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?
There are a 100 lightbulbs and the first one that will burn out is the minimum of the exponential random variables. Sum the lambdas (rate parameters) to get this minimum’s lambda. All these lambdas have the same value so you can do 100 times lambda, which will give you the lambda for this minimum random variable.
\[ \lambda = 1000 \] hours \[ 100 * \lambda \] \[ 100 * 1/1000 = 1/10 \] But this quantity is not in hours, it is the reciprocal. So invert it to get the expected time for the first bulb to burn out, 10 hours.
Since lambda is the reciprocal of the mean, invert it to get the expected time.
PDFs of the random variables \(X_1\) and \(X_2\)
\[ f_{X_1}(x_1) = \lambda e^{-\lambda x_1} \] \[ f_{X_2}(x_2) = \lambda e^{-\lambda x_2} \]
Then we use convolution to find $ f_Z(z) $
\[ f_Z(z) = \int_{-\infty}^{\infty} f_{X_1}(x_1) \cdot f_{X_2}(z - x_1) \, dx_1 \]
Now you can plug those PDFs we got earlier into this equation.
\[ f_Z(z) = \int_{-\infty}^{\infty} \lambda e^{-\lambda x_1} \cdot \lambda e^{-\lambda (z - x_1)} \, dx_1 \]
\[ = \lambda^2 \int_{-\infty}^{\infty} e^{-\lambda x_1} e^{-\lambda z} e^{\lambda x_1} \, dx_1 \]
\[ = \lambda^2 e^{-\lambda z} \int_{-\infty}^{\infty} dx_1 \]
It’s from negative infinity to positive infinity, but we need to adjust these limits. There are two cases.
If \(z < 0\), then \(z - x_1 < 0\) for all \(x_1\) because both \(x_1\) and \(x_2\) are non-negative.
\[ f_Z(z) = \lambda^2 e^{-\lambda z} \int_{0}^{\infty} dx_1 \]
\[ = \lambda^2 e^{-\lambda z} \left[ -\frac{1}{\lambda} e^{-\lambda x_1} \right]_{0}^{\infty} \]
\[ = \frac{\lambda}{2} e^{-\lambda z} \]
The other case is when, if \(z \geq 0\), then \(z - x_1 \geq 0\) for all \(x_1\).
\[ f_Z(z) = \lambda^2 e^{-\lambda z} \int_{0}^{z} dx_1 \]
\[ = \lambda^2 e^{-\lambda z} \left[ -\frac{1}{\lambda} e^{-\lambda x_1} \right]_{0}^{z} \]
\[ = \lambda^2 e^{-\lambda z} \left( -\frac{1}{\lambda} e^{-\lambda z} + \frac{1}{\lambda} e^{0} \right) \]
\[ = \frac{\lambda}{2} e^{-\lambda z} \]
But they’re the same regardless of whether z is non-negative or negative, so we can use the absolute value operation on z.
\[ f_Z(z) = \frac{\lambda}{2} e^{-\lambda |z|} \]
\[ P(|X - \mu| \geq k\sigma) \leq \frac{1}{k^2} \]
\(k = \frac{\sqrt{3}}{5}\)
\[ P(|X - 10| \geq 2) \leq (\frac{5}{\sqrt{3}})^2 = \frac{25}{3}\]
\(k = \frac{\sqrt{3}}{2}\)
\[ P(|X - 10| \geq 5) \leq (\frac{2}{\sqrt{3}})^2 = \frac{4}{3}\]
\(k = \frac{{9}\sqrt{3}}{10}\)
\[ P(|X - 10| \geq 5) \leq (\frac{10}{9\sqrt{3}})^2 = \frac{100}{243}\]
\(k = {2}\sqrt{3}\)
\[ P(|X - 10| \geq 20) \leq (\frac{1}{2\sqrt{3}})^2 = \frac{1}{12}\]