91Ó°ÊÓ

First, let's consider \(f(x)\) as the arbitrary continuous density function. We will need to understand the conditions for a record value to occur, and then use that to characterize the process \(\\{ N(t), t \geqslant 0]\). To have a better understanding, let's take a look at the hint given: "There will be a record whose value is between \(t\) and \(t + dt\) if the first \(X_{i}\) that is greater than \(t\) lies between .....". This sentence brings our attention to the fact that for a record to occur between \(t\) and \(t + dt\), all previous observations should be less than or equal to \(t\). Hence, we can consider the probability of having each observation less than or equal to \(t\). Let's denote the cumulative distribution function (CDF) of the density function \(f(x)\) as \(F(x)\). Then, the probability of having each observation less than or equal to \(t\) would be \(F(t)\). As the random variables are independent, the joint probability of having the first \(n - 1\) observations less than or equal to \(t\) would be \((F(t))^{n-1}\). On the other hand, we want the nth observation to fall between \(t\) and \(t+dt\), which could be represented by the probability density function \(f(x)\) evaluated at \(x=t\), i.e., \(f(t)dt\). Hence, for a record value to occur between \(t\) and \(t+dt\), the probability would be given by \((F(t))^{n-1}f(t)dt\).

Now, we'll characterize the process \(\\{N(t), t \geqslant 0]\) in terms of probabilities. The desired probability that there are \(n\) record values less than or equal to \(t\) is the sum of probabilities for \(n\) record values lying between \(0\) and \(t\) with respect to all possible positions in \(X_1, X_2, ..., X_n\). Therefore, the function \(N(t)\) can be expressed as: \[N(t)= \sum_{n=1}^{\infty}P(T_n \leq t)\] where \(T_n\) represents the time when the \(n\)-th record value occurs. To find the probability that \(T_n \leq t\), we can use the probability we calculated in Step 1, i.e., \((F(t))^{n-1}f(t)dt\). Therefore, we can now express \(N(t)\) as: \[N(t) = \sum_{n=1}^{\infty} \int_0^t (F(x))^{n-1} f(x) dx\] This expression characterizes the process of \(N(t)\) for a case with an arbitrary continuous density function \(f(x)\).

Now, let's analyze the case where the density function is given by \(f(x)=\lambda e^{-\lambda x}\). The corresponding CDF would be given by \(F(x)= 1 - e^{-\lambda x}\). We can substitute this density function and CDF in the expression of \(N(t)\), which we derived in Step 2: \[N(t) = \sum_{n=1}^{\infty} \int_0^t ((1 - e^{-\lambda x}))^{n-1} (\lambda e^{-\lambda x}) dx\] To further analyze and simplify this expression, we can differentiate \(N(t)\) with respect to \(t\) and compute its expected value. It's left as an exercise to reach the final expression for \(N(t)\) in terms of \(t\) and \(\lambda\).