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The article "Error Distribution in Navigation"\({\rm{(J}}\). of the Institute of Navigation, \({\rm{1971: 429 442}}\)) suggests that the frequency distribution of positive errors (magnitudes of errors) is well approximated by an exponential distribution. Let \({\rm{X = }}\) the lateral position error (nautical miles), which can be either negative or positive. Suppose the pdf of \({\rm{X}}\) is

\(f(x) = (.1){e^{ - .2k1}} - 楼 < x < 楼\)

a. Sketch a graph of \({\rm{f(x)}}\) and verify that \({\rm{f(x)}}\) is a legitimate pdf (show that it integrates to\({\rm{1}}\)).

b. Obtain the cdf of \({\rm{X}}\) and sketch it.

c. Compute \(P(X拢0),P(X拢2),P( - 1拢X拢2)\), and the probability that an error of more than \({\rm{2}}\) miles is made.

Short Answer

Expert verified

(a) \({\rm{f(x)}}\)is a legitimate pdf

(b) \(F(x) = \left\{ {\begin{array}{*{20}{l}}{(0.5){e^{0.2x}}}&{x < 0} \\{1 - (0.5) \times {e^{ - 0.2x}}}&{x0}\end{array}}\right.\)

(c) \(P(x拢0) = 0.5\)

\(P(x拢2) = 0.6648\)

\(P( - 1拢x拢2) = 0.2555\)

\(P(x <- 2{\text{ }}or{\text{ }}x > 2) = 0.6704\)

Step by step solution

01

Definition

Probability simply refers to the likelihood of something occurring. We may talk about the probabilities of particular outcomes鈥攈ow likely they are鈥攚hen we're unclear about the result of an event. Statistics is the study of occurrences guided by probability.

02

Sketch a graph of \({\rm{f(x)}}\) and verify that \({\rm{f(x)}}\) is a legitimate pdf

(a) It is given that \({\rm{X}}\) is the rv denoting the lateral position error (nautical miles), which can be either negative or positive. The pdf \({\rm{f(x)}}\)is given to us as:

\({\rm{f(x) = (0}}{\rm{.1)}}{{\rm{e}}^{{\rm{ - 0}}{\rm{.2|x|}}}}\)

If we remove modulus sign then it can also be written as:

\(f(x) = \left\{ {\begin{aligned}{}{(0.1){e^{0.2x}}}&{x < 0} \\{(0.1){e^{ - 0.2x}}}&{x0}\end{aligned}} \right.\)

It will be a valid pdf if it integrates to\({\rm{\;1}}\). Upon integrating pdf\({\rm{f(x)}}\), we obtain:

\(\begin{aligned}\int_{ - 楼}^楼 f (x) \times dx &= \int_{ - 楼}^0 {(0.1)} {e^{0.2x}} \times dx + \int_0^楼 {(0.1)} {e^{ - 0.2x}} \times dx \\&= (0.1)\left[ {\frac{{{e^{0.2x}}}}{{0.2}}} \right]_{ - 楼}^0 + (0.1)\left[ {\frac{{ - {e^{ - 0.2x}}}}{{0.2}}} \right]_0^楼 \\&= (0.1)\left[ {\frac{1}{{0.2}} - 0} \right] + (0.1)\left[ {0 - \left( {\frac{{ - 1}}{{0.2}}} \right)} \right]_0^楼\\&= \frac{{0.1}}{{0.2}} + \frac{{0.1}}{{0.2}}\int_{ - 楼}^楼 f (x) \times dx \\&= 1 \\\end{aligned}\)

Hence \({\rm{f(x)}}\) is a legitimate pdf.

03

Step 3: The graph

It is plotted in the figure given below.

04

The cdf of \({\rm{X}}\) and sketch it

(b) We recall the definition of cdf of a continuous variable.

Definition: The cumulative distribution function \({\rm{F(x)}}\) for a continuous rv \({\rm{X}}\) is defined for every number \({\rm{x}}\) by

\(F(x) = P(X拢x) = \int_{ - 楼}^x f (y) \times dy\)

pdf \({\rm{f(x)}}\) is given to us as:

\(f(x) = \left\{ {\begin{aligned}{{}{}}{(0.1){e^{0.2x}}}&{x < 0} \\{(0.1){e^{ - 0.2x}}}&{x30}\end{aligned}} \right.\)

For any number \({\rm{x}}\) in the interval \(( - 楼,0)\)

\(\begin{aligned}F(x) &= \int_{ - 楼}^x f (y)\times dy= \int_{ - 楼}^x {(0.1)} {e^{0.2y}} \times dy \\&= (0.1)\left[ {\frac{{{e^{0.2y}}}}{{0.2}}} \right]_{ - 楼}^x \\&= (0.1)\left[ {\frac{{{e^{0.2x}}}}{{0.2}} - 0} \right] \\F(x)\left. {= (0.5){e^{0.2x}}} \right]\\\end{aligned} \)

For any number \({\rm{x}}\) in the interval \((0,\infty )\)

05

Calculation for cdf of \({\rm{X}}\)

\(\begin{aligned}F(x) &= \int_{ - 楼}^x f (y) \times dy \\&= \int_{ - 楼}^0 {(0.1)} {e^{0.2y}} \times dy + \int_0^x {(0.1)} {e^{ - 0.2y}} \times dy \\&= (0.1)\left[ {\frac{{{e^{0.2y}}}}{{0.2}}} \right]_{ - 楼}^0 + (0.1)\left[ {\frac{{ - {e^{ - 0.2y}}}}{{0.2}}} \right]_0^x \\&= (0.1)\left[ {\frac{1}{{0.2}} - 0} \right] + (0.1)\left[ {\frac{{ - {e^{ - 0.2x}}}}{{0.2}} - \left( {\frac{{ - 1}}{{0.2}}} \right)} \right]_0^楼 \\&= \frac{{0.1}}{{0.2}} - \frac{{0.1}}{{0.2}} \times {e^{ - 0.2x}} + \frac{{0.1}}{{0.2}}F(x) \\&= 1 - (0.5) \times {e^{ - 0.2x}} \\\end{aligned}\)

Thus finally \({\rm{F(X)}}\) can be given as:

06

Step 6: The graph

\({\rm{F(x)}}\)is plotted in the figure given below:

07

Calculating \(P(X£0),P(X£2),P( - 1£X£2)\)

(c) Here we will make use of following propositions:

Proposition: Let \({\rm{X}}\) be a continuous rv with pdf \({\rm{f(x)}}\) and cdf\({\rm{F(x)}}\). Then for any number a,

\(P(X拢a) = F(a)\)

Proposition: Let \({\rm{X}}\) be a continuous rv with pdf \({\rm{f(x)}}\) and\({\rm{cdfF(x)}}\). Then for any two numbers a and \({\rm{b}}\) with\({\rm{a < b}}\),

\(P(a拢X拢b) = F(b) - F(a)\)

To compute \(P(X拢0)\), we can write:

\(\begin{aligned}P(X拢0)&= F(0) \\&= 1 - (0.5){e^{ - 0.2(0)}} \\&= 1 - 0.5P(X拢0) \\&= 0.5 \\\end{aligned}\)

To compute \(P(X拢2)\) , we can write:

\(\begin{aligned}P(X拢2)&= F(2) \\&= 1 - (0.5){e^{ - 0.2(2)}} \\&= 1 - (0.5){e^{ - 0.4}}P(X拢2) \\&= 0.6648 \\\end{aligned} \)

08

Calculation for \(P(X£0),P(X£2),P( - 1£X£2)\) 

To compute ([P( - 1拢X拢2)\), we can write:

\(\begin{aligned}P( - 1拢X拢2)&= F(2) - F( - 1) \\&= \left[ {1 - (0.5){e^{ - 0.2(2)}}} \right]-\left[ {(0.5){e^{0.2( - 1)}}} \right] \\&= 1 - (0.5){e^{ - 0.4}} - (0.5){e^{ - 0.2}}P( - 1拢X拢2) \\&= 0.2555 \\\end{aligned} \)

The probability that an error of more than \({\rm{2}}\)miles is made can be represented as\({\rm{P(x < - 2 or x > 2)}}\).

This event is the complement of the event that an error of less than or equal to \({\rm{2}}\) miles is made, hence using the complement rule of probability, we can write:

\(\begin{aligned}P(x < - 2{\text{ }}or{\text{ }}x > 2)&= 1 - P( - 2拢X拢2) \\&= 1 - [F(2) - F( - 2)] \\&= 1 - \left[ {\left( {1 - (0.5){e^{ - 0.2(2)}}} \right) - \left( {(0.5){e^{0.2( - 2)}}} \right)} \right] \\&= 1 - \left[ {\left( {1 - (0.5){e^{ - 0.4}}} \right) - \left( {(0.5){e^{ - 0.4)}}} \right)} \right] \\&= 1 - [0.6648 - 0.3352]P(x < - 2{\text{ }}or{\text{ }}x > 2) \\&= 0.6704 \\ \end{aligned} \)

Definition (Complement rule of probability): If \({\rm{\bar A}}\) is the complement of an event A, then

\({\rm{P(\bar A) = 1 - P(A)}}\)

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