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?
In this question we are interested in the lifetime of an individual lightbulb.
If \(\lambda\) represents the rate parameter of the distribution then the mean of the exponential distribution is \(\frac{1}{\lambda}\) or the inverse. This means we expect 1 lightbulb to last 1000 hours before burning out.
We are only interested in how long it would take for the first bulb to burn out. The likelyhood of a lightbulb burning out is not affected by time passed nor is it affected by the burning out of other bulbs, so we can add up the mean of all of the bulbs to get the number of hours on average for a the first bulb to burn out.
\[ 100 * \frac{1}{1000} \]
\(10\) hours
Assume that X1 and X2 are independent random variables, each having an exponential density with parameter λ. Show that \(Z = X1 − X2\) has density \(fZ(z) = \frac{1}{2}\lambda e^{-\lambda|z|}\)
\[Z = X + (-Y) \]
Convolutional formula:
\[ f_z(Z) = \int_{-\infty}^{\infty} f_x(X)f_{-y}(Z-X) \,dx \] We can rewrite \(f_{-y}\) as:
\[ f_{-y}(Z-X) = f_y(X-Z)\] So we get:
\[ f_z(Z) = \int_{-\infty}^{\infty} f_x(X)f_y(X-Z) \,dx \]
We have two cases to consider: \(Z<0\) and \(Z>=0\)
Integrating:
\[ f_z(Z) = \int_{0}^{\infty} \lambda e^{-\lambda x} \lambda e^{-\lambda (x-z)} \,dx \]
\[ = \lambda e^{\lambda z} \int_{0}^{\infty} \lambda e^{-2\lambda x} \,dx \] \[ = \lambda e^{\lambda z} (\frac{1}{2} e^{-2\lambda x}) = \frac{1}{2}\lambda e^{\lambda z} \\z<0\]
For the case where z >= 0 we can, instead of integrating again, use the fact that if \(Z=X-Y\) then \(-Z=Y-X\)
Since X and Y are independent and identically distributed, then Z and -Z have symmetric ditributions around 0.
Thus \(f(z)\) = \(f(-z)\)
And so: \[ f_z(Z) = \frac{1}{2}\lambda e^{\lambda z}, z<0 \\ \frac{1}{2}\lambda e^{-\lambda z}, z>=0 \]
Thus:
\[ f_z(Z) = \frac{1}{2}\lambda e^{-\lambda |z|} \] ## Question 1 p320
Let X be a continuous random variable with mean µ = 10 and variance σ^2 = 100/3. Using Chebyshev’s Inequality, find an upper bound for the following probabilities: