Q11 and Q14 on page 303 of probability text Q1 on page 320-321
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 \(X_j\) . Show that the density for M is exponential with mean \(\frac{\mu }{ n }\). Hint: Use cumulative distribution functions.
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.)
For exponential dstribution, \(\mu = \frac{1}{\lambda }\) so that pdf for each lightbulb is \({ f }_{ T }(t)\quad =\quad \lambda { e }^{ -\lambda t }\) for t>0. Each exponential distribution has lack of memory of property: \(Pr(T>t+s\quad |\quad T>t\quad )\quad =Pr(T>s)\)
For sum of independen exponential distributions is still exponential distribution (prof by example 7.4 on page 292).
Let T=min(T1,T2,…..T100), then
\(Pr(T>t\quad )\quad =Pr(min(T1,T2....T100)>t)\quad \\ \quad \quad \quad \quad \quad \quad \quad \quad =Pr(\{ T1>t\} \cap \{ T2>t\} \cap .....\cap \{ T100>t\} )\\ \quad \quad \quad \quad \quad \quad \quad \quad ={ e }^{ -\lambda t }\quad *\quad { e }^{ -\lambda t }\quad *\quad ....*{ e }^{ -\lambda t }\\ \quad \quad \quad \quad \quad \quad \quad \quad ={ e }^{ -100\lambda t }\)
\(E_{ T }[\mu ]=\frac { 1 }{ 100*\lambda } =\frac { 1 }{ 100*{ \frac { 1 }{ \mu } } } =\frac { 1 }{ 100*{ \frac { 1 }{ 1000 } } } \quad =\quad 10\)
14 Assume that X1 and X2 are independent random variables, each having an exponential density with parameter \(\lambda\). Show that Z = X-Y has density \({ f }_{ Z }(z)\quad =\quad \left( \frac { 1 }{ 2 } \right) \lambda { e }^{ -\lambda \left| z \right| }\).
Using convolutions: \(P\left( Z=z \right) =\sum _{ k=-\infty }^{ \infty }{ P\left( X=z-k \right) \quad *\quad P\left( Y=k \right) \quad \quad }\)
For exponential function: \({ f }_{ X }\left( k \right) ={ f }_{ Y }\left( k \right) \quad =\lambda { e }^{ -\lambda k }\)
\({ f }_{ Z }(z)\quad =\quad \int _{ y=0 }^{ y=+\infty }{ { f }_{ X }\left( z+y \right) } { f }_{ Y }\left( y \right) dy\quad \quad (y\ge 0,\quad z\ge 0)\\ \quad \quad \quad \quad \quad =\quad \int _{ 0 }^{ +\infty }{ \lambda { e }^{ -\lambda (z+y) }\lambda { e }^{ -\lambda y } } dy\\ \quad \quad \quad \quad \quad ={ \lambda }^{ 2 }{ e }^{ -\lambda z }\int _{ 0 }^{ +\infty }{ { e }^{ -2\lambda y } } dy\\ \quad \quad \quad \quad \quad ={ \lambda }^{ 2 }{ e }^{ -\lambda z }\int _{ 0 }^{ +\infty }{ { e }^{ -2\lambda y } } d\left( -2\lambda y \right) \quad *\quad \frac { -1 }{ 2\lambda } \\ \quad \quad \quad \quad \quad =\frac { 1 }{ 2 } \lambda { e }^{ -\lambda z }\quad *\quad \left( { e }^{ -2\lambda y } \right) \overset { +\infty }{ \underset { 0 }{ | } } \\ \quad \quad \quad \quad \quad =\frac { 1 }{ 2 } \lambda { e }^{ -\lambda z }\quad *(1-0)\\ \quad \quad \quad \quad \quad =\frac { 1 }{ 2 } \lambda { e }^{ -\lambda z }\)
Therefore, \({ f }_{ Z }(z)\quad =\quad \left( \frac { 1 }{ 2 } \right) \lambda { e }^{ -\lambda \left| z \right| }\).
Q1 Let X be a continuous random variable with mean \({\mu = 10}\) and variance \({\sigma^2 = \frac{100}{3}}\). Using Chebyshev’s Inequality, find an upper bound for the following probabilities.
Apply Theorem 8.3 Chebyshev Inequality for solving the following problems: \({P \left( \left| X-\mu \right| \ge \epsilon \right) \le \quad \frac { { \sigma }^{ 2 } }{ { \epsilon }^{ 2 } } }\)
a.\({P \left( \left| X-10 \right| \ge 2 \right) \quad \le \quad \frac { { \sigma }^{ 2 } }{ { 2 }^{ 2 } } \quad =\quad 8.3 }\)
b.\({P \left( \left| X-10 \right| \ge 5 \right) \quad \le \quad \frac { { \sigma }^{ 2 } }{ { 5 }^{ 2 } } \quad =\quad 1.3 }\)
c.\({P \left( \left| X-10 \right| \ge 9 \right) \quad \le \quad \frac { { \sigma }^{ 2 } }{ { 9 }^{ 2 } } \quad =\quad 0.41152 }\)
d.\({P \left( \left| X-10 \right| \ge 20 \right) \quad \le \quad \frac { { \sigma }^{ 2 } }{ { 20 }^{ 2 } } \quad =\quad 0.08325}\)