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Chapter 5: Problem 90

A circular sampling region with radius \(X\) is chosen by a biologist, where \(X\) has an exponential distribution with mean value \(10 \mathrm{ft}\). Plants of a certain type occur in this region according to a (spatial) Poisson process with "rate" \(.5\) plant per square foot. Let \(Y\) denote the number of plants in the region. a. Find \(E(Y \mid X=x)\) and \(V(Y \mid X=x)\) b. Use part (a) to find \(E(Y)\). c. Use part (a) to find \(V(Y)\).

Short Answer

Expert verified
a. \(E(Y \mid X=x) = 0.5\pi x^2\), \(V(Y \mid X=x) = 0.5\pi x^2\). b. \(E(Y) = 100\pi\). c. \(V(Y) = 100\pi + 100\pi^2\).

Step by step solution

01

Understand the given information

We are given that the radius \(X\) of a circular sampling region follows an exponential distribution with a mean of 10 ft. The number of plants \(Y\) in this region follows a Poisson process with a rate of 0.5 plants per square foot.
02

Derive the formula for the area of the circle

The area \(A\) of a circle with radius \(X\) is given by \(A = \pi X^2\).
03

Calculate the conditional expectation \(E(Y \mid X=x)\)

For a Poisson process with rate \(\lambda = 0.5\) plants per square foot, the expected number of plants is the product of the area and \(\lambda\): \[E(Y \mid X = x) = 0.5 \cdot \pi x^2.\]
04

Calculate the conditional variance \(V(Y \mid X=x)\)

In a Poisson distribution, the variance equals the mean. Thus, \[V(Y \mid X = x) = 0.5 \cdot \pi x^2.\]
05

Find \(E(Y)\) using the Law of Total Expectation

According to the law of total expectation, \(E(Y) = E[E(Y \mid X)]\). Substituting from Step 3, we need \(E[0.5 \pi X^2]\). Since \(X\) is exponentially distributed with mean 10, and the expected value of \(X^2\) for an exponential distribution with rate 0.1 is \(1/(0.1^{2})\): \[E(X^2) = \frac{2}{0.1^2} = 200,\] resulting in \[E(Y) = 0.5 \pi \times 200 = 100 \pi.\]
06

Find \(V(Y)\) using the Law of Total Variance

The law of total variance states: \(V(Y) = E[V(Y \mid X)] + V[E(Y \mid X)]\). From Step 4, \(E[V(Y \mid X)] = E[0.5 \pi X^2] = 100\pi\). Let's compute \(V[E(Y \mid X)] = V[0.5 \pi X^2]\), using \(V(X^2) = \frac{2}{0.1^4}\): \[V(0.5\pi X^2) = (0.5\pi)^2 \cdot 400 = 100\pi^2,\] resulting in \[V(Y) = 100\pi + 100\pi^2.\]

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Key Concepts

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

Poisson Process
A Poisson process is a fundamental concept in probability theory used to model events occurring randomly over time or space. Imagine it like raindrops falling on a surface or phone calls arriving at a call center.
  • The key aspect of a Poisson process is that events occur independently of each other.
  • It is used to describe the likelihood of a given number of events happening within a predetermined space or time interval.
  • The process is characterized by a constant average rate, denoted by \( \lambda \). This rate is the expected number of events per unit space or time.
In the exercise above, plants occur in a region at a rate of 0.5 plants per square foot. This means, on average, you'd expect to see 0.5 plants in each square foot of the chosen area.
Conditional Expectation
Conditional expectation combines the concept of a random variable dependent on another variable. It provides the expected value of a variable given some conditions are met. This essentially means, predicting an average outcome based on a scenario already determined.
In the problem, we calculate the conditional expectation of the number of plants \( Y \) given the radius \( X \) of the circular region.
  • Given \( X = x \), the area of the circle is \( \pi x^2 \).
  • For the Poisson rate of \( 0.5 \), the expected number of plants in this area can be calculated as \( E(Y \mid X = x) = 0.5 \cdot \pi x^2 \).
This step is crucial because it sets the stage for understanding how the number of plants scales with the size of the region sampled.
Law of Total Expectation
The Law of Total Expectation is all about breaking down complex expectations into simpler parts. Think of it as using smaller chunks to solve a larger puzzle, making things manageable and straightforward.
  • This law states that the expected value of a random variable can be found by averaging the expectations conditioned on another variable.
  • In the problem, we use it to find \( E(Y) \) by taking \( E[E(Y \mid X)] \).
  • After substituting known values and calculations from previous steps, we find \( E(Y) = 0.5 \cdot \pi \cdot 200 = 100 \pi \).
This principle is particularly powerful in terms of simplifying calculations over complex random scenarios by breaking them down into more manageable pieces.
Variance Calculation
Understanding variance is essential to grasping the notion of data spread or variability. Variance helps us understand how much the number of plants deviates from the expected value.
  • In a Poisson distribution, the variance equals the mean, which was found to be \( 0.5 \cdot \pi x^2 \) when conditioned on \( X = x \).
  • The Law of Total Variance allows us to calculate \( V(Y) \) by using both the conditional variance and the variance of the conditional expectations.
  • This results in \( V(Y) = E[V(Y \mid X)] + V[E(Y \mid X)] = 100\pi + 100\pi^2 \).
The process emphasizes both the contribution of the inherent randomness of the Poisson process and the variability introduced by the random selection of the radius.

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Most popular questions from this chapter

Let \(X\) represent a measurement error. It is natural to assume that the pdf \(f(x)\) is symmetric about 0 , so that the density at a value \(-c\) is the same as the density at \(c\) (an error of a given magnitude is equally likely to be positive or negative). Consider a random sample of \(n\) measurements, where \(n=\) \(2 k+1\), so that \(Y_{k+1}\) is the sample median. What can be said about \(E\left(Y_{k+1}\right)\) ? If the \(X\) distribution is symmetric about some other value, so that value is the median of the distribution, what does this imply about \(E\left(Y_{k+1}\right)\) ? [Hints: For the first question, symmetry implies that \(1-F(x)=P(X>x)=P(X<-x)=F(-x) .\) For the second question, consider \(W=X-\tilde{\mu}\); what is the median of the distribution of \(W ?]\)

You have two lightbulbs for a particular lamp. Let \(X=\) the lifetime of the first bulb and \(Y=\) the lifetime of the second bulb (both in 1000 's of hours). Suppose that \(X\) and \(Y\) are independent and that each has an exponential distribution with parameter \(\lambda=1\). a. What is the joint pdf of \(X\) and \(Y\) ? b. What is the probability that each bulb lasts at most \(1000 \mathrm{~h}\) (i.e., \(X \leq 1\) and \(Y \leq 1\) )? c. What is the probability that the total lifetime of the two bulbs is at most 2? [Hint: Draw a picture of the region \(A=\\{(x, y): x \geq 0\), \(y \geq 0, x+y \leq 2\\}\) before integrating.] d. What is the probability that the total lifetime is between 1 and 2 ?

Let \(X_{1}\) denote the time (hr) it takes to perform a first task and \(X_{2}\) denote the time it takes to perform a second one. The second take always takes at least as long to perform as the first task. The joint pdf of these variables is \(f\left(x_{1}, x_{2}\right)=\left\\{\begin{array}{cl}2\left(x_{1}+x_{2}\right) & 0 \leq x_{1} \leq x_{2} \leq 1 \\ 0 & \text { otherwise }\end{array}\right.\) a. Obtain the pdf of the total completion time for the two tasks. b. Obtain the pdf of the difference \(X_{2}-X_{1}\) between the longer completion time and the shorter time.

Suppose that \(X\) and \(Y\) are independent rv's with moment generating functions \(M_{X}(t)\) and \(M_{Y}(t)\), respectively. If \(Z=X+Y\), show that \(M_{Z}(t)=M_{X}(t) M_{Y}(t)\). [Hint: Use the proposition on the expected value of a product.]

The number of individuals arriving at a post office to mail packages during a certain period is a Poisson random variable \(X\) with mean value 20 . Independently of the others, any particular customer will mail either \(1,2,3\), or 4 packages with probabilities \(.4, .3, .2\), and .1, respectively. Let \(Y\) denote the total number of packages mailed during this time period. a. Find \(E(Y \mid X=x)\) and \(V(Y \mid X=x)\). b. Use part (a) to find \(E(Y)\). c. Use part (a) to find \(V(Y)\).
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