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Determine the value of \(c\) that makes the function \(f(x, y)=c(x+y)\) a joint probability density function over the range \(01)\) (d) \(P(X<2, Y<2)\) (e) \(E(X)\) (f) \(E(Y)\) (g) Marginal probability distribution of \(X\) (h) Conditional probability distribution of \(Y\) given \(X=1\) (i) \(E(Y \mid X=1)\) (j) \(P(Y>2 \mid X=1)\) (k) Conditional probability distribution of \(X\) given \(Y=2\)

Short Answer

Expert verified
Ensure the PDF integrates to 1 with \( c = \frac{2}{27} \). For each part, apply the integration over specified ranges to find probabilities or expectations.

Step by step solution

01

Verify Joint PDF

For a function to be a joint probability density function, the integral over its entire range should be 1. We need to find the constant \( c \) for which \( f(x, y) = c(x+y) \) is a joint PDF over the specified region. Integrate it twice over the region where \( 0 < y < x < 3 \).
02

Integrate over y

Integrate \( c(x+y) \) with respect to \( y \) from \( 0 \) to \( x \). The integral is: \[\int_{0}^{x} c(x+y) \, dy = c\left[ xy + \frac{y^2}{2} \right]_0^x = c \left( x^2 + \frac{x^2}{2} \right) = c \frac{3x^2}{2}.\]
03

Integrate over x

Integrate the result from Step 2 with respect to \( x \) from \( 0 \) to \( 3 \): \[\int_{0}^{3} c \frac{3x^2}{2} \, dx = c \cdot \frac{3}{2} \left( \frac{x^3}{3} \right)_0^3 = c \cdot \frac{3}{2} \cdot \frac{27}{3} = 27c/2.\]
04

Solve for c

Set the integral equal to 1 to solve for \( c \): \[\frac{27c}{2} = 1 \quad \Rightarrow \quad c = \frac{2}{27}.\]
05

Calculate (a) P(X

Calculate the probability \( P(X<1, Y<2) \) by integrating \( f(x, y) = \frac{2}{27}(x+y) \) over the region \( 0<y<x<1 \): \[\int_{0}^{1} \int_{0}^{x} \frac{2}{27}(x+y) \, dy \, dx.\]
06

Calculate (b) P(1

Calculate the probability \( P(1<X<2) \) by integrating the joint PDF over the specified range: \[\int_{1}^{2} \int_{0}^{x} \frac{2}{27}(x+y) \, dy \, dx.\]
07

Calculate (c) P(Y>1)

Calculate \( P(Y>1) \) by integrating the joint PDF over \( 1 < y < x < 3 \): \[\int_{1}^{3} \int_{y}^{3} \frac{2}{27}(x+y) \, dx \, dy.\]
08

Calculate (d) P(X

Calculate \( P(X<2, Y<2) \) over \( 0 < y < x < 2 \): \[\int_{0}^{2} \int_{0}^{x} \frac{2}{27}(x+y) \, dy \, dx.\]
09

Calculate (e) E(X)

The expected value \( E(X) \) is computed using \( \int \int x \cdot f(x, y) \, dy \, dx \) within the specified limits: \[E(X) = \int_{0}^{3} \int_{0}^{x} x \cdot \frac{2}{27}(x+y) \, dy \, dx.\]
10

Calculate (f) E(Y)

Calculate \( E(Y) \) using the same limits: \[E(Y) = \int_{0}^{3} \int_{0}^{x} y \cdot \frac{2}{27}(x+y) \, dy \, dx.\]
11

Marginal Distribution of X

Find the marginal distribution of \( X \) by integrating the joint PDF over \( y \): \[f_X(x) = \int_{0}^{x} \frac{2}{27}(x+y) \, dy.\]
12

Conditional Distribution of Y given X=1

Conditional distribution \( f_{Y|X}(y|X=1) \) is found by integrating over the range where \( 0<y<1 \): \[f_{Y|X}(y|1) = \frac {\frac{2}{27}(1+y)}{\frac{1}{6}}\]
13

Calculate (i) E(Y | X=1)

Use the conditional PDF from Step 12 to calculate \( E(Y | X=1) \): \[E(Y|X=1) = \int_{0}^{1} y \cdot f_{Y|X}(y|1) \, dy.\]
14

Calculate (j) P(Y>2 | X=1)

Under the condition \( X=1 \), \( y \) ranges from \( 0 \) to \( 1 \). Since \( P(Y>2|X=1) = 0 \) after normalized integration for a single point as range is wrong.
15

Conditional Distribution of X given Y=2

Find \( f_{X|Y}(x|2) \) using the conditional formula with limits \( 2<x<3 \): \[f_{X|Y}(x|2) = \frac{\frac{2}{27}(x+2)}{\frac{5}{81}}\]

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Key Concepts

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

Marginal Probability Distribution
In probability theory, marginal probability distributions help determine the probability of a single random variable occurring, independent of other variables. When dealing with joint probability density functions (PDFs), marginal distributions are found by integrating over the other variables.

For example, if you have a joint PDF such as \( f(x, y) \), to find the marginal probability density function of \( X \), denoted as \( f_X(x) \), you integrate \( f(x, y) \) over all possible values of \( y \).

This ensures that you're considering every possible value for \( y \) and focusing solely on \( X \). By doing so, you effectively 'ignore' the influence of \( Y \) and concentrate on the distribution of \( X \) alone. This process can greatly simplify analyses when one is only interested in particular variables, not their joint behaviors.
Conditional Probability
Conditional probability allows us to determine the probability of an event occurring given that another event has already occurred. This is often represented as \( P(A|B) \), which reads "the probability of \( A \) given \( B \)."

In the context of continuous variables and joint PDFs, we use conditional probability density functions. Suppose \( f(x,y) \) is a joint PDF, then the conditional probability density function of \( Y \) given \( X = x \) is obtained by dividing the joint PDF by the marginal PDF of \( X \):

\[ f_{Y|X}(y|x) = \frac{f(x,y)}{f_X(x)}. \]

This reflects how likely \( y \) is when we know the specific condition of \( x \) occurring, thus revealing dependencies between the variables. Understanding this can be critical in fields like statistics, data analysis, and artificial intelligence.
Expected Value
An expected value is essentially the average outcome you expect if an experiment, or scenario, was repeated many times. This is sometimes referred to as the "center" or "mean" of a distribution.

For a continuous probability distribution, such as the ones dealt with in joint PDFs, the expected value of a random variable \( X \), denoted \( E(X) \), is found by integrating the product of the variable and its probability density function. The formula is given by:
\[ E(X) = \int x f_X(x) \, dx. \]

For joint PDFs, the expected value of \( X \) is given by:
\[ E(X) = \int \int x \, f(x,y) \, dy \, dx. \]

The integration takes into account the distribution of \( X \) across its entire range, thus weighing each possible value of \( X \) by its probability. Expected values play a fundamental role in decision-making areas such as finance and risk management.
Integration in Probability
Integration in probability is vital for solving various problems involving continuous random variables. It is used to find probabilities, marginal distributions, and expected values.

Put simply, integration is the mathematical process that helps sum up infinitely small probabilities over a range of values. When working with continuous variables, you often deal with probability densities rather than discrete probabilities. Therefore, integrating over certain ranges on these densities allows us to compute the probability of events within those ranges.

Here’s how it applies:
  • **Probability Calculation**: To find the probability of a random variable within an interval, integrate the PDF over that interval.
  • **Marginal Distribution**: To determine the marginal PDF of one variable, integrate the joint PDF over the potential values of the other variable(s).
  • **Expected Value**: Calculate the expected value by integrating the product of the value and its PDF.
Developing strong integration skills within the context of probability will enable you to solve complex real-world problems more effectively.

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Most popular questions from this chapter

Determine the value of \(c\) that makes the function \(f(x, y)=c(x+y)\) a joint probability mass function over the nine points with \(x=1,2,3\) and \(y=1,2,3\) Determine the following: (a) \(P(X=1, Y<4)\) (b) \(P(X=1)\) (c) \(P(Y=2)\) (d) \(P(X<2, Y<2)\) (e) \(E(X), E(Y), V(X)\), and \(V(Y)\) (f) Marginal probability distribution of \(X\) (g) Conditional probability distribution of \(Y\) given that \(X=1\) (h) Conditional probability distribution of \(X\) given that \(Y=2\) (i) \(E(Y \mid X=1)\) (j) Are \(X\) and \(Y\) independent?

An article in Clinical Infectious Diseases ["Strengthening the Supply of Routinely Administered Vaccines in the United States: Problems and Proposed Solutions" (2006, Vol.42(3), pp. S97-S103)] reported that recommended vaccines for infants and children were periodically unavailable or in short supply in the United States. Although the number of doses demanded each month is a discrete random variable, the large demands can be approximated with a continuous probability distribution. Suppose that the monthly demands for two of those vaccines, namely measles-mumps-rubella (MMR) and varicella (for chickenpox), are independently, normally distributed with means of 1.1 and 0.55 million doses and standard deviations of 0.3 and 0.1 million doses, respectively. Also suppose that the inventory levels at the beginning of a given month for MMR and varicella vaccines are 1.2 and 0.6 million doses, respectively. (a) What is the probability that there is no shortage of either vaccine in a month without any vaccine production? (b) To what should inventory levels be set so that the probability is \(90 \%\) that there is no shortage of either vaccine in a month without production? Can there be more than one answer? Explain.

Suppose that \(X\) and \(Y\) are independent continuous random variables. Show that \(\sigma_{X Y}=0 .\)

In the transmission of digital information, the probability that a bit has high, moderate, or low distortion is 0.01 , \(0.04,\) and \(0.95,\) respectively. Suppose that three bits are transmitted and that the amount of distortion of each bit is assumed to be independent. Let \(X\) and \(Y\) denote the number of bits with high and moderate distortion of the three transmitted, respectively. Determine the following: (a) Probability that two bits have high distortion and one has moderate distortion (b) Probability that all three bits have low distortion (c) Probability distribution, mean, and variance of \(X\) (d) Conditional probability distribution, conditional mean, and conditional variance of \(X\) given that \(Y=2\)

Suppose that \(X\) and \(Y\) have a bivariate normal distribution with joint probability density function \(f_{X Y}\left(x, y ; \sigma_{X}, \sigma_{Y}, \mu_{X}, \mu_{Y}, \rho\right)\) (a) Show that the conditional distribution of \(Y\) given that \(X=x\) is normal. (b) \(\quad\) Determine \(E(Y \mid X=x)\). (c) \(\quad\) Determine \(V(Y \mid X=x)\).

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