Each of the 100 lightbulbs has an exponential lifetime with a mean of 1000 hours (\(\mu = 1000\)), implying a rate \(\lambda = \frac{1}{1000} = 0.001\) hour\(^{-1}\).
We are interested in finding the expected time until the first lightbulb fails, which involves dealing with the minimum of a set of exponential random variables as in Exercise 10.
The key property of exponential distributions in this context is that if you have \(n\) independent exponential random variables \(X_1, X_2, \ldots, X_n\) with the same rate \(\lambda\), the minimum of these variables, \(M = \min(X_1, X_2, \ldots, X_n)\), is also exponentially distributed, but with rate \(n\lambda\).
The expected value (mean) of an exponential distribution with rate \(\lambda\) is \(\frac{1}{\lambda}\). Therefore, for the minimum \(M\) of \(n\) such random variables, its expected value would be:
\[ E(M) = \frac{1}{n\lambda} \]
Substituting the given values into the formula:
\[ E(M) = \frac{1}{100 \times 0.001} = \frac{1}{0.1} = 10 \text{ hours} \]
To show that \(Z = X_1 - X_2\) has the density \(f_Z(z) = \frac{1}{2}\lambda e^{-\lambda|z|}\) when \(X_1\) and \(X_2\) are independent exponential random variables with the same parameter \(\lambda\), let’s include some integration steps.
Given \(X_1, X_2 \sim \text{Exponential}(\lambda)\), their PDFs are \(f_{X_1}(x) = \lambda e^{-\lambda x}\) and \(f_{X_2}(x) = \lambda e^{-\lambda x}\), for \(x \geq 0\).
Since \(X_1\) and \(X_2\) are independent random variables, we use the formula for the convolution of their density functions:
\[ f_Z(z) = \int_{-\infty}^{\infty} f_{X_1}(x) f_{X_2}(x - z) \, dx \]
Given the exponential distribution’s support is only for \(x \geq 0\), we adjust our limits accordingly and consider the absolute value of \(z\) to account for the symmetry of the situation (since \(X_1 - X_2\) could be negative).
For \(z \geq 0\), \(X_1\) must be greater than \(X_2\) by at least \(z\). Considering the transformation and the range of integration where both densities contribute (\(x \geq z\) for \(X_1\) and \(x-z\) for \(X_2\)):
\[ f_Z(z) = \lambda^2 \int_{z}^{\infty} e^{-\lambda x} e^{-\lambda (x-z)} \, dx \] \[ = \lambda^2 e^{\lambda z} \int_{z}^{\infty} e^{-2\lambda x} \, dx \] \[ = \lambda^2 e^{\lambda z} \left[ -\frac{1}{2\lambda} e^{-2\lambda x} \right]_{z}^{\infty} \] \[ = \frac{1}{2}\lambda e^{-\lambda z} \]
For \(z < 0\), the situation is symmetric, but \(X_2\) is greater than \(X_1\). The absolute value of \(z\) accounts for this symmetry. Thus, integrating from \(0\) to infinity, considering \(x\) for \(X_1\) and \(x+|z|\) (or equivalently \(x-z\) for negative \(z\)) for \(X_2\):
\[ f_Z(z) = \lambda^2 \int_{0}^{\infty} e^{-\lambda x} e^{-\lambda (x-|z|)} \, dx \] \[ = \lambda^2 e^{\lambda |z|} \int_{0}^{\infty} e^{-2\lambda x} \, dx \] \[ = \lambda^2 e^{\lambda |z|} \left[ -\frac{1}{2\lambda} e^{-2\lambda x} \right]_{0}^{\infty} \] \[ = \frac{1}{2}\lambda e^{-\lambda |z|} \]
Both cases, \(z \geq 0\) and \(z < 0\), yield the same result after integration: Then,
\[ f_Z(z) = \frac{1}{2}\lambda e^{-\lambda |z|} \]