A company buys 100 light bulbs, each of which has an exponential lifetime of 1000 hours. What is the expected time for the first of these bulbs to burn out?
From Q10. we know that for \(n\) independent variables, which has an exponential density with mean \(\mu\), the mean of the minimum value of any of the independent variable is \(\frac{\mu}{n}\)
Where; \[ \begin{aligned} \mu &= 1000 \: hours \\ n &= 100 \: lightbulbs \end{aligned} \]
Therefore the expected time for the first bulb to burn out: \[ \begin{aligned} E(x) &= \frac{\mu}{n} \\ &= \frac{1000}{100} \\ &= 10 \: hours \end{aligned} \]
Assume that X1 and X2 are independent random variables, each having an exponential density with parameter \(\lambda\). Show that Z = X1 - X2 has density fZ(z) = (1/2)\(\lambda\)e^(\(-\lambda\)|z|) :
Density function of an exponential distribution for a random variable \(X\) and rate parameter \(\lambda\) is \[ \begin{aligned} f_X(x) = \begin{cases} \lambda e^{-\lambda x} \: &{for} \: x \geq 0 \\ 0 \: &for \: x < 0 \end{cases} \end{aligned} \]
Using convolution of the sum two independent variables: \[ \begin{aligned} f_{X_1 + X_2}(z) &= \int_{-\infty}^{\infty}f_{X_1}(x)f_{X_2}(z-x)dx \end{aligned} \]
Therefore for \(Z = X_1 - X_2\) \[ \begin{aligned} f_{Z}(z) &= \int_{-\infty}^{\infty}f_{X_1}(x)f_{X_2}(x-z)dx \end{aligned} \] Given the properties of exponential distributions we have to evaluate for \(z \geq 0\) and \(z < 0\)
For \(z \geq 0\): \[ \begin{aligned} f_{Z}(z) &= \int_{0}^{\infty}\lambda e^{-\lambda x}\lambda e^{-\lambda (x-z)} dx \\ &= \lambda^{2}e^{\lambda z} \int_{0}^{\infty} e^{-2\lambda x}dx \\ &= \lambda^{2}e^{\lambda z}\begin{bmatrix}\frac{-1}{2 \lambda}e^{-2\lambda x} \end{bmatrix}_{0}^{\infty} \\ &= \frac{\lambda}{2}e^{\lambda z} \end{aligned} \]
For \(z < 0\): \[ \begin{aligned} f_{Z}(z) &= \int_{0}^{\infty}\lambda e^{-\lambda x}\lambda e^{-\lambda (x+z)} dx \\ &= \lambda^{2}e^{-\lambda z} \int_{0}^{\infty} e^{-2\lambda x}dx \\ &= \lambda^{2}e^{-\lambda z}\begin{bmatrix}\frac{-1}{2 \lambda}e^{-2\lambda x} \end{bmatrix}_{0}^{\infty} \\ &= \frac{\lambda}{2}e^{-\lambda z} \end{aligned} \]
Therefore: \[ \begin{aligned} f_Z(z) &= \frac{\lambda}{2}e^{-\lambda |z|} \end{aligned} \]
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.
Note, from Chebyshev’s Inequality: \[ \begin{aligned} P(|X - \mu| \geq k) \leq \frac{\sigma^2}{k^2} \end{aligned} \]
\[ \begin{aligned} P(|X - 10| \geq 2) & \leq \frac{\frac{100}{3}}{2^2} \\ &= \frac{100}{12} \\ &= \frac{25}{3} \end{aligned} \]
\[ \begin{aligned} P(|X - 10| \geq 5) & \leq \frac{\frac{100}{3}}{5^2} \\ &= \frac{100}{75} \\ &= \frac{4}{3} \end{aligned} \]
\[ \begin{aligned} P(|X - 10| \geq 9) & \leq \frac{\frac{100}{3}}{9^2} \\ &= \frac{100}{243} \end{aligned} \]
\[ \begin{aligned} P(|X - 10| \geq 20) & \leq \frac{\frac{100}{3}}{20^2} \\ &= \frac{100}{1200} \\ &= \frac{1}{12} \end{aligned} \]