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.)
We know from problem 10 that the density of the distribution \(M\), where \(M\) is the minimum value of each in a series of exponential functions, is itself exponential with a mean of \(\frac{\mu}{n}\). Therefore, the expected value (or mean) of the failure of the earliest lightbulb is \(\frac{\mu}{n}=\frac{1000}{100}=10\:hours\).
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 \(f_Z(z) = (1/2)\lambda e−\lambda|z|\)
The exponential probability density function is given by the following formula:
\[ f(x;\lambda)=\lambda e^{-\lambda x} \] In order to find the distribution for \(X_1 − X_2\), we must combine the independent probabilities of \(X_1\) and \(X_2\) using convolution. Convolution for the difference between two distributions is similar to that of addition, and is given by (I’ll be using \(X\) instead of \(X_1\) and \(Y\) in place of \(X_2\) for simplicity):
\[ f_Z(z) = \int_{-\infty}^{\infty}f_X(x)f_Y(z+x)dx \] Plugging in our exponential probability functions (using \(\phi\) as the rate perameter for the \(Y\) distribution), we get the following. Also note that we change the lower limit from \(-\infty\) to \(0\), as exponential functions are defined only defined for \(0\) and above.
\[ f_Z(z) = \int_{0}^{\infty}\lambda e^{-\lambda x} * \phi e^{-\phi (x+z)} \] Truth be told, I am not yet sure how to simplify an integral beyond this point.
Let \(X\) be a 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 \(P(∣X−\mu∣ ≥ kσ)≤ \frac{1}{k^2}\). That means the upper bound in each case is equal to \(\frac{1}{k^2}\), where \(k\) is a positive number.
Since the right side of the inequality is expressed as \(kσ\), or \(k\) times the standard deviation, it will be useful to determine the standard deviation (since finding \(k\), or the number of standard deviations away from the mean, will then be a matter of division). The standard deviation \(\sigma\) is given by the square root of the variance.
\[ \sigma = \sqrt{\frac{100}{3}} = \frac{10}{\sqrt{3}} = \frac{10\sqrt{3}}{3} \approx 5.77 \]
Given that \(2 = \sigma k\) and \(\sigma = 5.77\), \(k=2/\sigma=0.347\). Therefore, the probability in this case is given by \(\frac{1}{k^2}=\frac{1}{0.347^2}=1/0.12=8.33\). The probability should be less than \(1\), so this may be invalid.
By the same logic above, \(5 = \sigma k\) and \(\sigma = 5.77\), so \(k=5/\sigma=0.867\). The probabilty is given by \(\frac{1}{k^2}=\frac{1}{0.867^2}=1/0.75=1.33\). As above, the probability must be less than or equal to \(1\) to be valid.
Here, \(9 = \sigma k\) and \(\sigma = 5.77\), so \(k=9/\sigma=1.56\). The probabilty is given by \(\frac{1}{k^2}=\frac{1}{1.567^2}=1/2.43=0.412\). So the upper bound for the probability is around 41.2%.
As above, \(20 = \sigma k\) and \(\sigma = 5.77\), so \(k=20/\sigma=3.47\). The probabilty is given by \(\frac{1}{k^2}=\frac{1}{3.47^2}=1/12.04=0.083\). So the upper bound for the probability is around 8.3%.