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Chapter 9: Problem 12

The joint probability density function \(f\) of the pair \((X, Y)\) is given by $$ f(x, y)=K\left(3 x^{2}+8 x y\right) \quad \text { for } 0 \leq x \leq 1 \text { and } 0 \leq y \leq 2, $$ and \(f(x, y)=0\) for all other values of \(x\) and \(y\). Here \(K\) is some positive constant. a. Find \(K\). b. Determine the probability \(\mathrm{P}(2 X \leq Y)\).

Short Answer

Expert verified
K = 1/10; P(2X 鈮� Y) = 1/4.

Step by step solution

01

Set up the equation to find K

To find the constant \(K\), we must ensure that the integral of the joint probability density function over its entire range equals 1. Since \(f(x, y)\) is defined for \(0 \leq x \leq 1\) and \(0 \leq y \leq 2\), we need to integrate \(f(x, y)\) over this region: \[ \int_{0}^{1} \int_{0}^{2} K(3x^2 + 8xy) \, dy \, dx = 1. \]
02

Evaluate the inner integral with respect to y

First, we compute the inner integral over \(y\): \[ \int_{0}^{2} (3x^2 + 8xy) \, dy = \left[ 3x^2 y + 4xy^2 \right]_{0}^{2} = 6x^2 + 16x. \]
03

Evaluate the outer integral with respect to x

Next, we substitute the result from Step 2 into the outer integral: \[ \int_{0}^{1} K(6x^2 + 16x) \, dx. \] Calculating this gives: \[ K \left[ 2x^3 + 8x^2 \right]_{0}^{1} = K(2 + 8) = 10K. \]
04

Solve for K

Setting the result of the outer integral equal to 1 to solve for \(K\): \[ 10K = 1 \quad \Rightarrow \quad K = \frac{1}{10}. \]
05

Set up the probability equation for 2X 鈮� Y

For the probability \(P(2X \leq Y)\), substitute the condition into the density function. The new limits for \(y\) are \(y \geq 2x\), and still \(y \leq 2\). So we integrate over this region: \[ P(2X \leq Y) = \int_{0}^{1} \int_{2x}^{2} \frac{1}{10}(3x^2 + 8xy) \, dy \, dx. \]
06

Evaluate the inner integral with updated limits

The inner integral with respect to \(y\) is computed as follows: \[ \int_{2x}^{2} (3x^2 + 8xy) \, dy = \left[ 3x^2 y + 4xy^2 \right]_{2x}^{2} = 6x^2 + 16x - (6x^3 + 16x^3) = 6x^2 + 16x - 22x^3. \]
07

Evaluate the final outer integral with respect to x

Finally, we solve the outer integral: \[ \int_{0}^{1} \frac{1}{10} (6x^2 + 16x - 22x^3) \, dx. \] This computes to: \[ \frac{1}{10} \left[ 2x^3 + 8x^2 - \frac{22}{4}x^4 \right]_{0}^{1} = \frac{1}{10}(2 + 8 - \frac{11}{2}) = \frac{1}{10}(\frac{5}{2}) = \frac{1}{4}. \]

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Key Concepts

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

Probability
Probability is the mathematical concept that measures the likelihood of an event happening. To express it numerically, probability ranges from 0 to 1. A probability of 0 means the event cannot happen, while 1 indicates certainty. If we visualize the probability range as a scale, 0.5 would mean the event is as likely to happen as not, providing a fair level of uncertainty. When dealing with continuous variables, such as in the problem, we move from simple probability to more complex concepts like density functions.
  • Understanding Variances: Probabilities help in measuring how varied outcomes can be in events.
  • Additive Law: The total probability of all possible outcomes in a situation always adds up to 1.
Differentiating between discrete and continuous outcomes is key in higher probability topics. Continuous outcomes often require advanced mathematics, like calculus, to accurately model events, unlike their discrete counterparts.
Probability Density Function
A probability density function (PDF) is a core part of statistics that describes the likelihood of a continuous random variable. With continuous data, possible outcomes occupy an infinite number of points within a range. The joint probability density function, like the one in the exercise, deals with two random variables, giving a picture of how likely particular pairs of outcomes are.
Key points about PDFs:
  • PDFs must be non-negative, ensuring that they don't imply negative likelihoods for outcomes.
  • The total area under the PDF curve equals 1, signifying all possibilities are accounted for.
In the exercise, the function's role is to represent how the variables X and Y behave collectively, allowing for the calculation of probabilities for various events, like \(2X \le Y\).
Integral Calculus
Integral calculus is an essential branch of mathematics used extensively in probability and statistics. It enables us to find areas under curves, which is crucial for calculating probabilities with continuous random variables. The joint probability density function we discussed involves integration to find cumulative probabilities, such as determining a normalization constant or specific event probability.
How integral calculus applies:
  • Inner and Outer Integrals: In problems like ours, integrations are performed sequentially for each variable, an approach known as iterated integration.
  • Definite Integrals: These are used to calculate a specific probability range, like finding the constant K or solving for the conditions \(2X \le Y\).
By practicing problems like these, where you have defined limits of integration, one can control the region of interest and correctly identify probabilities, reinforcing the powerful utility of calculus in probability.

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Most popular questions from this chapter

Let \(X\) and \(Y\) be two continuous random variables with joint probability density function $$ f(x, y)=\frac{12}{5} x y(1+y) \quad \text { for } 0 \leq x \leq 1 \text { and } 0 \leq y \leq 1 $$ and \(f(x, y)=0\) otherwise. a. Find the probability \(\mathrm{P}\left(\frac{1}{4} \leq X \leq \frac{1}{2}, \frac{1}{3} \leq Y \leq \frac{2}{3}\right)\). b. Determine the joint distribution function of \(X\) and \(Y\) for \(a\) and \(b\) between 0 and \(1 .\) c. Use your answer from \(\mathbf{b}\) to find \(F_{X}(a)\) for \(a\) between 0 and \(1 .\) d. Apply the rule on page 122 to find the probability density function of \(X\) from the joint probability density function \(f(x, y)\). Use the result to verify your answer from \(\mathbf{c}\). e. Find out whether \(X\) and \(Y\) are independent.

Let \(U_{1}\) and \(U_{2}\) be two independent random variables, both uniformly distributed over \([0, a] .\) Let \(V=\min \left\\{U_{1}, U_{2}\right\\}\) and \(Z=\max \left\\{U_{1}, U_{2}\right\\} .\) Show that the joint distribution function of \(V\) and \(Z\) is given by $$ F(s, t)=\mathrm{P}(V \leq s, Z \leq t)=\frac{t^{2}-(t-s)^{2}}{a^{2}} \quad \text { for } 0 \leq s \leq t \leq a . $$

Let \(X\) and \(Y\) be independent random variables with probability distributions given by $$ \mathrm{P}(X=0)=\mathrm{P}(X=1)=\frac{1}{2} \quad \text { and } \quad \mathrm{P}(Y=0)=\mathrm{P}(Y=2)=\frac{1}{2} . $$ a. Compute the distribution of \(Z=X+Y\). b. Let \(\tilde{Y}\) and \(\tilde{Z}\) be independent random variables, where \(\tilde{Y}\) has the same distribution as \(Y\), and \(\tilde{Z}\) the same distribution as \(Z\). Compute the distribution of \(\tilde{X}=\tilde{Z}-\tilde{Y}\).

Suppose a vase contains balls numbered \(1,2, \ldots, N\). We draw \(n\) balls without replacement from the vase. Each ball is selected with equal probability, i.e., in the first draw each ball has probability \(1 / N\), in the second draw each of the \(N-1\) remaining balls has probability \(1 /(N-1)\), and so on. For \(i=\) \(1,2, \ldots, n\), let \(X_{i}\) denote the number on the ball in the \(i\) th draw. We have shown that the marginal probability mass function of \(X_{i}\) is given by $$ p_{X_{i}}(k)=\frac{1}{N}, \quad \text { for } \quad k=1,2, \ldots, N . $$ a. Show that $$ \mathrm{E}\left[X_{i}\right]=\frac{N+1}{2} $$ b. Compute the variance of \(X_{i}\). You may use the identity $$ 1+4+9+\cdots+N^{2}=\frac{1}{6} N(N+1)(2 N+1) $$

Let \(X\) and \(Y\) be two independent \(\operatorname{Ber}\left(\frac{1}{2}\right)\) random variables. Define random variables \(U\) and \(V\) by: $$ U=X+Y \quad \text { and } \quad V=|X-Y| $$ a. Determine the joint and marginal probability distributions of \(U\) and \(V\). b. Find out whether \(U\) and \(V\) are dependent or independent.
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