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Suppose that the range of the continuous variables \(X\) and \(Y\) is \(0

Short Answer

Expert verified
Since the joint PDF can be written as a product of two marginal PDFs, \(X\) and \(Y\) are independent.

Step by step solution

01

Understand Independence of Random Variables

For two continuous random variables \(X\) and \(Y\), they are independent if and only if their joint probability density function (PDF) \(f_{XY}(x,y)\) can be expressed as a product of their marginal PDFs, \(f_X(x)\) and \(f_Y(y)\). So, we need to prove that \( f_{XY}(x, y) = f_X(x) \cdot f_Y(y) \).
02

Use Given Joint PDF Form

We are given that the joint PDF is \( f_{XY}(x, y) = g(x) h(y) \). This is already in the product form where \( g(x) \) depends only on \( x \) and \( h(y) \) depends only on \( y \). Our aim is now to compute and match this form with the marginal PDFs.
03

Find Marginal PDF of X

To find the marginal PDF \( f_X(x) \) of \(X\), we integrate the joint PDF over all values of \(Y\): \[ f_X(x) = \int_0^b f_{XY}(x, y)\, dy = \int_0^b g(x) h(y)\, dy.\]Since \(g(x)\) does not depend on \(y\), it comes out of the integral: \[ f_X(x) = g(x) \int_0^b h(y)\, dy.\]
04

Find Marginal PDF of Y

Similarly, to find the marginal PDF \( f_Y(y) \) of \(Y\), integrate the joint PDF over all values of \(X\): \[ f_Y(y) = \int_0^a f_{XY}(x, y)\, dx = \int_0^a g(x) h(y)\, dx.\]Here, \(h(y)\) does not depend on \(x\), so:\[f_Y(y) = h(y) \int_0^a g(x)\, dx.\]
05

Show the Independence

From Steps 3 and 4, plug back the integrals:\[f_X(x) \cdot f_Y(y) = \left( g(x) \int_0^b h(y)\, dy \right) \cdot \left( h(y) \int_0^a g(x)\, dx \right).\]Simplify:\[f_X(x) \cdot f_Y(y) = g(x) h(y) \left( \int_0^b h(y)\, dy \right) \left( \int_0^a g(x)\, dx \right).\]The product \( g(x) h(y) \) is the joint PDF, and we can assert that the constant integrals do not affect independence. Therefore, \( f_{XY}(x, y) = f_X(x) \cdot f_Y(y) \), showing \(X\) and \(Y\) are independent.

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Key Concepts

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

Joint Probability
Joint probability is a concept used to describe the probability of two events occurring simultaneously. In the context of continuous random variables, the joint probability is expressed through a joint probability density function (PDF). If we have two random variables, say \( X \) and \( Y \), the joint PDF is denoted as \( f_{XY}(x, y) \) and it describes the likelihood of \( X \) being around a specific value \( x \) and \( Y \) being around a value \( y \). It's like getting both random variables under one roof!
The joint PDF should not be confused with individual probabilities or marginal PDFs, which describe just one random variable at a time. Instead, the joint PDF gives us a holistic view of how \( X \) and \( Y \) interact or relate with each other. It's defined over a range of values for each variable, such as \( 0 < x < a \) and \( 0 < y < b \).
The exercise challenges us by starting with an expression of the joint PDF in a special form: \( f_{XY}(x, y) = g(x) h(y) \). Here, \( g(x) \) depends only on \( x \) and \( h(y) \) only on \( y \), hinting at the independence of the variables, but more on that later.
Marginal PDF
A marginal probability density function, or marginal PDF, is like looking at an individual piece of a larger puzzle. When you have a joint PDF representing a pair of random variables, the marginal PDF isolates one of the variables by considering the entire range of the other variable.
For instance, to find the marginal PDF of \( X \), denoted \( f_X(x) \), we integrate the joint PDF over all possible values of \( Y \). This integration is expressed as:

\[ f_X(x) = \int_0^b f_{XY}(x, y) \, dy = \int_0^b g(x) h(y) \, dy. \]
Since \( g(x) \) does not depend on \( y \), it comes out of the integral, simplifying our task to evaluating the remaining integral of \( h(y) \). Similarly, finding the marginal PDF of \( Y \), \( f_Y(y) \), involves integrating over all values of \( X \):

\[ f_Y(y) = \int_0^a f_{XY}(x, y) \, dx = \int_0^a g(x) h(y) \, dx. \]
The marginal PDF is crucial because it allows us to understand each variable individually within the joint distribution framework. The simpler parts of solving the puzzle! When dealing with independence, marginal PDFs help verify if two random variables can be considered separately or if they influence each other.
Independence of Random Variables
When we say two random variables \( X \) and \( Y \) are independent, it means the occurrence of one does not affect the probability of the occurrence of the other. In terms of probability density functions, \( X \) and \( Y \) are independent if their joint PDF can be expressed as the product of their marginal PDFs.
In mathematical terms, \( X \) and \( Y \) are independent if:

\[ f_{XY}(x, y) = f_X(x) \cdot f_Y(y). \]
In our problem, the joint PDF is \( f_{XY}(x, y) = g(x) h(y) \), where \( g(x) \) is only about \( x \) and \( h(y) \) only about \( y \). This already denotes a form suitable for independence. We demonstrate independence by showing:

- We can derive each marginal PDF and then
- Multiply those marginal PDFs to obtain the joint PDF.
Since the given joint PDF \( g(x) h(y) \) matches the form required for independence without further adjustments (except constant factors resulting from integration), we confirm \( X \) and \( Y \) are independent. They do not influence each other - they are like two separate channels broadcasting without interference.

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Most popular questions from this chapter

Suppose the random variables \(X, Y\), and \(Z\) have the following joint probability distribution. $$ \begin{array}{cccc} \hline x & y & z & f(x, y, z) \\ \hline 1 & 1 & 1 & 0.05 \\ 1 & 1 & 2 & 0.10 \\ 1 & 2 & 1 & 0.15 \\ 1 & 2 & 2 & 0.20 \\ 2 & 1 & 1 & 0.20 \\ 2 & 1 & 2 & 0.15 \\ 2 & 2 & 1 & 0.10 \\ 2 & 2 & 2 & 0.05 \\ \hline \end{array} $$ Determine the following: (a) \(P(X=2)\) (b) \(P(X=1, Y=2)\) (c) \(P(Z < 1.5)\) (d) \(P(X=1\) or \(Z=2)\) (e) \(E(X)\) (f) \(P(X=1 \mid Y=1)\) (g) \(P(X=1, Y=1 \mid Z=2)\) (h) \(P(X=1 \mid Y=1, Z=2)\) (i) Conditional probability distribution of \(X\) given that \(Y=1\) and \(Z=2\)

The joint probability distribution is $$ \begin{array}{llcll} x & -1 & 0 & 0 & 1 \\ y & 0 & -1 & 1 & 0 \\ f_{X Y}(x, y) & 1 / 4 & 1 / 4 & 1 / 4 & 1 / 4 \end{array} $$ Show that the correlation between \(X\) and \(Y\) is zero, but \(X\) and \(Y\) are not independent.

Four electronic ovens that were dropped during shipment are inspected and classified as containing either a major, a minor, or no defect. In the past, \(60 \%\) of dropped ovens had a major defect, \(30 \%\) had a minor defect, and \(10 \%\) had no defect. Assume that the defects on the four ovens occur independently. (a) Is the probability distribution of the count of ovens in each category multinomial? Why or why not? (b) What is the probability that, of the four dropped ovens, two have a major defect and two have a minor defect? (c) What is the probability that no oven has a defect? Determine the following: (d) The joint probability mass function of the number of ovens with a major defect and the number with a minor defect (e) The expected number of ovens with a major defect (f) The expected number of ovens with a minor defect (g) The conditional probability that two ovens have major defects given that two ovens have minor defects (h) The conditional probability that three ovens have major defects given that two ovens have minor defects (i) The conditional probability distribution of the number of ovens with major defects given that two ovens have minor defects (j) The conditional mean of the number of ovens with major defects given that two ovens have minor defects.

The conditional probability distribution of \(Y\) given \(X=x\) is \(f_{Y \mid x}(y)=x e^{-x y}\) for \(y > 0,\) and the marginal probability distribution of \(X\) is a continuous uniform distribution over 0 to 10 . (a) Graph \(f_{Y \mid X}(y)=x e^{-x y}\) for \(y > 0\) for several values of \(x\) Determine: (b) \(P(Y < 2 \mid X=2)\) (c) \(E(Y \mid X=2)\) (d) \(E(Y \mid X=x)\) (e) \(f_{X Y}(x, y)\) (f) \(f_{Y}(y)\)

A small company is to decide what investments to use for cash generated from operations. Each investment has a mean and standard deviation associated with the percentage gain. The first security has a mean percentage gain of \(5 \%\) with a standard deviation of \(2 \%,\) and the second security provides the same mean of \(5 \%\) with a standard deviation of \(4 \%\). The securities have a correlation of \(-0.5,\) so there is a negative correlation between the percentage returns. If the company invests two million dollars with half in each security, what are the mean and standard deviation of the percentage return? Compare the standard deviation of this strategy to one that invests the two million dollars into the first security only.

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