Question 2.10: The number of light switch turn-ons at which the first failu......
The number of light switch turn-ons at which the first failure occurs is a r.v. X whose p.d.f. is given by: f(x)=c({\textstyle{\frac{9}{10}}})^{x-1},x=1,2,\dots (and 0 otherwise).
(i) Determine the constant c.
(ii) Calculate the probability that the first failure will not occur until after the 100th turn-on.
(iii) Determine the corresponding d.f. F.
Hint: At this point, recall how we sum geometric series; namely, \textstyle\sum_{n=k}^{\infty}t^{n}= \textstyle{\frac{t^{k}}{1-t}},\,|t|\lt 1,\,k=0,\,1,\,….
(i) The constant c is determined through the relationship: ~\textstyle\sum_{x=1}^{\infty}f(x)\ = 1~ or ~\textstyle\sum_{x=1}^{\infty}c({\frac{{9}}{10}})^{x-1}\;=\;1.~ However, \textstyle\sum_{x=1}^{\infty}c({\frac{9}{10}})^{x-1}\;=\;c\sum_{x=1}^{\infty}({\frac{9}{10}})^{x-1}\;=\; c[1+(\frac{9}{10})+(\frac{9}{10})^{2}+\cdots]=c\frac{1}{1-\frac{9}{10}}=10c, so that c={\frac{1}{10}}.
(ii) Here ~P(X\gt10)=P(X\geq1{1})=c\textstyle\sum_{x=11}^{\infty}(\frac{9}{10})^{x-1}=c[(\frac{9}{10})^{10}+(\frac{9}{10})^{11}+\cdots]= c{\frac{(\frac{9}{10})^{10}}{1-{\frac{9}{10}}}}=c\cdot10({\frac{9}{10}})^{10}={\frac{1}{10}}\cdot10({\frac{{{9}}}{10}})^{10}=(0.9)^{10}\simeq0.349.
(iii) First, for {x}\,\lt \,1,F(x)\,=\,0.~ Next, for ~{{x}}\geq{1},{{F}}({{x}})\;=\;\textstyle\sum_{t=1}^{x}c({\frac{9}{10}})^{t-1}\;=\; 1-\textstyle\sum_{t=x+1}^{\infty}c\cdot({\frac{9}{10}})^{t-1}=1-c\sum_{t=x+1}^{\infty}({\frac{9}{10}})^{t-1}=1-{\frac{1}{10}}\cdot{\frac{\langle{\frac{9}{10}}\rangle^{x}}{1-{\frac{9}{10}}}}=1-({\frac{9}{10}})^{x}.
Thus, F(x)=0\,{\mathrm{for}}\,x\lt 1,\,\mathrm{and}\,F(x)=1-({\frac{9}{10}})^{x}\,{\mathrm{for}}\,x\geq{{1}}.