Page 303-11 A company buys 100 lightbulbs, each of which has an exponential lifetime of 1000 hours. What is the expected time for the first of these bulbs to burn out? (See Exercise 10: Let X1, X2, . . . , Xn be n independent random variables each of which has an exponential density with mean \(\mu\). Let M be the minimum value of the Xj . Show that the density for M is exponential with mean \(\mu\)/n. Hint: Use cumulative distribution functions.)
Answer:The probability density function (pdf) of an exponential distribution is
\(let\quad { X }_{ 1 },\quad { X }_{ 2 },\quad ...\quad { X }_{ 100 }\quad be\quad 100\quad lifetimes\quad of100\quad lightbulbs\quad which\quad each\quad has\quad f\left( x \right) =\lambda { e }^{ -\lambda x },\quad mean\quad 1000\quad hours\\ \lambda =\frac { 1 }{ 1000 } ,\quad set\quad y=min({ X }_{ 1 },X_{ 2 },...{ X }_{ 100 })\quad as\quad the\quad shortest\quad lifetime\)
\(P({ X }_{ i })\quad =\int _{ 0 }^{ y }{ \frac { 1 }{ 1000 } { e }^{ -\frac { 1 }{ 1000 } x }dx } =1-{ e }^{ -\frac { y }{ 1000 } }\quad \quad \quad ({ X }_{ i }<y)\)
\(then\quad P({ X }_{ i })\quad =\quad 1-\quad (1-{ e }^{ -\frac { y }{ 1000 } })\quad =\quad { e }^{ -\frac { y }{ 1000 } }\quad \quad ({ X }_{ i }>y)\)
\(F({ x }<y)\quad =\quad 1\quad -\quad P({ X }_{ 1 }>y)P({ X }_{ 2 }>y)...P({ X }_{ 100 }>y)\\ \qquad \qquad =\quad 1\quad -\quad { e }^{ -\frac { y }{ 1000 } }*{ \quad e }^{ -\frac { y }{ 1000 } }*...*\quad { e }^{ -\frac { y }{ 1000 } }\\ \qquad \quad \quad =\quad 1-{ e }^{ -\frac { 100y }{ 1000 } }\)
\(\lambda \quad of\quad f(y)\quad =\quad \quad \frac { 100 }{ 1000 } =\frac { 1 }{ 10 }\)
\(So\quad E(y)\quad =\quad \frac { 1 }{ \lambda } =10\)
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Page 303-14 Assume that X1 and X2 are independent random variables, each having an exponential density with parameter \(\lambda\). Show that \(Z\quad =\quad { X }_{ 1 }-{ X }_{ 2 }\) has density
\(f\left( z \right) \quad =(1/2)\lambda { e }^{ -\lambda |z| }\)
Answer: \(let\quad { X }_{ 1 }=X,\quad { X }_{ 2 }=Y\quad\) \(let\quad { X }_{ 1 }=X\quad (x\ge 0),\quad { X }_{ 2 }=Y\quad (y\ge 0),\quad Z={ X }_{ 1 }-{ X }_{ 2 }=X-Y,\quad X=Z+Y\\ f\left( z \right) \quad =\int _{ 0 }^{ +\infty }{ f\left( z+y \right) f\left( y \right) dy } \qquad \quad \\ \qquad \quad =\int _{ 0 }^{ +\infty }{ \lambda { e }^{ -\lambda (z+y) }\lambda { e }^{ -\lambda (y) }dy } \\ \quad \qquad =\int _{ 0 }^{ +\infty }{ { \lambda }^{ 2 }{ e }^{ -\lambda (z+2y) }\frac { 1 }{ 2\lambda } d(z+2y) } \\ \qquad \quad =\frac { 1 }{ 2 } \lambda { e }^{ -\lambda |z| }\quad (z\quad in\quad (-\infty ,+\infty ))\)
Page 321-1 Let X be a continuous random variable with mean \(\mu\)= 10 and variance \({ \sigma }^{ 2 }\) = 100/3. Using Chebyshev’s Inequality, find an upper bound for the following probabilities.
Answer: By Theorem 8.3 (Chebyshev Inequality), $P(|X-|) $
(100/3)/(2*2)## [1] 8.333333(100/3)/(5*5)## [1] 1.333333(100/3)/(9*9)## [1] 0.4115226(100/3)/(20*20)## [1] 0.08333333