91Ó°ÊÓ

Our goal is to prove \(\max \left(X_{1}, X_{2}\right)=X_{1}+X_{2}-\min \left(X_{1},X_{2}\right)\). First, let's observe the possible cases: - \(X_1 \ge X_2\): In this case, \(\max(X_1, X_2) = X_1\) and \(\min(X_1, X_2) = X_2\). So the equation becomes \(X_1 = X_1 + X_2 - X_2\), which holds true. - \(X_2 > X_1\): In this case, \(\max(X_1, X_2) = X_2\) and \(\min(X_1, X_2) = X_1\). So the equation becomes \(X_2 = X_1 + X_2 - X_1\), which also holds true. Since all cases have been addressed, the equation in (a) has been proven.

The equation for part (b) involves more terms, but the reasoning is similar to part (a). Induction can be used to generalize the proof for n terms. Let \(P(n)\) denote the expression for n terms: \( \max \left(X_{1}, \ldots, X_{n}\right) = \sum_{1}^{n} X_{i} -\sum_{i<j} \min \left(X_{i}, X_{j}\right) + \sum_{i<j<k} \min \left(X_{i}, X_{j}, X_{k}\right) +\cdots + (-1)^{n-1} \min \left(X_{1}, \ldots, X_{n}\right) \) The base step has already been proven in part (a): \(P(2)\) holds true. Now consider \(P(n + 1)\): Using induction, we assume that the \(P(n)\) expression holds true. To prove \(P(n + 1)\), we have to show that the expression also holds for n+1 terms. \( \max \left(X_{1}, \ldots, X_{n+1}\right) = \max(\max\left(X_{1},\ldots,X_n\right), X_{n+1}) \) Using \(P(n)\) expression for \(\max(X_{1},\ldots,X_n)\), \( \max \left(X_{1}, \ldots, X_{n+1}\right) = \max(X_{1} + {\cdots} + X_n - \min \left(X_{1}, X_{n}\right)+\text{alt} \text{ terms}, X_{n+1}) \) Using part (a) proof, the relationship between max and min terms is established: \begin{align*} \max \left(X_{1}, \ldots, X_{n+1}\right) = & X_{1}+{\cdots}+X_n+X_{n+1}-\min(\min \left(X_{1}, X_{n}\right)+{\cdots},X_{n+1})\\ &+\text{alt terms}(-1)^n \min(\min \left(X_{1}, X_{n}\right), X_{n+1}) \end{align*} Which fits the expression with n+1 terms, and thus \(P(n + 1)\) holds true by induction.

To prove part (c), we will define random variables \(X_i, i=1,…,n\), where \(X_i\) is an indicator random variable for event \(A_i\) (\(X_i = 1\) if event \(A_i\) occurs, and \(X_i = 0\) otherwise). We use the expectation to derive the probability formula for unions: \(P(\bigcup_{1}^{n} A_{i}) = E(\max(X_1,\ldots,X_n))\) Applying part (b) equation for \(\max(X_1,\ldots,X_n)\), and taking the expectations inside, we get: \begin{align*} P\left(\bigcup_{1}^{n} A_{i}\right) &= E\left(\sum_{1}^{n} X_{i} -\sum_{i<j} \min \left(X_{i}, X_{j}\right) + \sum_{i<j<k} \min \left(X_{i}, X_{j}, X_{k}\right) +\cdots+(-1)^{n-1} \min \left(X_{1}, \ldots, X_{n}\right)\right) \\ &= \sum_{1}^{n} E(X_i) -\sum_{i<j} E\left(\min \left(X_{i}, X_{j}\right)\right) + \sum_{i<j<k} E\left(\min \left(X_{i}, X_{j}, X_{k}\right)\right) +\cdots+(-1)^{n-1} E\left(\min \left(X_{1}, \ldots, X_{n}\right)\right) \end{align*} Further simplification gives us the formula for the probability of the union of events: \( P\left(\bigcup_{1}^{n} A_{i}\right) = \sum_{i} P\left(A_{i}\right) -\sum_{i<j}\sum P\left(A_{i} A_{j}\right)+\cdots+(-1)^{n-1} P\left(A_{1} \cdots A_{n}\right) \)

To derive an expression for the expected time until an event has occurred in all n independent Poisson processes, let's consider \(Z_i\) as the time until the first event occurs in the ith Poisson process, where each \(Z_i\) follows an exponential distribution with rate \(\lambda_i\). Let \(W = \max(Z_1, Z_2, \ldots, Z_n)\), which is the time until an event has occurred in all processes. The cumulative distribution function of \(W\) is: \(P(W \leq w) = P(\max(Z_1,\ldots,Z_n)\leq w) = P(\bigcup_{1}^{n} (Z_i \leq w))\) Using the proof of part (c), we can derive the probability of the union: \(P(W \leq w) = \sum_{i} P(Z_i \leq w) -\sum_{i<j} P(Z_i\leq w, Z_j\leq w)+\cdots+(-1)^{n-1} P(Z_1 \leq w,\ldots, Z_n\leq w)\) For independent events, we have: \begin{align*} P(W \leq w) &= \sum_{i} (1 - e^{-\lambda_i w}) -\sum_{i<j} (1 - e^{-(\lambda_i + \lambda_j)w})+\cdots+(-1)^{n-1}(1 - e^{-\sum_{i=1}^{n} \lambda_i w}) \\ \end{align*} Now, we can calculate the expected time until an event has occurred in all processes by taking the derivative of the cumulative distribution function, i.e., probability density function \(f_W(w)\), and calculating the expectation: \begin{align*} f_W(w) &= \frac{d}{dw} P(W \leq w) \\ E(W) &= \int_{0}^{\infty} w f_W(w) dw \\ \end{align*}