11)
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? (See Exercise 10.).
Answer:
Since the problem is interested in the “first of these bulbs” I am interpreting this as the minimum of these exponential RVs \(\min\{X_1, X_2, ...,X_n\}\) where \(X_i\) exponential RV’s with CDF and PDF:
\[F_{X_i}(x) = P(X_i \leq x)= 1-\lambda e^{-\lambda x}\] \[f_{X_i}(x) = \lambda e^{-\lambda x}\]
The probability density function \(f_{min}(x)\) is then given by using the pdf and the cdf via:
\[f_{min}(x)=nf_{X_i}(x)(1-F_{X_i}(x))^{n-1}\]
Then \(f_{min}(x)=n\lambda e^{-\lambda x}(1-(1-\lambda e^{-\lambda x}))^{n-1}\)
\(f_{min}(x)=n\lambda e^{-\lambda x}(1-(1-\lambda e^{-\lambda x}))^{n-1}=n\lambda e^{-n\lambda x}\)
The last form: \(f_{min}(x)=n\lambda e^{-n\lambda x}\) appears exponential in nature if we rearrange and let \(n\lambda = \lambda_{min}\) gives: \(f_{min}(x)=\lambda_{min} e^{-\lambda_{min} x}\).
Since we know the expectation of \(f_{X_i}(x) = \lambda e^{-\lambda x}\) is \(\frac{1}{\lambda}\) the expectation of \(f_{min}(x)\) would be: \(\frac{1}{\lambda_{min}}=\frac{1}{n \lambda}=\frac{1}{n\frac{1}{\mu}}=\frac{\mu}{n}\)
The expectation for the minimum of the exponential RV’s is then:
\(\therefore E[X_i]=\frac{\mu}{n}=1000/100=10\) hours is the mean failure of the minimums.
14)
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:
\[f_Z(z)=\frac{\lambda}{2} e^{-\lambda|z|}\]
Answer:
Given two random exponential variables: \(X_1\) and \(X_2\), the sum of these two variables is then \(Z=X_1+X_2\) which has density:
\[f_Z(z)=\int_{-\infty}^\infty f_{X_1}(x)f_{X_2}(z-x)dx\]
but this problem asks to find \(Z=X_1-X_2\) so I’ll rearrange the convolution formula such that \(Z=X_1+{(-X_2)}\) to yield:
\[f_Z(z)=\int_{-\infty}^\infty f_{X_1}(x)f_{(-X_2)}(z-x)dx=\]
(Note that \(f_{(-X_2)}(z-x)=f_{X_2}(x-z)\) since \(P(-X_2 = -1) = P(X_2 = 1)\) then I get):
\[\int_{-\infty}^\infty f_{X_1}(x)f_{X_2}(x-z)dx\]
Since the value of the \(z\) is absolute, I’m confident I don’t have to check both ranges of values but let’s see how this breaks down:
For \(z < 0\):
\[=\int_{0}^\infty \lambda e^{-\lambda x} \lambda e^{-\lambda (x-z)} dx\]
\[= \lambda e^{-\lambda z} \int_{0}^\infty \lambda e^{-2\lambda x} dx\]
\[= \lambda e^{-\lambda z} \Big(-\frac{1}{2} e^{-2\lambda x}\Big \rvert_{0}^{\infty} )\]
Then this yields: \(f_Z(z)= \frac{\lambda}{2} e^{-\lambda z}\) for \(z < 0\).
Before checking \(z \geq 0\), I note that since \(Z=X_1-X_2 \implies -Z=X_2-X_1\) because both have identical distributions. This tells me that the distribution of \(Z\) is symmetric about zero which would imply, for example, that: \(f_Z(1)=f_Z(-1)\). This tells me that for \(z \geq 0\), \(f_Z(z)=f_Z(-z)\)
\(f_{Z}(x) = \begin{cases} \frac{\lambda}{2} e^{\lambda z}, & z < 0 \\ \frac{\lambda}{2} e^{-\lambda z}, & z \geq 0 \end{cases} \implies f_{Z}(x) =\frac{\lambda}{2} e^{-\lambda |z|}\)
The wikipedia page and has a good discussion of the Convolution. MIT OpenCourseWare. discussion of using the convolution here.
1)
1 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.
- \(P(|X − 10| ≥ 2)\)
- \(P(|X − 10| ≥ 5)\)
- \(P(|X − 10| ≥ 9)\)
- \(P(|X − 10| ≥ 20)\)
Theorem 8.3 (Chebyshev Inequality) Let \(X\) be a continuous random variable with density function \(f(x)\). Suppose \(X\) has a finite expected value \(\mu = E(X)\) and finite variance \(\sigma^2 = V(X)\). Then for any positive number \(\epsilon > 0\) we have
\[P(|X-\mu|\geq \epsilon) \leq \frac{\sigma^2}{\epsilon^2}\]
Answer:
a):
\(P(|X − 10| ≥ 2) \leq \frac{\sigma^2}{\epsilon^2} \rightarrow \frac{(100/3)}{2^2}= 25/3\)
b):
\(P(|X − 10| ≥ 5) \leq \frac{\sigma^2}{\epsilon^2} \rightarrow \frac{(100/3)}{5^2}= 4/3\)
c):
\(P(|X − 10| ≥ 9) \leq \frac{\sigma^2}{\epsilon^2} \rightarrow \frac{(100/3)}{9^2}= 100/243\)
c):
\(P(|X − 10| ≥ 20) \leq \frac{\sigma^2}{\epsilon^2} \rightarrow \frac{(100/3)}{20^2}= 1/12\)