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?
Based on gamma distribution, the expected value = alpha * beta
= number of exponentials * rate
= 1 / (100 * 1/1000)
= 1/(1/10)
= 10
Assume that \(X_{1}\) and \(X_{2}\) are independent random variables, each having an exponential density with parameter λ. Show that Z = \(X_{1}\) - \(X_{2}\) has density fZ(z) = (1/2)λe−λ|z|
Let \(X_{1}\) and \(X_{2}\) are two independent exponential Random VariableS(rvs) with parameter \(\lambda\)
Probability Density Formula(pdf) of \(x_{1}\), \(f(x_{1})\) = \(\lambda e^-\lambda x_{1}\) where \(x_{1}\) >= 0
Probability Density Formula(pdf) of \(x_{2}\), \(f(x_{2})\) = \(\lambda e^-\lambda x_{2}\) where \(x_{2}\) >= 0
Formula for Joint density function of \(X_{1}\) and \(X_{2}\) is \(f(x_{1}, x_{2})\) = \(f(x_{1})\) \(f(x_{2})\) = \(\lambda e^-\lambda x_{1}\) x \(\lambda e^- \lambda x_{2}\) = $ ^2 e^ - (x_{1} + x_{2})$
Let Z = \(X_{1}\) - \(X_{2}\)
V = \(X_{2}\)
In other way
\(x_{1}\) = z + v
\(x_{2}\) = v
Jacobian value (J) = \(\partial(x_{1}, x_{2})\) / \(\partial(x_{1}, x_{2})\)
= \(| ^1 _0 \ ^1 _1 |\)
= 1
Joint pdf of Z and V becomes: \(g(z,v)\) = \(\lambda ^2\) x \(e^-\lambda (x + 2v)\), v > 0, $-$ < z < \(\infty\)
z = \(x_{1}\) - v
v = \(x_{1}\) - z
v > -z, if \(-\infty\) < z < 0
v > 0, if z > 0
For \(-\infty\) < x < 0,
\(f(z)\) = \(\int_{-\infty}^{\infty}\) g(z,v) \(dv\)
= \(\int_{-\infty}^{\infty}\) \(\lambda^2\) x \(e^-\lambda(z+2v)\) \(dv\)
= \(\lambda^2\) x \(e^-\lambda z\) $ | (e^-2v)/$ \(-2\lambda _x ^\infty\)
= \((\lambda / 2)\) \(e^(\lambda z)\)
#1) 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
(a) P(|X − 10| ≥ 2).
(b) P(|X − 10| ≥ 5).
(c) P(|X − 10| ≥ 9).
(d) P(|X − 10| ≥ 20).
Given \(\mu = 10\), \(\sigma^2\) = 100/3
\(\sigma = 5.77\)
Using Chebyshev’s inequality,
a) P(\(|x - \mu| >= k \sigma\) ) <= 1/\(k^2\)
\(k\sigma = 2\)
k = \((2/\sigma)\)
= 2/5.77 = 0.3466
Therefore, \(P(|x-10| >= 2)\) <= 1/\(0.3466^2\) = 8.3242
\(k\sigma = 5\) k= 5/5.77 = 0.87
P(|x-10| >= 25) \(<=\) 1/\(0.87^2\)
= 1.3212
p(|x-10| >= 9) \(<=\) \(1/k^2\)
\(k\sigma = 9\) k = 9/5.77
= 1.56
P(|x-10|>= 9) \(<=\) \(1/1.56^2\)
= 0.4109
P(|x-10| >= 20) \(<=\) \(1/k^2\)
\(k\sigma\) = 20
k = 20/5.77 = 3.47
P(|x-10|>= 20) \(<=\) \(1/3.47^2\) = 0.0831