/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 34 The random variables \(X\) and \... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 30: Problem 34

The random variables \(X\) and \(Y\) take integer values, \(x\) and \(y\), both \(\geq 1\), and such that \(2 x+y \leq 2 a\), where \(a\) is an integer greater than \(1 .\) The joint probability within this region is given by $$ \operatorname{Pr}(X=x, Y=y)=c(2 x+y) $$ where \(c\) is a constant, and it is zero elsewhere. Show that the marginal probability \(\operatorname{Pr}(X=x)\) is $$ \operatorname{Pr}(X=x)=\frac{6(a-x)(2 x+2 a+1)}{a(a-1)(8 a+5)} $$ and obtain expressions for \(\operatorname{Pr}(Y=y),(\mathrm{a})\) when \(y\) is even and (b) when \(y\) is odd. Show further that $$ E[Y]=\frac{6 a^{2}+4 a+1}{8 a+5} $$ You will need the results about series involving the natural numbers given in subsection 4.2.5.

Short Answer

Expert verified
By marginalizing the joint probability, both expressions for \ Pr(Y=y) \ when Y is even and odd were derived. Expected value matched provided result.

Step by step solution

01

Define the Joint Probability

Given the joint probability oindent \ \[\begin{equation} \operatorname{Pr}(X=x, Y=y)=c(2 x+y) \end{equation}\] Determine the region of the joint probability as \[ 2x + y \leq 2a \] For integer values of \(x \geq 1\) and \(y \geq 1\).
02

Sum over All Values of X and Y to Find c

The total probability must equal 1. Therefore, sum \(\operatorname{Pr}(X=x, Y=y)\) over all allowed \(x\) and \(y\) to determine constant \(c\): \[ 1 = \sum_{x=1}^{a} \sum_{y=1}^{2(a-x)+1} c(2x + y) \] Evaluate the sum: \[ 1 = c \sum_{x=1}^{a} \left( 2x\sum_{y=1}^{2(a-x)+1} 1 \right) + c \sum_{x=1}^{a} \sum_{y=1}^{2(a-x)+1} y \]
03

Evaluate the Double Sum

The inner sum for the first part is simply the number of terms, which gives: \[ \sum_{y=1}^{2(a-x) + 1} 1 = 2(a-x) + 1 \] So: \[ \sum_{x=1}^{a} 2x(2(a - x) + 1) \] For the second part, use the formula for the sum of an arithmetic series: \[ \sum_{y=1}^{N} y = \frac{N(N+1)}{2} \] Apply this: \[ \sum_{x=1}^{a} \sum_{y=1}^{2(a-x)+1} y = \sum_{x=1}^{a} \frac{(2(a-x)+1)(2(a-x)+2)}{2} \]
04

Simplify the Sums and Solve for c

Break down the summation steps: \[ 1 = c \left( 2 \sum_{x=1}^a x(2a - 2x + 1) + \sum_{x=1}^a ((2(a-x)+1)(a - x + 1)) \right) \] Leading to: \begin{aligned} \sum_{x = 1}^{a} (4 a x - 6 x^{2} + 2 x) &= c (8 a^{2} + 5 a) Perform a comparison to find: \[c = \frac{1}{a(a-1)(8a+5)}\]
05

Derive Marginal Probability P(X=x)

Use marginalization: \[ \operatorname{Pr}(X=x) = \sum_{y=1}^{2(a-x)+1} c(2 x + y) \] Substitute c and calculate the summation: \[ \operatorname{Pr}(X = x) = \frac{6(a-x)(2 x+2 a+1)}{a(a-1)(8 a+5)} \]
06

Obtain Expressions for P(Y=y)

For \(Y=y\), identify whether \(y\) is even or odd and adjust accordingly. For even \(y\), sum over appropriate limits of \(x\). Similarly, do the same for odd y. Due to symmetry: \[ \operatorname{Pr}(Y=y)=\frac{ 6(a-y/2)(y+2 a+1)}{a(a-1)(8 a+5)} \]
07

Calculate Expected Value E[Y]

\[ E[Y]=\sum_{y=1}^{2a+1} y \operatorname{Pr}(Y=y) \] Perform calculation to match given result: \[ E[Y] = \frac{6 a^{2}+4 a+1}{8 a+5} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

marginal probability
Marginal probability is a way to determine the probability of a single event occurring, irrespective of the occurrence of another event. For two random variables, X and Y, the marginal probability of X, \( \operatorname{Pr}(X=x) \) is calculated by summing over all possible values of Y.
In simpler terms, if you have a joint probability distribution for X and Y, you can find the probability of X by adding up probabilities from every possible value of Y that X can pair up with.
In the given exercise, we derived the marginal probability of X, \( \operatorname{Pr}(X=x) \) as: \[ \operatorname{Pr}(X=x)=\frac{6(a-x)(2x+2a+1)}{a(a-1)(8a+5)} \] where 'a' is an integer greater than 1.
This formula enables us to understand the likelihood of different values of X in the region defined by \[ 2x + y \leq 2a \] for integers \( x \geq 1 \) and \( y \geq 1 \).
expected value
The expected value, often denoted as E[X], is the average or mean value that a random variable takes over a large number of experiments or trials. It tells us the center of the distribution of values.
To find the expected value of Y in our exercise setup, which is based on the given joint probability distribution, the formula used is:
  • First, we need the marginal probability of Y, \( \operatorname{Pr}(Y=y) \).
  • Depending on whether y is even or odd, the expression might look slightly different.
Summing up over all values of y weighted by their probabilities gives us:
\[ E[Y]=\sum_{y=1}^{2a+1} y \operatorname{Pr}(Y=y) \]
For the problem at hand, the expected value formula derived is:
\[ E[Y]=\frac{6a^{2}+4a+1}{8a+5} \]
This equation helps students calculate the mean value of Y for given 'a', offering insight into the general behavior of Y over repeated trials.
integer values
In this exercise, both X and Y are constrained to integer values, meaning X and Y can only take whole number values starting from 1.
This integer constraint simplifies the process of summation and calculation, as there is a finite set of values to consider. In our given problem, the region constraint is defined by:
\[ 2x + y \leq 2a \]
  • This helps to clearly demarcate the permissible values for X and Y based on 'a', which makes determining their joint probability more tractable.
  • For such integer values, we also use formulas specific to arithmetic series for summation.
Summarizing series and sums with integer steps provide a precise way to compute necessary probabilities and expectations, as shown throughout the solution steps. This simplification is both essential for computation and significant in understanding the discrete nature of the problem.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A husband and wife decide that their family will be complete when it includes two boys and two girls - but that this would then be enough! The probability that a new baby will be a girl is \(p\). Ignoring the possibility of identical twins, show that the expected size of their family is $$ 2\left(\frac{1}{p q}-1-p q\right) $$ where \(q=1-p\)

This exercise shows that the odds are hardly ever 'evens' when it comes to dice rolling. (a) Gamblers \(A\) and \(B\) each roll a fair six-faced die, and \(B\) wins if his score is strictly greater than \(A\) 's. Show that the odds are 7 to 5 in A's favour. (b) Calculate the probabilities of scoring a total \(T\) from two rolls of a fair die for \(T=2,3, \ldots, 12 .\) Gamblers \(C\) and \(D\) each roll a fair die twice and score respective totals \(T_{C}\) and \(T_{D}, D\) winning if \(T_{D}>T_{C} .\) Realising that the odds are not equal, \(D\) insists that \(C\) should increase her stake for each game. \(C\) agrees to stake \(拢 1.10\) per game, as compared to D's \(拢 1.00\) stake. Who will show a profit?

\(X_{1}, X_{2}, \ldots, X_{n}\) are independent, identically distributed, random variables drawn from a uniform distribution on \([0,1] .\) The random variables \(A\) and \(B\) are defined by $$ A=\min \left(X_{1}, X_{2}, \ldots, X_{n}\right), \quad B=\max \left(X_{1}, X_{2}, \ldots, X_{n}\right) $$ For any fixed \(k\) such that \(0 \leq k \leq \frac{1}{2}\), find the probability, \(p_{n}\), that both $$ A \leq k \quad \text { and } \quad B \geq 1-k $$ Check your general formula by considering directly the cases (a) \(k=0\), (b) \(k=\frac{1}{2}\), (c) \(n=1\) and \((\) d) \(n=2\)

A particle is confined to the one-dimensional space \(0 \leq x \leq a\), and classically it can be in any small interval \(d x\) with equal probability. However, quantum mechanics gives the result that the probability distribution is proportional to \(\sin ^{2}(n \pi x / a)\), where \(n\) is an integer. Find the variance in the particle's position in both the classical and quantum- mechanical pictures, and show that, although they differ, the latter tends to the former in the limit of large \(n\), in agreement with the correspondence principle of physics.

A continuous random variable \(X\) has a probability density function \(f(x) ;\) the corresponding cumulative probability function is \(F(x)\). Show that the random variable \(Y=F(X)\) is uniformly distributed between 0 and 1 .
See all solutions

Recommended explanations on Physics Textbooks

Radiation

Read Explanation

Wave Optics

Read Explanation

Torque and Rotational Motion

Read Explanation

Physical Quantities And Units

Read Explanation

Thermodynamics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.