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Chapter 4: Q19E (page 155)

Let \({\rm{X}}\) be a continuous \({\rm{rv}}\) with cdf

\({\rm{F(x) = }}\left\{ {\begin{array}{*{20}{c}}{\rm{0}}&{{\rm{x}} \le {\rm{0}}}\\{\frac{{\rm{x}}}{{\rm{4}}}\left( {{\rm{1 + ln}}\left( {\frac{{\rm{4}}}{{\rm{x}}}} \right)} \right)}&{{\rm{0 < x}} \le {\rm{4}}}\\{\rm{1}}&{{\rm{x > 4}}}\end{array}} \right.\)

(This type of cdf is suggested in the article 鈥淰ariability in Measured Bedload Transport Rates鈥� (Water 91影视 Bull., \({\rm{1985:39 - 48}}\)) as a model for a certain hydrologic variable.) What is a. \({\rm{P(X}} \le {\rm{1)}}\)? b. \({\rm{P(1}} \le {\rm{X}} \le {\rm{3)}}\)? c. The pdf of \({\rm{X}}\)?

Short Answer

Expert verified

(a) The value is \({\rm{0}}{\rm{.5966}}\).

(b) The value is\({\rm{0}}{\rm{.3692}}\).

(c) The function is \({\rm{f(x) = }}\left\{ {\begin{array}{*{20}{l}}{\frac{{\rm{1}}}{{\rm{4}}}\left( {{\rm{ln}}\frac{{\rm{4}}}{{\rm{x}}}} \right)}&{{\rm{0 < x}} \le {\rm{4}}}\\{\rm{0}}&{{\rm{ otherwise }}}\end{array}} \right.\).

Step by step solution

01

Define variable

An unknown number, unknown value, or unknown quantity is represented by a variable, which is an alphabet or word. In the context of algebraic expressions or algebra, the variables are particularly useful.

02

Explanation

(a) The format of cdf is as follows:

\({\rm{F(x) = }}\left\{ {\begin{array}{*{20}{l}}{\rm{0}}&{{\rm{x < 0}}}\\{\frac{{\rm{x}}}{{\rm{4}}}\left( {{\rm{1 + ln}}\frac{{\rm{4}}}{{\rm{x}}}} \right){\rm{0 < x}} \le {\rm{4}}}&{}\\{\rm{1}}&{{\rm{x > 4}}}\end{array}} \right.\)

We can write the following using the proposition and the given cdf:

\(\begin{aligned}{\rm{P(X}} \le 1) &= F(1)\\ &= \frac{{\rm{1}}}{{\rm{4}}}\left( {{\rm{1 + ln}}\frac{{\rm{4}}}{{\rm{1}}}} \right)\\&= \frac{{\rm{1}}}{{\rm{4}}}{\rm{(1 + 1}}{\rm{.3863)}}\\&= \frac{{{\rm{2}}{\rm{.3863}}}}{{\rm{4}}}\\{\rm{P(X}} \le 1) &= 0{\rm{.5966}}\end{aligned}\)

Let X be a continuous\({\rm{rv}}\)with pdf\({\rm{f(x)}}\)and cdf\({\rm{F}}\)as parameters\({\rm{(x)}}\). After that, for any number a,

\({\rm{P(X}} \le {\rm{a) = F(a)}}\)

Therefore, the value is \({\rm{0}}{\rm{.5966}}\).

03

Explanation

(b) Using the following proposition and the provided cdf, we can write:

\(\begin{aligned}{\rm{P(1}} \le {\rm{X}} \le 3) &= F(3) - F(1)\\ & = \frac{{\rm{3}}}{{\rm{4}}}\left( {{\rm{1 + ln}}\frac{{\rm{4}}}{{\rm{3}}}} \right){\rm{ - }}\frac{{\rm{1}}}{{\rm{4}}}\left( {{\rm{1 + ln}}\frac{{\rm{4}}}{{\rm{1}}}} \right)\\& = \frac{{\rm{3}}}{{\rm{4}}}{\rm{(1 + 0}}{\rm{.2877) - }}\frac{{\rm{1}}}{{\rm{4}}}{\rm{(1 + 1}}{\rm{.3863)}}\\ & = \frac{{{\rm{3}}{\rm{.8631}}}}{{\rm{4}}}{\rm{ - }}\frac{{{\rm{2}}{\rm{.3863}}}}{{\rm{4}}}\\ & = 0 {\rm{.9658 - 0}}{\rm{.5966}}\\{\rm{P(1}} \le {\rm{X}} \le 3) &= 0 {\rm{.3692}}\end{aligned}\)

Let X be a continuous\({\rm{rv}}\)with pdf\({\rm{f(x)}}\)and cdf\({\rm{F}}\)as a proposition\({\rm{(x)}}\). Then, using\({\rm{a < b}}\), for any two numbers a and b,

\({\rm{P(a}} \le {\rm{X}} \le {\rm{b) = F(b) - F(a)}}\)

Therefore, the value is \({\rm{0}}{\rm{.3692}}\).

04

Explanation

(c) The proposition presented below is used to determine \({\rm{f(x)}}\):

Proposition: If\({\rm{X}}\)is a continuous\({\rm{rv}}\)with pdf\({\rm{f(x)}}\)and cdf\({\rm{F(x)}}\), the derivative\({{\rm{F}}^{\rm{'}}}{\rm{(x)}}\)exists at every x.

\({\rm{f(x) = F'(x)}}\)

In the intervals\({\rm{x}} \le {\rm{0}}\)and\({\rm{x > 4}}\),\({\rm{F(x)}}\)is constant, hence\({\rm{f(x) = 0}}\)in these intervals.

For any x in the\({\rm{0 < x}} \le {\rm{4}}\)interval:

\(\begin{aligned}f(x) &= {{\rm{F}}^{\rm{'}}}{\rm{(X)}}\\ &= \frac{{\rm{1}}}{{\rm{4}}}\left( {{\rm{1 + ln}}\frac{{\rm{4}}}{{\rm{x}}}} \right){\rm{ + }}\frac{{\rm{x}}}{{\rm{4}}}\left( {\frac{{\rm{x}}}{{\rm{4}}}} \right){\rm{ \times }}\left( {\frac{{{\rm{ - 4}}}}{{{{\rm{x}}^{\rm{2}}}}}} \right)\\ &= \frac{{\rm{1}}}{{\rm{4}}}\left( {{\rm{1 + ln}}\frac{{\rm{4}}}{{\rm{x}}}} \right){\rm{ - }}\frac{{\rm{1}}}{{\rm{4}}}\\f(x) &= \frac{{\rm{1}}}{{\rm{4}}}\left( {{\rm{ln}}\frac{{\rm{4}}}{{\rm{x}}}} \right)\end{aligned}\)

Finally,\({\rm{f(x)}}\)can be written as:

\({\rm{f(x) = }}\left\{ {\begin{array}{*{20}{l}}{\frac{{\rm{1}}}{{\rm{4}}}\left( {{\rm{ln}}\frac{{\rm{4}}}{{\rm{x}}}} \right)}&{{\rm{0 < x}} \le {\rm{4}}}\\{\rm{0}}&{{\rm{ otherwise }}}\end{array}} \right.\)

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