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.)
We have 100 Light bulbs with an expoential lifetime of 1000 hours.
This means that the Expected lifetime of the light bulbs is 1000 hours: \(E[X]=1000\) hours.
Then, since \(E[X]=1000\) this means that \(\lambda= \frac{1}{1000}\)
From problem 10 we know that the mean of the minimum of the independent random variables of an exponential is \(\frac{\mu}{n}\) or \(\frac{\lambda}{n}\)
We know that \(n = 100\) and \(\lambda=1000\)
So, by solving for this with the mean we will figure out the expected lifetime for the first bulb to run out.
\(E[firstLightBulbBurnOut]=\frac{1000}{10}=10\) 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 \(fZ(z)=(\frac{1}{2})\lambda e^{-\lambda |z|}\)
Sine we know \(X_1,X_2\) are exponential we know that the pdfs are:
\(X_1 = f(x_1) = \lambda e^{-\lambda x_1}\)
\(X_2 = f(x_2) = \lambda e^{-\lambda x_2}\)
Then to find the pdf of Z we can use convolution \(f_z(z)=\int_{-\infty}^{\infty} f_1(x_1) f_2(x_1-z)dx\)
Substituting in \(X_1,X_2\) we get \(f_z(z)= \int_{-\infty}^{\infty} \lambda e^{-\lambda x_1} * \lambda e^{-2\lambda x_1 - \lambda z}\)
This can be simplified to \(\lambda^{2}e^{\lambda z} \int_{-\infty}^{\infty} e^{-2\lambda x_1} dx_1\)
For when \(z \geq 0\)
\(f_z(z) = \lambda^{2}e^{-\lambda z} \int_{0}^{\infty} e^{-2\lambda x_1} dx_1\) \(= \lambda^{2}e^{\lambda z} * \frac{1}{2 \lambda} = \frac{1}{2} * \lambda e ^{\lambda z}\)
Then for \(z < 0\):
\(f_z(z) = \lambda^{2}e^{-\lambda z} \int_{-z}^{\infty} e^{-2\lambda x_1} dx_1\) \(= \lambda^{2}e^{\lambda z} * \frac{1}{2 \lambda} = \frac{1}{2} * \lambda e ^{\lambda -z}\)
So, \(f_z(z) = \frac{1}{2} * \lambda e ^{\lambda |z|}\)
Let X be a continuous random variable with mean μ = 10 and variance σ2 = 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).
Since \(\sigma ^{2} = \frac{100}{3}\) then, \(\sigma = \sqrt{\sigma^{2}}=\frac{10}{\sqrt{3}}\)
Chebyshev’s: \(P(|X-\mu| \leq k\sigma)= \frac{1}{k^2}\)
\(P(|X-10| \geq 2)\)
\(k = \frac{2}{\frac{10}{\sqrt{3}}}=\frac{2\sqrt{3}}{10}\)
\(P(|X-10| \geq 2) = \frac{1}{(\frac{2\sqrt{3}}{10})^2}=\frac{100}{12}\)
\(P(|X-10| \geq 5)\)
\(k = \frac{5}{\frac{10}{\sqrt{3}}}=\frac{5\sqrt{3}}{10}\)
\(P(|X-10| \geq 5)= \frac{1}{(\frac{5\sqrt{3}}{10})^2}=\frac{100}{75}\)
\(P(|X-10| \geq 9)\)
\(k = \frac{9}{\frac{10}{\sqrt{3}}}=\frac{9\sqrt{3}}{10}\)
\(P(|X-10| \geq 9)= \frac{1}{(\frac{9\sqrt{3}}{10})^2}=\frac{100}{243}\)
\(P(|X-10| \geq 20)\)
\(k = \frac{20}{\frac{10}{\sqrt{3}}}=2\sqrt{3}\)
\(P(|X-10| \geq 20)= \frac{1}{(2\sqrt{3})^2}=\frac{1}{12}\)