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Chapter 6: Problem 5

Let \(Y_{(1)}<\cdots0, \lambda>0 .\) Show that for \(r=2, \ldots, n\) $$ \operatorname{Pr}\left(Y_{(r)}>y \mid Y_{(1)}, \ldots, Y_{(r-1)}\right)=\exp \left\\{-\lambda r\left(y-y_{(r-1)}\right)\right\\}, \quad y>y_{(r-1)} $$ and deduce that the order statistics from a general continuous distribution form a Markov process.

Short Answer

Expert verified
The conditional probability is \( \exp(-\lambda r(y - y_{(r-1)})) \), showing the order statistics form a Markov process.

Step by step solution

01

Understand the Exponential Distribution

Recall that the PDF of an exponential random variable with rate \( \lambda \) is given by \( f(y) = \lambda e^{-\lambda y} \) for \( y > 0 \). This implies that the exponential distribution is memoryless, and each time interval is independent of the elapsed time.
02

Define the Condition in Terms of Order Statistics

We are asked to find the conditional probability \( \operatorname{Pr}(Y_{(r)} > y \mid Y_{(1)}, \ldots, Y_{(r-1)}) \), where \( Y_{(i)} \) are the order statistics. The conditions imply we have information on the first \( r-1 \) smallest observations and need the probability regarding the \( r^{th} \) smallest.
03

Use the Property of Exponential Variables

Since the exponential distribution is memoryless, for \( y > y_{(r-1)} \), the distribution of \( Y_{(r)} - y_{(r-1)} \) given \( Y_{(1)}, \ldots, Y_{(r-1)} \) is still exponential with rate \( \lambda (n-r+1) \). Here the sample size isn鈥檛 changing, but the effective sample size does because of the conditioning on order statistics.
04

Calculate the Conditional Probability

The conditional density of \( Y_{(r)} \) given \( Y_{(r-1)} \) is given by \( f(y \mid y_{(r-1)}) = \lambda (r)(e^{-\lambda r(y - y_{(r-1)})}) \). Thus, the conditional probability is calculated as the survival function: \[\operatorname{Pr}(Y_{(r)} > y \mid Y_{(r-1)}) = \int_{y}^{\infty} \lambda r e^{-\lambda r(t - y_{(r-1)})} \, dt = e^{-\lambda r(y - y_{(r-1)})}.\]
05

Deduce the Markov Property

The Markov property implies that future states depend only on the current state, not the sequence of events that preceded it. Here, \( Y_{(r)} \) only depends on \( Y_{(r-1)} \) and not on \( Y_{(1)}, \ldots, Y_{(r-2)} \). This shows that the sequence of order statistics forms a Markov chain because each \( Y_{(r)} \) only depends on the immediately preceding \( Y_{(r-1)} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Exponential Distribution
The exponential distribution is a continuous probability distribution often used to model the time between events in a Poisson process. It is characterized by a constant rate or rate parameter \( \lambda \), meaning that events occur continuously and independently at a constant average rate.
- The probability density function (PDF) of an exponential distribution is given by \( f(y) = \lambda e^{-\lambda y} \) for \( y > 0 \).
- One important property is its memorylessness, which means the probability of an event occurring in the future is independent of the past, given the present. This is formally described as: \[ \operatorname{Pr}(Y > t + s \mid Y > s) = \operatorname{Pr}(Y > t). \]
- Due to this property, exponential distributions are often used in scenarios where waiting time is required between independent events, such as in telecommunications or service systems.
Markov Process
A Markov process is a stochastic process that satisfies the Markov property, which states that the future state of the process only depends on the present state, not on how it arrived at that state.
- This "memorylessness" aligns closely with the similar feature of the exponential distribution.
- In the context of order statistics, each subsequent statistic \( Y_{(r)} \) depends only on the immediate prior statistic \( Y_{(r-1)} \). This was demonstrated in our solution where the probability \( \operatorname{Pr}(Y_{(r)} > y \mid Y_{(1)}, \ldots, Y_{(r-1)}) \) simplified to depend solely on \( Y_{(r-1)} \).
- This property is fundamental, as it simplifies many computations in various fields such as queueing theory, and financial mathematics where complex systems can be reduced to simpler components.
Conditional Probability
Conditional probability refers to the likelihood of an event or outcome occurring based on the occurrence of a previous event or condition.
- In mathematical terms, the conditional probability of \( A \) given \( B \) is \( \operatorname{Pr}(A \mid B) = \frac{\operatorname{Pr}(A \cap B)}{\operatorname{Pr}(B)} \), provided \( \operatorname{Pr}(B) > 0 \).
- In the solved exercise, we computed the conditional probability of an order statistic \( Y_{(r)} \) exceeding a value, conditioned on the previous order statistics \( Y_{(1)}, \ldots, Y_{(r-1)} \).
- This illustrates how conditional probability is used extensively in statistical methodologies to make predictions or explanations based on known conditions, from reliability engineering to weather forecasting.

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Most popular questions from this chapter

Consider two binary random variables with local characteristics $$ \begin{aligned} &\operatorname{Pr}\left(Y_{1}=1 \mid Y_{2}=0\right)=\operatorname{Pr}\left(Y_{1}=0 \mid Y_{2}=1\right)=1 \\ &\operatorname{Pr}\left(Y_{2}=0 \mid Y_{1}=0\right)=\operatorname{Pr}\left(Y_{2}=1 \mid Y_{1}=1\right)=1 \end{aligned} $$ Show that these do not determine a joint density for \(\left(Y_{1}, Y_{2}\right) .\) Is the positivity condition satisfied?

A Poisson process of rate \(\lambda(t)\) on the set \(\mathcal{S} \subset \mathbb{R}^{k}\) is a collection of random points with the following properties (among others): \- the number of points \(N_{\mathcal{A}}\) in a subset \(\mathcal{A}\) of \(\mathcal{S}\) has the Poisson distribution with mean \(\Lambda(\mathcal{A})=\int_{\mathcal{A}} \lambda(t) d t\) \- given \(N_{\mathcal{A}}=n\), the positions of the points are sampled randomly from the density \(\lambda(t) / \int_{\mathcal{A}} \lambda(s) d s, t \in \mathcal{A}\) (a) Assuming that you have reliable generators of \(U(0,1)\) and Poisson variables, show how to generate the points of a Poisson process of constant rate \(\lambda\) on the interval \(\left[0, t_{0}\right]\). (b) Let \(t=(x, y) \in \mathbb{R}^{2}, \eta, \xi \in \mathbb{R}, \tau>0, \lambda(x, y)=\tau^{-1}\\{1+\xi(y-\eta) / \tau\\}^{-1 / \xi-1}\). Give an algorithm to generate realisations from the Poisson process with rate \(\lambda(x, y)\) on $$ \mathcal{S}=\\{(x, y): 0 \leq x \leq 1, y \geq u, \lambda(x, y)>0\\}. $$ $$ \begin{array}{rrrrrrrrrrrrrrrrr} \hline 9 & 12 & 11 & 4 & 7 & 2 & 5 & 8 & 5 & 7 & 1 & 6 & 1 & 9 & 4 & 1 & 3 \\ 3 & 6 & 1 & 11 & 33 & 7 & 91 & 2 & 1 & 87 & 47 & 12 & 9 & 135 & 258 & 16 & 35 \\\ \hline \end{array} $$

Consider a Poisson process of intensity \(\lambda\) in the plane. Find the distribution of the area of the largest disk centred on one point but containing no other points.

Let \(Y_{1}, \ldots, Y_{n}\) represent the trajectory of a stationary two-state discrete-time Markov chain, in which $$ \operatorname{Pr}\left(Y_{j}=a \mid Y_{1}, \ldots, Y_{j-1}\right)=\operatorname{Pr}\left(Y_{j}=a \mid Y_{j-1}=b\right)=\theta_{b a}, \quad a, b=1,2 $$ note that \(\theta_{11}=1-\theta_{12}\) and \(\theta_{22}=1-\theta_{21}\), where \(\theta_{12}\) and \(\theta_{21}\) are the transition probabilities from state 1 to 2 and vice versa. Show that the likelihood can be written in form \(\theta_{12}^{n_{12}}\left(1-\theta_{12}\right)^{n_{11}} \theta_{21}^{n_{21}}\left(1-\theta_{21}\right)^{n_{22}}\), where \(n_{a b}\) is the number of \(a \rightarrow b\) transitions in \(y_{1}, \ldots, y_{n}\). Find a minimal sufficient statistic for \(\left(\theta_{12}, \theta_{21}\right)\), the maximum likelihood estimates \(\widehat{\theta}_{12}\) and \(\widehat{\theta}_{21}\), and their asymptotic variances.

Let \(Y^{\mathrm{T}}=\left(Y_{1}, \ldots, Y_{3}\right)\) be a multivariate normal variable with $$ \Omega=\left(\begin{array}{ccc} 1 & m^{-1 / 2} & \frac{1}{2} \\ m^{-1 / 2} & \frac{2}{m} & m^{-1 / 2} \\ \frac{1}{2} & m^{-1 / 2} & 1 \end{array}\right). $$ Find \(\Omega^{-1}\) and hence write down the moral graph for \(Y\). If \(m \rightarrow \infty\), show that the distribution of \(Y\) becomes degenerate while that of \(\left(Y_{1}, Y_{3}\right)\) given \(Y_{2}\) remains unchanged. Is the graph an adequate summary of the joint limiting distribution? Is the Markov property stable in the limit?
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