91Ó°ÊÓ

Since we know that \(X_0=x\), we can condition the expected value on \(X_1\). That is, we will calculate the expected value based on two cases: 1. If \(X_1 > X_0\) 2. If \(X_1 < X_0\) We have: \[ f(x) = E[N] = \int_0^x E[N | X_1 = u] du \] Notice that: E[N | X_1 = u] = \begin{cases} 1, & \text{if } u < x \\ E[N], & \text{if } u > x \end{cases} Thus: \begin{aligned} f(x) &= \int_0^x E[N | X_1 = u] du \\ & = \int_0^x 1 du + \int_x^1 E[N] du \\ & = x + (1-x)f(x) \end{aligned} This is the integral equation for \(f(x)\).

Now we need to differentiate both sides of this equation with respect to \(x\). We have: \begin{aligned} \frac{d}{dx}f(x) &= \frac{d}{dx}(x + (1-x)f(x)) \\ f'(x) &= 1 + (1-x)f'(x) - f(x) \end{aligned}

In order to solve this equation, we will rearrange it: \[ f'(x) - (1-x)f'(x) = 1 - f(x) \] Now, we can express this equation as a first-order linear differential equation: \[ f'(x) = \frac{1-f(x)}{1-x} \] This equation can be solved by multiplying both sides by the integrating factor, which is \(e^{\int (1-x)dx} = e^{x^2/2}\). Then, we have: \[ e^{x^2/2} f'(x) = \frac{e^{x^2/2} (1-f(x))}{1-x} \] Now, integrate both sides with respect to \(x\): \[ f(x) = e^{-x^2/2} \int \frac{e^{x^2/2}(1-f(x))}{1-x} dx + C \] The constant C can be determined by using the boundary condition: \(f(0) = 1\) \[ 1 = e^0 \int \frac{e^{0}(1-f(0))}{1-0} d0 + C \Rightarrow C = 1 \] Thus, the function \(f(x)\) can be expressed as: \[ f(x) = e^{-x^2/2} \int \frac{e^{x^2/2}(1-f(x))}{1-x} dx + 1 \]

We are given that: \[ P\{N \geqslant k\}=\frac{(1-x)^{k-1}}{(k-1) !} \] This probability represents the probability that there have been at least \(k-1\) values in the sequence that are greater than their preceding values. Since we have i.i.d. uniform random variables in the sequence, the probability that \(k-1\) are greater than their predecessors is a direct consequence of the memoryless property.

To find \(f(x)\), we know that the expected value is given by: \[ f(x) = E[N] = \sum_{k=1}^{\infty}kP\{N=k\} \] Since we have \(P\{N \geqslant k\}\) and we know that probabilities sum to 1, we can find \(P\{N=k\}\): \[ P\{N=k\} = P\{N \geqslant k\} - P\{N \geqslant k+1\} = \frac{(1-x)^{k-1}}{(k-1) !} - \frac{(1-x)^{k}}{k !} \] Then, we compute f(x): \begin{aligned} f(x) &= \sum_{k=1}^{\infty}k \left( \frac{(1-x)^{k-1}}{(k-1) !} - \frac{(1-x)^{k}}{k !} \right) \\ & = \sum_{k=1}^{\infty} \left( k \frac{(1-x)^{k-1}}{(k-1) !} - (k-1) \frac{(1-x)^{k-1}}{(k-1) !} \right) \\ & = \sum_{k=1}^{\infty} \frac{(1-x)^{k-1}}{(k-1) !} \left( k - (k-1) \right) \\ & = \sum_{k=1}^{\infty} \frac{(1-x)^{k-1}}{(k-1) !} \\ & = e^{-x^2/2}+1 \end{aligned} Thus, the expected value of \(N\), given by \(f(x)\), is: \[ f(x) = e^{-x^2/2}+1 \]