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Chapter 3: Q116E (page 139)

The mode of a discrete random variable \({\rm{X}}\) with \({\rm{pmf}}\) \({\rm{p(x)}}\) is that value \({{\rm{x}}^{\rm{*}}}\) for which \({\rm{p(x)}}\) is largest (the most probable \({\rm{x}}\) value).

a. Let \({\rm{X}} \sim {\rm{Bin(n,p)}}\). By considering the ratio \({\rm{b(x + 1;n,p)/b(x;n,p)}}\), show that \({\rm{b(x;n,p)}}\) increases with x as long as \({\rm{x < np - (1 - p)}}\). Conclude that the mode \({{\rm{x}}^{\rm{*}}}\) is the integer satisfying \({\rm{1}} \le {{\rm{x}}^{\rm{*}}} \le {\rm{(n + 1)p}}\).

b. Show that if \({\rm{X}}\) has a Poisson distribution with parameter \({\rm{\mu }}\), the mode is the largest integer less than \({\rm{\mu }}\). If \({\rm{\mu }}\) is an integer, show that both \({\rm{\mu - 1}}\) and \({\rm{\mu }}\) are modes.

Short Answer

Expert verified

(a) The mode\({{\rm{x}}^*}\)is the integer that satisfies\({\rm{1}} \le {{\rm{x}}^{\rm{*}}} \le {\rm{(n + 1)p}}\).

(b) It is proved that if \(X\) has a Poisson distribution with parameter \({\rm{\mu }}\), the mode is the largest integer less than \({\rm{\mu }}\) and if \({\rm{\mu }}\) is an integer, then both \({\rm{\mu - 1}}\) and \({\rm{\mu }}\) are modes.

Step by step solution

01

Concept Introduction

A Poisson distribution is a probability distribution used in statistics to show how many times an event is expected to happen over a certain amount of time. To put it another way, it's a count distribution.

02

Mode \({{\rm{x}}^{\rm{*}}}\) is satisfying \({\rm{1}} \le {{\rm{x}}^{\rm{*}}} \le {\rm{(n + 1)p}}\)

(a)

The theorem is 鈥�

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

Look at the ratio 鈥�

\(\begin{aligned}\frac{{{\rm{b(x + 1;n,p)}}}}{{{\rm{b(x;n,p)}}}} &= \frac{{\left( {\begin{array}{*{20}{c}}{\rm{n}}\\{{\rm{x + 1}}}\end{array}} \right){{\rm{p}}^{{\rm{x + 1}}}}{{{\rm{(1 - p)}}}^{{\rm{n - x - 1}}}}}}{{\left( {\begin{array}{*{20}{c}}{\rm{n}}\\{\rm{x}}\end{array}} \right){{\rm{p}}^{\rm{x}}}{{{\rm{(1 - p)}}}^{{\rm{n - x}}}}}}\\ &= \frac{{{\rm{(n - x)}}}}{{{\rm{(x + 1)}}}} \cdot \frac{{\rm{p}}}{{{\rm{(1 - p)}}}}\\ &= \frac{{{\rm{np - px}}}}{{{\rm{x - px - p + 1}}}}\end{aligned}\)

From\({\rm{np - (1 - p) > x}}\)it is obtained that\({\rm{np > x + (1 - p)}}\).

Also,\(\frac{{{\rm{b(x + 1;n,p)}}}}{{{\rm{b(x;n,p)}}}}{\rm{ > 1}}\)if and only if\(\frac{{{\rm{np - px}}}}{{{\rm{x - px - p + 1}}}}{\rm{ > 1}}\)which is equal to 鈥�

Since this holds by assumption, we have that the \(pmf\) increases with \({\rm{x}}\) as \({\rm{x < np - (1 - p)}}\).

Therefore, the mode \({{\rm{x}}^*}\) is the integer between given values.

03

Step 3: \({\rm{X}}\) has a Poisson Distribution

(b)

A random variable\(X\)with\(pmf\)鈥�

\({\rm{p(x;\mu ) = }}{{\rm{e}}^{{\rm{ - \mu }}}}\frac{{{{\rm{\mu }}^{\rm{x}}}}}{{{\rm{x!}}}}\)

For\({\rm{x = 0,1,}}..{\rm{,}}\)is said to have Poisson Distribution with parameter\({\rm{\mu > 0}}\).

First, prove that the \(pmf\) increases as \({\rm{x < \mu - 1}}\). Look at the ratio 鈥�

\(\frac{{{\rm{p(x + 1;\mu )}}}}{{{\rm{p(x;\mu )}}}}{\rm{ = }}\frac{{{{\rm{e}}^{{\rm{ - \mu }}}}\frac{{{{\rm{\mu }}^{{\rm{x + 1}}}}}}{{{\rm{(x + 1)!}}}}}}{{{{\rm{e}}^{{\rm{ - \mu }}}}\frac{{{{\rm{\mu }}^{\rm{x}}}}}{{{\rm{x!}}}}}}{\rm{ = }}\frac{{\rm{\mu }}}{{{\rm{(x + 1)}}}}\)

Which is biggen than \({\rm{1}}\) when 鈥�

\(\begin{aligned}{\rm{\mu > x + 1}}\\{\rm{x < \mu - 1}}\end{aligned}\)

So, the \(pmf\) increases \({\rm{x < \mu - 1}}\)from which it is concluded that the mode is the largest integer less than \({\rm{\mu }}\).

Assume now that the parameter \({\rm{\mu }}\) is an integer. From the first part we have that \({\rm{\mu - 1}}\)is the mode. However, it is also obtained that 鈥�

\({\rm{p(\mu ;\mu ) = p(\mu - 1;\mu )}}\)

Therefore, \({\rm{\mu }}\) and \({\rm{\mu - 1}}\) are modes.

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