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.)
n=100
lifetime=1000
lifetime/n## [1] 10Assume 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
\(f_Z(z) = (1/2)λe^{−λ|z|}\)
Expanding the solution
\[f_Z(z) = λ/2e^{−λ|z|} \; \begin{cases} λ/2e^{−λz} & z >= 0 \\ λ/2e^{λz} & z < 0 \\ \end{cases}\]
let \(X = X_1 \& Y = X_2 \rightarrow Z=X-Y \rightarrow Y = X - Z\)
For \(z >= 0 \rightarrow x \in (z , \infty)\)
\[ \int_{z}^{\infty}\lambda e ^{-\lambda x} \lambda e ^ {-\lambda(x-z)}dx = \lambda^2\int_{z}^{\infty} e ^{\lambda(z - 2x)} dx \] Using u substitution
\(u = \lambda * (z - 2x) \rightarrow du=-2\lambda dx\)
Simplify \[(-\frac{\lambda}{2}e^ {z - 2x}|_z^{\infty}) = \frac{\lambda}{2}e^{-\lambda z}\] For \(z < 0 \rightarrow x \in (0 , \infty)\)
\[(-\frac{\lambda}{2}e^ {z - 2x}|_0^{\infty}) = \frac{\lambda}{2}e^{\lambda z}\]
Let X be a continuous random variable with mean \(\mu = 10\) and variance \(σ^2 = 100/3\). Using Chebyshev’s Inequality, find an upper bound for the following probabilities
\(\sigma = 5.77\)
\(P (|X − 10| ≥ 2) \rightarrow 2 = k\sigma \rightarrow k=0.346 \rightarrow 1/k^2 = 0.12\)
P (|X − 10| ≥ 5)
std = sqrt(100/3)
k = 5/std
1/k^2## [1] 1.333333k = 9/std
1/k^2## [1] 0.4115226k = 20/std
1/k^2## [1] 0.08333333