SUMS OF CONTINUOUS RANDOM VARIABLES (7.2) - Question 11

Week08 Discussion
Solution:

Let \(X_1\), \(X_2\), …., \(X_n\) be \(n\) independent random variables each of which has an exponential density with mean \(\mu\).

\(P(X > x)\) = \(e^{-\frac{x}{\mu}}\)

\(P\)(\(min\) (\(X_1\), \(X_2\), … \(X_n\) > \(x\)) = \(P(X_i > x))^n\) \(for\) \(i\)=\(1\),..\(n\)

= \(e^{-\frac{x}{\mu_1}}\). \(e^{-\frac{x}{\mu_2}}\) … \(e^{-\frac{x}{\mu_n}}\) = \(e^{-(\frac{1}{\mu_1} + \frac{1}{\mu_2} + \frac{1}{\mu_3}).x}\)

Since this distribution is also a exponential distribution

Therefore, \(E(X)\) = \(\frac{1}{\displaystyle\sum_{i=1}^{n} \mu_i}\)

\(n\) = \(100\), \(\mu_i\) = \(\frac{1}{1000}\) \(for\) \(i\)=\(1\),\(2\), … \(100\)

Therefore, \(E(X)\) = \(\frac{1}{\mu_1+\mu_2 + ... + \mu_{100}}\) = \(\frac{1}{\frac{1}{1000}+\frac{1}{1000} + ... + \frac{1}{1000}}\) = \(\frac{1}{\frac{100}{1000}}\)
= \(\frac{1000}{100}\) = \(10\)
Therefore \(E(X)\) = \(10\) hours

n   <- 100
mu  <- 1/1000
sum <- 0

for (i in 1:n) {
  sum <- sum + mu
}

expected_value = 1/sum
print (expected_value)
## [1] 10