\[ U = 1000 \] \[ \lambda = \frac{1}{u} = \frac{1}{1000} \]
\[ If\ X_i ∼ exponential(λi), for\ i = 1, 2, . . . , n,\ and\ X_1, X_2, . . . , X_n\ are\ mutually\ independent\ random\ variables,\ then \\ min\lbrace{X_1, X_2, . . . , X_n}\rbrace ~ ∼ exponential \left( \displaystyle\sum\limits_{i=0}^n \lambda_i\right) \]
\[ \lambda = \displaystyle\sum\limits_{i=0}^{100} \frac{1}{1000} = \frac{1}{10} = 0.1 \]
\[ Expected\ time = \frac{1}{\lambda} = \frac{1}{0.1} = 10\ hours\]
\[ Given\ \int_{-\infty}^\infty f_X(x)f_Y(z-x)\mathrm{d}x\] \[Z = X + (-Y)\] \[f_{-Y}(z-x) = f_{Y}(x-z)\]
\[f_Z(z) = \int_{-\infty}^\infty f_X(x)f_{-Y}(z-x)\mathrm{d}x \\= \int_{-\infty}^\infty f_X(x)f_{Y}(x-z)\mathrm{d}x\]
\[ Case 1: Z<0\\ f_Z(z) = \int_{0}^\infty \lambda \mathrm{e}^{-\lambda x} \lambda \mathrm{e}^{-\lambda (x-z)} \mathrm{d}x \\ = \lambda \mathrm{e}^{\lambda z} \int_{0}^\infty \lambda \mathrm{e}^{-2\lambda x} \mathrm{d}x \\ = \lambda \mathrm{e}^{\lambda z} \left(\frac{-1}{2}\mathrm{e}^{-2\lambda x}|_0^\infty\right) \\ = \frac{\lambda}{2}\mathrm{e}^{\lambda z} \]
\[ Case 2: Z \geqslant 0 \\ f_{Z}(z) = f_{Z}(-z)\\ Therefore, \frac{\lambda}{2}\mathrm{e}^{-\lambda z}\]
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.
\[ Using\ Chebyshev’s\ Inequality\ P ((X - \mu)^2 \geqslant t^2) = P(|X - \mu| \geqslant t ) \leqslant \frac{\sigma}{t^2} \]
\[ Var(X) = \frac{100}{3} \]
\[ P(|X − 10| ≥ 2) \leqslant \frac{100}{3*4} = \frac{100}{12} > 1 \]
\[ P(|X − 10| ≥ 5) \leqslant \frac{100}{3*25} = \frac{100}{75} > 1 \]
\[ P(|X − 10| ≥ 9) \leqslant \frac{100}{3*81} = \frac{100}{243} \approx 0.41 \]
\[ P(|X − 10| ≥ 20) \leqslant \frac{100}{3*400} = \frac{100}{1200} \approx 0.083 \]