following statement is correct:

\(\begin{array}{c}{\rm{B(x;n,1 - p) = P(X}} \le {\rm{x)}}\\{\rm{ = 1 - P(X > x)}}\\{\rm{ = 1 - B(n - x - 1;n,p)}}\end{array}\)

Values larger than \({\rm{0}}{\rm{.5}}\) do not need to be included in Appendix Table A.\({\rm{1}}\). As can be seen, we can always utilise the findings from (a) and (b) to change the probability to be less than or equal to \({\rm{0}}{\rm{.5}}\) in order to get the desired result. So, if p > \({\rm{0}}{\rm{.5}}\), \({\rm{1 - }}\)p is less than \({\rm{0}}{\rm{.5}}\), and we may utilise it.

Therefore, it is not necessary.