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- 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?
Expected Value of Exponential Distribution: \[EX=\frac{1}{\lambda}\] Also, if \(X_i\sim Exp(\lambda_i)\) then \(minX_i\sim Exp(\sum\lambda_i)\)
n <- 100
lambda <- 1/1000
sum_lambda <- n * lambda
ex <- 1/sum_lambda
print(paste("The expected time for the first of these bulbs to burn out is", ex, "hours."))
## [1] "The expected time for the first of these bulbs to burn out is 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 \[f_z(z)=\frac{1}{2}\lambda e^{-\lambda\mid z\mid}\]
In general, \[f(x_1)=\lambda e^{\lambda x_1} \] \[f(x_2)=\lambda e^{\lambda x_2} \] Convolution of two exponential densities with \(\lambda = 1\), \[f_z(z)=\lambda^{2} e^{\lambda (x_1+x_2)} \]
Since \(Z=X_1+X_2\), or \(X_1=Z+X_2\), \[f_z(z)=\lambda^{2} e^{\lambda (z+2x_2)} \] So when \(Z\) is negative, \[\int_{-z}^{\infty}\lambda^{2} e^{\lambda (z+2x_2)}dx \] \[=\frac{\lambda}{2} e^{\lambda z} \] and when \(Z\) is positive, \[\int_{0}^{\infty}\lambda^{2} e^{\lambda (z+2x_2)}dx \] \[=\frac{\lambda}{2} e^{-\lambda z} \] \[\therefore f_z(z)=\frac{1}{2}\lambda e^{-\lambda\mid z\mid} \Box\]
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- Let \(X\) be a continuous random variable with mean \(\mu=10\) and variance \(\sigma^{2} = \frac{100}{3}\). Using Chebyshev’s Inequality \(P(\mid X-\mu \mid \ge k )\ge \frac{\sigma^2}{k^2})\), find an upper bound for the following probabilities.
- \[P(\mid X-10\mid)\ge2)\]
#We know
mu <- 10
sigma_sq <- 100/3
sigma_sq / 2^2
## [1] 8.333333
- \[P(\mid X-10\mid)\ge5)\]
sigma_sq <- 100/3
sigma_sq / 5^2
## [1] 1.333333
- \[P(\mid X-10\mid)\ge9)\]
sigma_sq <- 100/3
sigma_sq / 9^2
## [1] 0.4115226
- \[P(\mid X-10\mid)\ge20)\]
sigma_sq <- 100/3
sigma_sq / 20^2
## [1] 0.08333333