Question 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.)
M is the minimum value of Xj. M is an exponential with mean mu/n. We want the min distribution M since we’re interested in finding the first of the bulbs to burn out.
mu = 1000 #hours
n = 100 #bulbs
mu/n## [1] 10Or another way to see this is to use the cumulative distribution function of the min values of independent random variables, X1, X2,… Xn.
\[\text{The cumulative distribution function if Y when Y = min{X1,X2...Xn}} is...\] \[F_Y(y)=1-e^{\sum_{i=1}^{n}\lambda _iy}\] \[\text{The expected time is }\sum_{i=1}^{n=100}1/1000=100/1000\] \[\lambda=1/\mu\] \[\mu=10\]
Question 14:
Assume that X1 and X2 are independent random variables, each having an exponential density with parameter λ. Show that Z = X1 − X2 has density
\[f_z(z)=(1/2)\lambda e^{-\lambda |z|}\]
Textbook: If X and Y are independent, then the density of their sum is the convolution of their densities.
Definition 7.2: Let X and Y be two continuous random variables with density functions f(x) and g(y), respectively. Assume that both f(x) and g(y) are defined for all real numbers. Then the convolution f ∗ g of f and g is the function given by
\[(f\cdot g)(z)=\int_{-\infty}^{\infty}f(z-y)g(y)dy=\int_{-\infty}^{\infty}f(z-x)g(x)dx\]
\[\text{pairwise disjoint events (pg.285): }\] \[X_1=z+x_2 \text{ and } X_2=x_2\] \[x_1,x_2,z\text{ are some arbitrary numbers}\]
Finding the density function fZ(z) where FZ is the convultion of fX and fY. \[\text{Finding the density function } f_Z(z) \text{ where } f_Z \text{ is the convolution of } f_{X_1} and f_{X_2}\]
\[(f_Z)(z)=\int_{-\infty}^{\infty}f_{x_1}(x_2+z)f_{x_2}(x_2)dx_2\] \[=\int_{-\infty}^{\infty}\lambda e^{-\lambda (z+x_2)}\lambda e^{-\lambda x_2}dx_2\] \[=\int_{-\infty}^{\infty}\lambda ^2 e^{-2\lambda x_2-\lambda z}dx_2\] \[=\int_{-\infty}^{\infty}\lambda ^2 e^{-\lambda (2x_2+z)}dx_2\]
\[\text{When }z\ge0: \] \[x_1=z+x_2\ge0\] \[x_2\ge0\]
\[\text{When }z<0:\] \[x_1=z+x_2<0\] \[x_2<-z\]
\[(1/2)e^{-\lambda z} \text{ for }z\ge0\] \[=(1/2) e^{\lambda z} \text{ for }z<0\]
Equivalent to: \[f_Z(z)=(1/2)\lambda e^{-\lambda |z|} \]
Question 1: Let X be a continuous random variable with mean μ = 10 and variance σ2 = 100/3. Using Chebyshev’s Inequality, find an upper bound for the following probabilities.
We need to find the upper bound: No less than 1/k^2 data falls outside of this range.
\[\text{No less than what proportion of the data will wall outside of k standard devitions from the mean. }\] \[P(|X-10|\ge2)\]
mu=10
sd = sqrt(100/3)
k=2/sd
P = 1/(k^2)
print(paste0('No less than ',round(P*100,2),'% will fall outside of ', round(k,3),' standard deviations from the mean.'))## [1] "No less than 833.33% will fall outside of 0.346 standard deviations from the mean."print(paste0('This clearly does not make sense because chebychevs theorem only works for k >1. In this case, k = ',round(k,3)))## [1] "This clearly does not make sense because chebychevs theorem only works for k >1. In this case, k = 0.346"\[P(|X-10|\ge5)\]
mu=10
sd = sqrt(100/3)
k=5/sd
P = 1/(k^2)
print(paste0('No less than ',round(P*100,2),'% will fall outside of ', round(k,3),' standard deviations from the mean.'))## [1] "No less than 133.33% will fall outside of 0.866 standard deviations from the mean."print(paste0('This clearly does not make sense because chebychevs theorem only works for k >1. In this case, k = ',round(k,3)))## [1] "This clearly does not make sense because chebychevs theorem only works for k >1. In this case, k = 0.866"\[P(|X-10|\ge9)\]
mu=10
sd = sqrt(100/3)
k=9/sd
P = 1/(k^2)
print(paste0('No less than ',round(P*100,2),'% will fall outside of ', round(k,3),' standard deviations from the mean.'))## [1] "No less than 41.15% will fall outside of 1.559 standard deviations from the mean."\[P(|X-10|\ge20)\]
mu=10
sd = sqrt(100/3)
k=20/sd
P = 1/(k^2)
print(paste0('No less than ',round(P*100,2),'% will fall outside of ', round(k,3),' standard deviations from the mean.'))## [1] "No less than 8.33% will fall outside of 3.464 standard deviations from the mean."