91Ó°ÊÓ

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

There are several ways to prove E(X) = np, but possibly the simplest is to use an indicator random variable. We won't utilise it, but I invite you to locate the proof and examine it.

We may deduce the following from the exercise's hint:

\(\begin{array}{c}{\rm{E(X) }}\mathop {\rm{ = }}\limits^{{\rm{(1)}}} \sum\limits_{{\rm{x = 0}}}^{\rm{n}} {\rm{x}} {\rm{ \times }}\left( {\begin{array}{*{20}{l}}{\rm{n}}\\{\rm{x}}\end{array}} \right){{\rm{p}}^{\rm{x}}}{{\rm{(1 - p)}}^{{\rm{n - x}}}}\\\mathop {\rm{ = }}\limits^{{\rm{(2)}}} \sum\limits_{{\rm{x = 1}}}^{\rm{n}} {\rm{x}} {\rm{ \times }}\left( {\begin{array}{*{20}{l}}{\rm{n}}\\{\rm{x}}\end{array}} \right){{\rm{p}}^{\rm{x}}}{{\rm{(1 - p)}}^{{\rm{n - x}}}}\\\mathop {\rm{ = }}\limits^{{\rm{(3)}}} \sum\limits_{{\rm{x = 1}}}^{\rm{n}} {\rm{x}} {\rm{ \times }}\frac{{{\rm{n!}}}}{{{\rm{x!(n - x)!}}}}{{\rm{p}}^{\rm{x}}}{{\rm{(1 - p)}}^{{\rm{n - x}}}}\\{\rm{ = }}\sum\limits_{{\rm{x = 1}}}^{\rm{n}} {\frac{{{\rm{n!}}}}{{{\rm{(x - 1)!(n - x)!}}}}} {{\rm{p}}^{\rm{x}}}{{\rm{(1 - p)}}^{{\rm{n - x}}}}\\\mathop {\rm{ = }}\limits^{{\rm{(4)}}} {\rm{np}}\sum\limits_{{\rm{x = 1}}}^{\rm{n}} {\frac{{{\rm{(n - 1)!}}}}{{{\rm{(x - 1)!(n - x)!}}}}} {{\rm{p}}^{{\rm{x - 1}}}}{{\rm{(1 - p)}}^{{\rm{n - x}}}}\\\mathop {\rm{ = }}\limits^{{\rm{(5)}}} {\rm{np}}\sum\limits_{{\rm{y = 0}}}^{{\rm{n - 1}}} {\frac{{{\rm{(n - 1)!}}}}{{{\rm{y!(n - y - 1)!}}}}} {{\rm{p}}^{\rm{y}}}{{\rm{(1 - p)}}^{{\rm{n - y - 1}}}}\\\mathop {\rm{ = }}\limits^{{\rm{(6)}}} {\rm{np}}\left( {\sum\limits_{{\rm{y = 0}}}^{{\rm{n - 1}}} {\left( {\begin{array}{*{20}{c}}{{\rm{n - 1}}}\\{\rm{y}}\end{array}} \right)} {{\rm{p}}^{\rm{y}}}{{{\rm{(1 - p)}}}^{{\rm{n - y - 1}}}}} \right)\\\mathop {\rm{ = }}\limits^{{\rm{(7)}}} {\rm{np \times 1}}\\{\rm{ = np;}}\end{array}\)

(1): as defined in the following definition of anticipated value;

(2): because the product is zero for x = 0, we may eliminate it;

(3): we take use of the fact that

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

(4): we pick n from n!=n(\({\rm{n - 1}}\))and x from p = p.\({\rm{p - 1}}\);

(5): y = \({\rm{x - 1}}\) implies x = \({\rm{y + 1}}\); just replace;

(6): we take use of the fact that:

\(\left( {\begin{array}{*{20}{c}}{{\rm{n - 1}}}\\{\rm{y}}\end{array}} \right){\rm{ = }}\frac{{{\rm{(n - 1)!}}}}{{{\rm{y!(n - y - 1)!}}}}\)

(7): Because the values are from a binomial distribution based on \({\rm{n - 1}}\) trials, the total reflects the sum over all y = 0, \({\rm{1}}\),...,\({\rm{n - 1}}\), which must be 1, we may state that

\({\rm{Y}} \sim {\rm{Bin(n - 1,p)}}\)

As a result, the total must be \({\rm{1}}\) (for any discrete distribution, the sum of all potential distribution pmf values must equal \({\rm{1}}\)).

A discrete random variable X with a set of potential values S and pmf p(x) has an Expected Value (mean value).

\(\begin{array}{c}{\rm{E(X) = }}{{\rm{\mu }}_{\rm{X}}}\\{\rm{ = }}\sum\limits_{{\rm{x}} \in {\rm{S}}} {{\rm{x \bullet p(x)}}} \end{array}\)

Therefore, it is proved that: E(X) = np.