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Chapter 5: Q12E (page 211)

Two components of a minicomputer have the following joint pdf for their useful lifetimes \({\rm{X}}\)and \({\rm{Y}}\)

a. What is the probability that the lifetime \({\rm{X}}\) of the first component exceeds \({\rm{3}}\)?

b. What are the marginal pdf鈥檚 of \({\rm{X}}\)and \({\rm{Y}}\)? Are the two lifetimes independent? Explain.

c. What is the probability that the lifetime of at least one component exceeds\({\rm{3}}\)?

Short Answer

Expert verified

a. The probability is \({\rm{P(X > 3) = 0}}{\rm{.05}}\)that the lifetime X of the first component exceeds \({\rm{3}}\).

b. The marginal pdf鈥檚 of X and Y ,

\[{{\text{f}}_{\text{X}}}\text{(x)=}\left\{ \begin{aligned}{*{35}{l}}{{\text{e}}^{\text{-x}}} & \text{,x }\!\!{}^\text{3}\!\!\text{0}\\\text{0}&\text{,otherwise}\\\end{aligned}\right.\],\[{{\text{f}}_{\text{Y}}}\text{(y)=}\left\{\begin{aligned}{*{35}{l}}\frac{\text{1}}{{{\text{(1+y)}}^{\text{2}}}} &\text{,y}\!\!{}^\text{3}\!\!\text{0} \\\text{0}&\text{,otherwise}\\\end{aligned}\right.\]

c. The probability that the lifetime of at least one component exceeds \({\rm{3}}\) \({\rm{P(at least one exceeds 3) = 0}}{\rm{.3}}\)

Step by step solution

01

Definition of Probability

Probability is a metric for determining the possibility of an event occurring. Many things are impossible to forecast with\({\rm{100\% }}\)accuracy. Using it, we can only anticipate the probability of an event occurring, or how probable it is to occur. Probability can range from\({\rm{0}}\)to\({\rm{1}}\), with\({\rm{0}}\)indicating an improbable event and 1 indicating a certain event. Possibility of...

02

Calculating the probability

(a):

There are two ways to compute the probability. First is to compute marginal pdf of \({\rm{X}}\) and then find \({\rm{X > 3}}\) using the marginal pdf, or compute it immediately. Since we need to determine marginal pdf's of \({\rm{X}}\) and \({\rm{Y}}\) in part\({\rm{(b)}}\), we will show you the second way (in which you indirectly compute pdf of \({\rm{X}}\) ).

For every adequate set \({\rm{A}}\) the following holds

The following holds

\(\begin{aligned}{l}{\rm{P(X > 3) = }}\int_{\rm{3}}^{\rm{楼}} {\overbrace {\int_{\rm{0}}^{\rm{楼}} {\rm{x}} {{\rm{e}}^{{\rm{ - x(1 + y)}}}}{\rm{dy}}}^{{\rm{marginal pdf of X}}}} {\rm{dx}}\\{\rm{ = }}\int_{\rm{3}}^{\rm{楼}} {\int_{\rm{0}}^{\rm{楼}} {\rm{x}} } {{\rm{e}}^{{\rm{ - x}}}}{\rm{ \times }}{{\rm{e}}^{{\rm{ - xy}}}}{\rm{dydx}}\\{\rm{ = }}\left. {\int_{\rm{3}}^{\rm{楼}} {\rm{x}} {{\rm{e}}^{{\rm{ - x}}}}{\rm{ \times }}\left( {{\rm{ - }}\frac{{\rm{1}}}{{\rm{x}}}} \right){{\rm{e}}^{{\rm{ - xy}}}}} \right|_{\rm{0}}^{\rm{楼}}\\{\rm{ = }}\int_{\rm{3}}^{\rm{楼}} {{{\rm{e}}^{{\rm{ - x}}}}} {\rm{dx = - }}\left. {{{\rm{e}}^{{\rm{ - x}}}}} \right|_{\rm{3}}^{\rm{楼}}{\rm{ = }}{{\rm{e}}^{{\rm{ - 3}}}}{\rm{ = 0}}{\rm{.05}}\end{aligned}\)

03

 The marginal pdf’s of \({\rm{X}}\)

(b):

The marginal probability density function

of continuous random variable \({\rm{X}}\) is

\({{\rm{f}}_{\rm{X}}}{\rm{(x) = }}\int_{{\rm{ - 楼}}}^{\rm{楼}} {\rm{f}} {\rm{(x,y)dy, for - 楼< x < 楼}}\)

The marginal probability density function

of continuous random variable \({\rm{Y}}\) is

\({{\rm{f}}_{\rm{Y}}}{\rm{(y) = }}\int_{{\rm{ - 楼}}}^{\rm{楼}} {\rm{f}} {\rm{(x,y)dx, for - 楼< y < 楼}}{\rm{.}}\)

The marginal pdf of \({\rm{X}}\) is

We can write it as

04

 The marginal pdf’s of  \({\rm{Y}}\)  

Similarly, the marginal pdf of \({\rm{Y}}\) is

\({{\rm{f}}_{\rm{Y}}}{\rm{(y) = }}\int_{{\rm{ - 楼}}}^{\rm{楼}} {\rm{f}} {\rm{(x,y)dx = }}\int_{\rm{0}}^{\rm{楼}} {\rm{x}} {{\rm{e}}^{{\rm{ - x(1 + y)}}}}{\rm{dx}}\)

\({\rm{ = }}\left| {\begin{aligned}{*{20}{c}}{{\rm{u = x}}}&{\rm{n}}&{{\rm{du = dx}}}\\{{{\rm{e}}^{{\rm{ - x(1 + y)}}}}{\rm{dx = dv}}}&{\rm{n}}&{{\rm{v = - }}\frac{{\rm{1}}}{{{\rm{1 + y}}}}{{\rm{e}}^{{\rm{ - x(1 + y)}}}}}\\{{\rm{ integration by parts: }}}&{{\rm{uv}}}&{{\rm{ - }}\int {\rm{v}} {\rm{du}}}\end{aligned}} \right|\)

\({\rm{ = }}\left. {{\rm{x \times }}\left( {{\rm{ - }}\frac{{\rm{1}}}{{{\rm{1 + y}}}}{{\rm{e}}^{{\rm{ - x(1 + y)}}}}} \right)} \right|_{\rm{0}}^{\rm{楼}}{\rm{ + }}\int_{\rm{0}}^{\rm{楼}} {\frac{{\rm{1}}}{{{\rm{1 + y}}}}} {{\rm{e}}^{{\rm{ - x(1 + y)}}}}{\rm{dx}}\)

\(\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{0 + }}\frac{{\rm{1}}}{{{\rm{1 + y}}}}\int_{\rm{0}}^{\rm{楼}} {{{\rm{e}}^{{\rm{ - x(1 + y)}}}}} {\rm{dx}}\)

\({\rm{ = }}\left. {\frac{{\rm{1}}}{{{\rm{1 + y}}}}{\rm{ \times }}\left( {{\rm{ - }}\frac{{\rm{1}}}{{{\rm{1 + y}}}}{{\rm{e}}^{{\rm{ - x(1 + y)}}}}} \right)} \right|_{\rm{0}}^{\rm{楼}}\)

\({\rm{ = }}\frac{{\rm{1}}}{{{{{\rm{(1 + y)}}}^{\rm{2}}}}}\),\({{\rm{f}}_{\rm{Y}}}{\rm{(y) = 0}}\),,\({\rm{y < 0}}\).

05

 The two lifetimes independent

(1): here we used L'Hopital's rule to obtain the limit:

where \({\rm{c}}\) is constant we get when we derivative the denominator.

The marginal pdf of \({\rm{Y}}\) is

Two random variables \({\rm{X}}\) and \({\rm{Y}}\) are independent if and only if

1. \({\rm{p(x,y) = }}{{\rm{p}}_{\rm{X}}}{\rm{(x) \times }}{{\rm{p}}_{\rm{Y}}}{\rm{(y)}}\),

for every \({\rm{(x,y)}}\) and when \({\rm{X}}\) and \({\rm{Y}}\) discrete rv's,

2. \({\rm{f(x,y) = }}{{\rm{f}}_{\rm{X}}}{\rm{(x) \times }}{{\rm{f}}_{\rm{Y}}}{\rm{(y)}}\),

for every \({\rm{(x,y)}}\) and when \({\rm{X}}\) and \({\rm{Y}}\) continuous rv's,

otherwise, they are dependent.

It is obvious that the joint pdf is not the product of marginal pdf's for every\({\rm{(x,y)}}\), hence, random variables are dependent.

06

Calculating the probability

(c):

At least one component exceeds \({\rm{3}}\) means that we need to find probability of event

\({\rm{\{ X > 3\} `E \{ Y > 3\} }}\)

However, it is easier if we look at the complement of the mentioned event:

\({\rm{\{ X拢3\} {C}\{ Y拢3\} }}\)

Therefore, the following holds

\(\begin{aligned}{l}{\rm{P(\{ X > 3\} `E \{ Y > 3\} ) = 1 - P(\{ X拢3\} {C}\{ Y拢3\} )}}\\\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{1 - }}\int_{\rm{0}}^{\rm{3}} {\int_{\rm{0}}^{\rm{3}} {\rm{x}} } {{\rm{e}}^{{\rm{ - x(1 + y)}}}}{\rm{dydx}}\\{\rm{ = 1 - }}\int_{\rm{0}}^{\rm{3}} {\int_{\rm{0}}^{\rm{3}} {\rm{x}} } {{\rm{e}}^{{\rm{ - x}}}}{{\rm{e}}^{{\rm{ - xy}}}}{\rm{dydx}}\\{\rm{ = 1 - }}\int_{\rm{0}}^{\rm{3}} {\rm{x}} {{\rm{e}}^{{\rm{ - x}}}}\left( {{\rm{ - }}\frac{{\rm{1}}}{{\rm{x}}}\left( {{{\rm{e}}^{{\rm{ - 3x}}}}{\rm{ - }}{{\rm{e}}^{{\rm{ - x \times 0}}}}} \right)} \right){\rm{dx}}\\{\rm{ = 1 - }}\int_{\rm{0}}^{\rm{3}} {{{\rm{e}}^{{\rm{ - x}}}}} \left( {{\rm{1 - }}{{\rm{e}}^{{\rm{ - 3x}}}}} \right){\rm{dx}}\\{\rm{ = 1 - }}\left( {{\rm{ - }}\left. {{{\rm{e}}^{{\rm{ - x}}}}} \right|_{\rm{0}}^{\rm{3}}} \right){\rm{ + }}\left( {{\rm{ - }}\left. {\frac{{\rm{1}}}{{\rm{4}}}{{\rm{e}}^{{\rm{ - 4x}}}}} \right|_{\rm{0}}^{\rm{3}}} \right)\\{\rm{ = 0}}{\rm{.3}}\end{aligned}\)

(1): for every adequate set \({\rm{A}}\) the following holds

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Most popular questions from this chapter

Consider a random sample of size n from a continuous distribution having median \({\rm{0}}\)so that the probability of any one observation being positive is \({\rm{.5}}\). Disregarding the signs of the observations, rank them from smallest to largest in absolute value, and let \({\rm{w = }}\)the sum of the ranks of the observations having positive signs. For example, if the observations are \({\rm{ - }}{\rm{.3, + }}{\rm{.7, + 2}}{\rm{.1 and - 2}}{\rm{.5 }}\) then the ranks of positive observations are \({\rm{ 2and 3 }}\), so \({\rm{w = 5}}\). In Chapter \({\rm{15}}\), W will be called Wilcoxon鈥檚 signed-rank statistic. \({\rm{w}}\)can be represented as follows:

where the Yi 鈥檚 are independent Bernoulli rv鈥檚, each with \({\rm{ p = }}{\rm{.5 }}\) (\({{\rm{Y}}_{\rm{1}}}{\rm{ = 1}}\)corresponds to the observation with rank \({\rm{i }}\)being positive)

a. Determine \({\rm{E}}\left( {{{\rm{Y}}_{\rm{i}}}} \right){\rm{ }}\)and then \({\rm{E(W)}}\)using the equation for \({\rm{W}}\). (Hint: The first n positive integers sum to \({\rm{n(n + 1)/2}}{\rm{.)}}\)

b. Determine \({\rm{V(}}{{\rm{Y}}_{\rm{1}}}{\rm{)}}\)and then \({\rm{V(W)}}\). (Hint: The sum of the squares of the first n positive integers can be expressed as \({\rm{n(n + 1)(2n + 1)/6}}{\rm{.)}}\)

Compute the correlation coefficient \({\rm{\rho }}\) for \({\rm{X}}\) and \({\rm{Y}}\)(the covariance has already been computed).

a. Recalling the definition of \({{\rm{\sigma }}^{\rm{2}}}\) for a single \({\rm{rv X}}\), write a formula that would be appropriate for computing the variance of a function \({\rm{h(X,Y)}}\) of two random variables. (Hint: Remember that variance is just a special expected value.)

b. Use this formula to compute the variance of the recorded score\({\rm{h(X,Y)( = max(X,Y))}}\).

A shipping company handles containers in three different sizes:\(\left( 1 \right)\;27f{t^3}\;\left( {3 \times 3 \times 3} \right)\)\(\left( 2 \right) 125 f{t^3}, and \left( 3 \right)\;512 f{t^3}\). Let \({X_i}\left( {i = \;1, 2, 3} \right)\)denote the number of type i containers shipped during a given week. With \({\mu _i} = E\left( {{X_i}} \right)\)and\(\sigma _i^2 = V\left( {{X_i}} \right)\), suppose that the mean values and standard deviations are as follows:

\(\begin{array}{l}{\mu _1} = 200 {\mu _2} = 250 {\mu _3} = 100 \\{\sigma _1} = 10 {\sigma _2} = \,12 {\sigma _3} = 8\end{array}\)

a. Assuming that \({X_1}, {X_2}, {X_3}\)are independent, calculate the expected value and variance of the total volume shipped. (Hint:\(Volume = 27{X_1} + 125{X_2} + 512{X_3}\).)

b. Would your calculations necessarily be correct if \({X_i} 's\)were not independent?Explain.

Two different professors have just submitted final exams for duplication. Let \({\rm{X}}\) denote the number of typographical errors on the first professor鈥檚 exam and \({\rm{Y}}\) denote the number of such errors on the second exam. Suppose \({\rm{X}}\) has a Poisson distribution with parameter \({{\rm{\mu }}_{\rm{1}}}\), \({\rm{Y}}\) has a Poisson distribution with parameter \({{\rm{\mu }}_{\rm{2}}}\), and \({\rm{X}}\) and \({\rm{Y}}\) are independent.

a. What is the joint pmf of \({\rm{X}}\) and\({\rm{Y}}\)?

b. What is the probability that at most one error is made on both exams combined?

c. Obtain a general expression for the probability that the total number of errors in the two exams is m (where \({\rm{m}}\) is a nonnegative integer). (Hint: \({\rm{A = }}\left\{ {\left( {{\rm{x,y}}} \right){\rm{:x + y = m}}} \right\}{\rm{ = }}\left\{ {\left( {{\rm{m,0}}} \right)\left( {{\rm{m - 1,1}}} \right){\rm{,}}.....{\rm{(1,m - 1),(0,m)}}} \right\}\)Now sum the joint pmf over \({\rm{(x,y)}} \in {\rm{A}}\)and use the binomial theorem, which says that

\({\rm{P(X + Y = m)}}\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\sum\limits_{{\rm{k = 0}}}^{\rm{m}} {\left( {\begin{array}{*{20}{c}}{\rm{m}}\\{\rm{k}}\end{array}} \right){{\rm{a}}^{\rm{k}}}{{\rm{b}}^{{\rm{m - k}}}}{\rm{ = }}\left( {{\rm{a + b}}} \right)} ^{\rm{m}}}\)
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