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?
Based on Ex.10, let \(X_1,X_2,...,X_n\) be \(n\) independent random variables each of which has an exponential density with mean \(\mu\) and \(M\) is exponential with mean \(\frac{\mu}{n}\)
In another word, if \(X_i\sim Exp(\lambda)\) then \(min\{ X_{ 1 },X_{ 2 },...,X_{ n })\sim e ^{ \lambda n }\)
For an exponential distribution:
\(E[X]=1000=\frac{1}{\lambda}=\frac{1}{1000}\) \(n\) = 100 bulbs, \(n\lambda= 100 (\frac{1}{1000})=\frac{1}{10}\)
Cumulative distribution function for exponential distribution: \(1-e ^{\lambda x}\) First to burnout= \(min(X_1,X_2,...,X-n)\)
\(P(min(X_1,X_2,...,X_n))=1-e ^{-\frac{1}{10}x}\) \(E[X]=\frac{1}{\frac{1}{10}}=10 hours\)
Assume that \(X_1\) and \(X_2\) are independent random variables, each having an exponential density with parameter \(\lambda\). Show that \(Z=X_1-X_2\) has density \(f(z)=(\frac{1}{2})\lambda e^{\lambda \left\lfloor z \right\rfloor }\)
\(fX_{ 1 }(x)=fX_{ 2 }(x)=\begin{cases}\lambda e^{-\lambda x} & x\ge0\\0 & otherwise\end{cases}\)
\(f_Z(z) = \int_{-\infty}^{+\infty}f_{X_1}(x)f_{X_2}(x-z) dx\)
\(\int_{0}^{+\infty}\lambda e^{-\lambda x} \lambda e^{-\lambda(x-z)} dx\)
\(\int_{0}^{+\infty}\lambda^2 e^{-2\lambda x +\lambda z} dx\)
\(\lambda e^{\lambda z}\int_{0}^{+\infty}e^{-2\lambda x} dx\)
\(\lambda e^{\lambda z}\int_{0}^{+\infty}e^{-2\lambda x} dx\)
\((\frac{1}{2}) \lambda e^{\lambda z}\)
Since \(X_1\) and \(X_2\) are independent random variable having an exponential density with the same partner \(\lambda\), \(Z=X_1-X_2=-(X_2-X_1)=-Z\), therefore
\(f_Z(z)=\begin{cases}\frac{1}{2} \lambda e^{\lambda z} & z < 0\\\frac{1}{2}\lambda e^{-\lambda z} & otherwise\end{cases} f_Z(z)=(\frac{1}{2}) \lambda e^{-\lambda|z|}\)
Let \(X\) be 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.
Chebyshev’s Inequality states that
Let \(X\) be any random variable with \(E[X]=\mu\) and \(V(X)=\sigma ^2\)
\(P(|X - \mu|\ge k\sigma) \le \frac{\sigma ^2}{k^2\sigma ^2} = \frac{1}{k^2}\)
Thus for any random vriable, the probability of a deviation from the mean of more than \(k\) standard deviation is \(\le \frac{1}{k^2}\)
For any of the following inequality. \(\mu=10\) and \(\sigma ^2= \frac{100}{3}\) \(k=\frac{k \sigma}{\sigma}\)
if (!require('knitr')) install.packages('knitr')## Loading required package: knitrci <- "Pr(|X-m|>= kv)"
output <- matrix(NA, nrow=4, ncol=4)
kv <- c(2,5,9,20)
m <- 10
sd <- sqrt(100/3)
for (i in 1:length(kv)){
k <- kv[i]/sd
ks <- 1/(k^2)
fun <- gsub("m", m, ci)
fun <- gsub("kv", kv[i], fun)
con <- paste("for any random variable X, the probability of a deviation from the mean of more than", round(k,2), sep=" ")
con <- paste(con, "standard deviations is", sep=" ")
con <- paste(con, round(ks,2), sep=" ")
if (k < 1){
con <- "Since k is less than 1, Chebyshev's Inequality is not accurate"
}
output[i,1] <- fun
output[i,2] <- round(k,2)
output[i,3] <- round(ks,2)
output[i,4] <- con
}
kable (output,col.names = c("Function", "k", "1/k^2","Conclusion"), caption = "Chebyshev's Inequality", escape=T)| Function | k | 1/k^2 | Conclusion |
|---|---|---|---|
| Pr(|X-10|>= 2) | 0.35 | 8.33 | Since k is less than 1, Chebyshev’s Inequality is not accurate |
| Pr(|X-10|>= 5) | 0.87 | 1.33 | Since k is less than 1, Chebyshev’s Inequality is not accurate |
| Pr(|X-10|>= 9) | 1.56 | 0.41 | for any random variable X, the probability of a deviation from the mean of more than 1.56 standard deviations is 0.41 |
| Pr(|X-10|>= 20) | 3.46 | 0.08 | for any random variable X, the probability of a deviation from the mean of more than 3.46 standard deviations is 0.08 |