Given that:
Exponential lifetime = 1000 hours and n = 100 (number of lightbulbs)
From Exercise 10, for n independent random variables that have an exponential density and mean μ, the minimum value M is exponential density with a mean as \(\frac{μ}{n}\).
For any of the bulbs i, let \(X_i\) = its independent random variable => \(E[X_i]=\frac{1}{\lambda_i}=1000\) = the lifetime expectancy of bulb i.
\(\lambda_i=\frac{1}{1000}\)
Also, we have \(min{X_1,X_2,...,X_100} \sim exponential(\sum _{i=1} ^ {100} \lambda_i)\) because \(X_i\) is given as exponential
=> \(\sum_{i=1} ^ {100} \lambda_i=100×\frac{1}{1000}=\frac{1}{10}\)
=> \(E[minXi]= \frac{1}{\frac{1}{10}} = \frac{10}{1}=10 hours\)
Show that \(Z = X_1 − X_2\) has density \(f_Z(z) = (\frac{1}{2})e^{−\lambda|z|}\)
First,
\(f(x_1)=\lambda e^{−\lambda x_1}\) \(f(x_2)=\lambda e^{−\lambda x_2}\)
can be used to compute the pdf for the variables \(X_1\) and \(X_2\)
=> \(\lambda e^{−\lambda x_1}* \lambda e^{−λ_{x2}} = \lambda ^2e^{−\lambda(x_1+x_2)}\)
Since \(Z=X_1−X_2\) => \(x_1=z+x_2\)
Substituting the above, we can get the joint density of Z and \(X_2\) as \(\lambda ^2 e^{−\lambda((z+x_2)+x_2)} = \lambda ^2 e^{−\lambda(z+2x_2)}\).
From Z as shown above, x2=x1−z => when z is negative, \(x_2 \gt -z\); if z is positive, \(x_2\) will also be positive
Therefore, if z is negative => \(\int^∞_−z λ^2 e^{−\lambda (z+2x_2)}dx =\frac {\lambda}{2} e^{\lambda z}\)
When z is positive, => \(\int ^∞_0 \frac{\lambda}{2}e^{−\lambda(z+2x_2)}dx=\frac{\lambda}{2}e^{−\lambda |z|}\).
=> \(f_Z(z)=\frac{1}{2}\lambda e^{λ|z|}\)
Let X be a continuous random variable with mean μ = 10 and variance \(\sigma ^2 = \frac{100}{3}\). Using Chebyshev’s Inequality, find an upper bound for the following probabilities.
Chebyshev’s Inequality is given by \(P(|X−\mu| \geq k\sigma) \leq \frac{1}{k^2}\)
Given that \(\sigma ^2 = \frac{100}{3}\) => \(\sigma = \sqrt{\frac{100}{3}} = \frac{10}{\sqrt{3}}\)
\((a) P(|X − 10| \geq 2)\)
=> \(k\sigma=2\)
Therefore \(k*\frac{10}{\sqrt{3}}=2 => k=\frac{2\sqrt{3}}{10}\) \(k^{-2}=\frac{1}{(\frac{2\sqrt{3}}{10})^2} = \frac{1}{0.12} = 10.12 = 8.333 = 1\) since highest value of probability is 1.
\((b) P(|X − 10| \geq 5)\) \(kσ=5\) => \(k\frac{10}{\sqrt{3}}=5 => k=\frac{\sqrt{3}}{2}\) \(k^{-2}=\frac{1}{(\frac{\sqrt{3}}{2})^2} = \frac{4}{3}=1.333 = 1\) since highest value of probability is 1
\((c) P(|X − 10| \geq 9)\)
=> \(k\sigma=9\) \(k\frac{10}{\sqrt{3}}=9 => k=\frac{9\sqrt{3}}{2}\) \(k^{-2}=\frac{1}{(\frac{9\sqrt{3}}{2})^2} = \frac{1}{2.43}=0.412\)
\((d) P(|X − 10| \geq 20)\)
\(kσ=20\) => \(k\frac{10}{\sqrt{3}}=20 => k=2\sqrt{3}\) \(k^{-2}=\frac{1}{(2\sqrt{3})^2} = \frac{1}{12}= 0.083\)