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Chapter 3: Q8SE (page 203)

Suppose that an electronic system comprises four components, and let\({X_j}\)denote the time until component j fails to operate (j = 1, 2, 3, 4). Suppose that\({X_1},{X_2},{X_3}\)and\({X_4}\)are i.i.d. random variables, each of which has a continuous distribution with c.d.f.\(F\left( x \right)\)Suppose that the system will operate as long as both component 1 and at least one of the other three components operate. Determine the c.d.f. of the time until the system fails to operate.

Short Answer

Expert verified

The cdf of the time until the system fails to operate is \(1 - G\left( x \right) = F\left( x \right)\left[ {1 + {F^2}\left( x \right) - {F^3}\left( x \right)} \right]\)

Step by step solution

01

Given information:

\({X_j}\) be the time until component j fails to operate

\({X_{1,}}{X_{2,}}{X_3}\) and \({X_4}\) are the random variables, each of which has a continuous distribution .

02

Calculating the cdf of the time until the system fails to operate.

Let\(G\left( x \right)\)devote the cdf of the time until the system fails, let A denote the event that component 1 is still operating at time x, and let B denote the event that at least one of the other three components is still operating at time x. Then

\(\begin{array}{c}1 - G\left( x \right) = {\rm P}\left( {system\;stillo\;operating\;at\;time\;x\;} \right)\\ = {\rm P}\left( {A \cap B} \right)\\ = {\rm P}\left( A \right){\rm P}\left( B \right)\end{array}\)

\(1 - G\left( x \right) = \left[ {1 - F\left( x \right)} \right]\left[ {1 - {F^3}\left( x \right)} \right]\)

Hence

\(1 - G\left( x \right) = F\left( x \right)\left[ {1 + {F^2}\left( x \right) - {F^3}\left( x \right)} \right]\)

The cdf of the time until the system fails to operate is\(1 - G\left( x \right) = F\left( x \right)\left[ {1 + {F^2}\left( x \right) - {F^3}\left( x \right)} \right]\)

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Most popular questions from this chapter

Suppose that three random variables X1, X2, and X3 have a continuous joint distribution with the following joint p.d.f.:

\({\bf{f}}\left( {{{\bf{x}}_{\bf{1}}}{\bf{,}}{{\bf{x}}_{\bf{2}}}{\bf{,}}{{\bf{x}}_{\bf{3}}}} \right){\bf{ = }}\left\{ {\begin{align}{}{{\bf{c}}\left( {{{\bf{x}}_{\bf{1}}}{\bf{ + 2}}{{\bf{x}}_{\bf{2}}}{\bf{ + 3}}{{\bf{x}}_{\bf{3}}}} \right)}&{{\bf{for0}} \le {{\bf{x}}_{\bf{i}}} \le {\bf{1}}\,\,\left( {{\bf{i = 1,2,3}}} \right)}\\{\bf{0}}&{{\bf{otherwise}}{\bf{.}}}\end{align}} \right.\)

Determine\(\left( {\bf{a}} \right)\)the value of the constant c;

\(\left( {\bf{b}} \right)\)the marginal joint p.d.f. of\({{\bf{X}}_{\bf{1}}}\)and\({{\bf{X}}_{\bf{3}}}\); and

\(\left( {\bf{c}} \right)\)\({\bf{Pr}}\left( {{{\bf{X}}_{\bf{3}}}{\bf{ < }}\frac{{\bf{1}}}{{\bf{2}}}\left| {{{\bf{X}}_{\bf{1}}}{\bf{ = }}\frac{{\bf{1}}}{{\bf{4}}}{\bf{,}}{{\bf{X}}_{\bf{2}}}{\bf{ = }}\frac{{\bf{3}}}{{\bf{4}}}} \right.} \right){\bf{.}}\)

Suppose that the joint p.d.f. of two random variables X and Y is as follows:

\(f\left( {x,y} \right) = \left\{ \begin{aligned}c\sin x\;\;\;\;\;for\;0 \le x \le \frac{\pi }{2}\;\;and\;0 \le y \le 3\\0\;\;\;\;\;\;\;\;\;\;\;\;\;\;Otherwise\end{aligned} \right.\)

Determine (a) the conditional p.d.f. of Y for every given value of X, and

(b)\({\rm P}\left( {1 < y < \frac{2}{x} = 0.73} \right)\)

A civil engineer is studying a left-turn lane that is long enough to hold seven cars. LetXbe the number of cars in the lane at the end of a randomly chosen red light. The engineer believes that the probability thatX=xis proportional to(x+1)(8鈭�x)forx=0, . . . ,7 (the possible values ofX).

a. Find the p.f. ofX.

b. Find the probability thatXwill be at least 5.

Let X be a random vector that is split into three parts,\(X = \left( {Y,Z,W} \right)\)Suppose that X has a continuous joint distribution with p.d.f.\(f\left( {y,z,w} \right)\).Let\({g_1}\left( {y,z|w} \right)\)be the conditional p.d.f. of (Y, Z) given W = w, and let\({g_2}\left( {y|w} \right)\)be the conditional p.d.f. of Y given W = w. Prove that\({g_2}\left( {y|w} \right) = \int {{g_1}\left( {y,z|w} \right)dz} \)

The definition of the conditional p.d.f. of X given\({\bf{Y = y}}\)is arbitrary if\({{\bf{f}}_{\bf{2}}}\left( {\bf{y}} \right){\bf{ = 0}}\). The reason that this causes no serious problem is that it is highly unlikely that we will observe Y close to a value\({{\bf{y}}_{\bf{0}}}\)such that\({{\bf{f}}_{\bf{2}}}\left( {{{\bf{y}}_{\bf{0}}}} \right){\bf{ = 0}}\). To be more precise, let\({{\bf{f}}_{\bf{2}}}\left( {{{\bf{y}}_{\bf{0}}}} \right){\bf{ = 0}}\), and let\({{\bf{A}}_{\bf{0}}}{\bf{ = }}\left( {{{\bf{y}}_{\bf{0}}}{\bf{ - }} \in {\bf{,}}{{\bf{y}}_{\bf{0}}}{\bf{ + }} \in } \right)\). Also, let\({{\bf{y}}_{\bf{1}}}\)be such that\({{\bf{f}}_{\bf{2}}}\left( {{{\bf{y}}_{\bf{1}}}} \right){\bf{ > 0}}\), and let\({{\bf{A}}_{\bf{1}}}{\bf{ = }}\left( {{{\bf{y}}_{\bf{1}}}{\bf{ - }} \in {\bf{,}}{{\bf{y}}_{\bf{1}}}{\bf{ + }} \in } \right)\). Assume that\({{\bf{f}}_{\bf{2}}}\)is continuous at both\({{\bf{y}}_{\bf{0}}}\)and\({{\bf{y}}_{\bf{1}}}\).

Show that

\(\mathop {{\bf{lim}}}\limits_{ \in \to {\bf{0}}} \,\frac{{{\bf{Pr}}\left( {{\bf{Y}} \in {{\bf{A}}_{\bf{0}}}} \right)}}{{{\bf{Pr}}\left( {{\bf{Y}} \in {{\bf{A}}_{\bf{1}}}} \right)}}{\bf{ = 0}}{\bf{.}}\)

That is, the probability that Y is close to\({{\bf{y}}_{\bf{0}}}\)is much smaller than the probability that Y is close to\({{\bf{y}}_{\bf{1}}}\).
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