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For each of the following joint pdfs, find \(F_{X, Y}(x, y)\). (a) \(f_{X, Y}(x, y)=\frac{3}{2} y^{2}, 0 \leq x \leq 2,0 \leq y \leq 1\) (b) \(f_{X, Y}(x, y)=\frac{2}{3}(x+2 y), 0 \leq x \leq 1,0 \leq y \leq 1\) (c) \(f_{X, Y}(x, y)=4 x y, 0 \leq x \leq 1,0 \leq y \leq 1\)

Short Answer

Expert verified
The joint CDFs for the three given pdfs are: (a) \(F_{X, Y}(x, y)=y^{3}x\), (b) \(F_{X, Y}(x, y)=\frac{x^{2}}{2}+x y^{2}\), (c) \(F_{X, Y}(x, y)=x^{2} y^{2}\).

Step by step solution

01

Case (a)

Given the pdf, \(f_{X, Y}(x, y)=\frac{3}{2} y^{2}\) defined over \(0 \leq x \leq 2\) and \(0 \leq y \leq 1\), the joint CDF is calculated as follows: \[F_{X, Y}(x, y)=\int_0^x\int_0^y f_{X, Y}(t_1, t_2) dt_2 dt_1\], substituting \( f_{X, Y}(t_1, t_2) \) and solving the integral, we get \(F_{X, Y}(x, y)=y^{3}x\).
02

Case (b)

Given the pdf, \(f_{X, Y}(x, y)=\frac{2}{3}(x+2 y)\) defined over \(0 \leq x \leq 1\) and \(0 \leq y \leq 1\), the joint CDF is calculated as follows: \[F_{X, Y}(x, y)=\int_0^x\int_0^y f_{X, Y}(t_1, t_2) dt_2 dt_1\], substituting \( f_{X, Y}(t_1, t_2) \) and solving the integral, we get \(F_{X, Y}(x, y)=\frac{x^{2}}{2}+x y^{2}\).
03

Case (c)

Given the pdf, \(f_{X, Y}(x, y)=4 x y\) defined over \(0 \leq x \leq 1\) and \(0 \leq y \leq 1\), the joint CDF is calculated as follows: \[F_{X, Y}(x, y)=\int_0^x\int_0^y f_{X, Y}(t_1, t_2) dt_2 dt_1\], substituting \( f_{X, Y}(t_1, t_2) \) and solving the integral, we get \(F_{X, Y}(x, y)=x^{2} y^{2}\).

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Key Concepts

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

Cumulative Distribution Function
A cumulative distribution function (CDF) is a fundamental concept in probability and statistics, serving as a tool to describe the probability that a real-valued random variable will take a value less than or equal to a certain threshold. It is used for both continuous and discrete random variables. In simple terms, the CDF starts at 0, increases as the value of the variable increases, and approaches 1 as it reaches the upper limit of the variable. For a joint distribution of two random variables, the joint CDF is denoted as \( F_{X, Y}(x, y) \), which represents the probability that the random variable \( X \) is less than or equal to \( x \) and the random variable \( Y \) is less than or equal to \( y \).

When dealing with continuous random variables, the joint CDF is obtained by integrating the joint probability density function (PDF) over the appropriate limits of the variables. For instance, in the given problem, we find the joint CDF \( F_{X, Y}(x, y) \) by integrating the joint PDF over the specified range of \( x \) and \( y \). The integration bounds depend on the provided ranges for \( X \) and \( Y \). It's a process similar to finding the area under the curve but extended to two dimensions for two variables.
Integration in Probability
Integration is a key technique used in probability theory to derive various probability distributions from their density functions. Specifically, when dealing with continuous random variables, the integration of the probability density function (PDF) over the given interval yields the cumulative distribution function (CDF), which represents the aggregate probability till that point.

In our joint probability scenario, double integration is often involved, as seen in the solution to find the joint CDF from the joint PDF. This is because we are dealing with two variables, and integration is performed first with respect to one variable and then the other. The limits of integration correspond to the range over which these variables are defined.

Let's break down the process:
  • First, integrate the joint PDF with respect to \( y \) from 0 to \( y \).
  • Next, integrate the result with respect to \( x \) from 0 to \( x \).
  • The result is the joint CDF \( F_{X, Y}(x, y) \), indicating the probability within the specified range for both \( X \) and \( Y \).
By following this integration approach, we can successfully transform a joint PDF into a joint CDF, gaining insight into the distribution of probabilities within a specific domain.
Continuous Random Variables
Continuous random variables differ from discrete random variables as they can take any value within a given range, unlike discrete variables which only take specific values. These variables are critical in understanding real-world phenomena where outcomes are not fixed and can vary broadly, such as temperature, height, or the time until an event occurs.

In the context of probability, continuous random variables are modeled using probability density functions (PDFs). The PDF describes the likelihood of the random variable assuming a particular range of values, but not exact values - because in continuous cases, the probability of any single point is essentially zero. Instead, probability is determined over intervals.

The cumulative distribution function (CDF) for continuous variables offers us a method to calculate the probability that a variable falls within a certain range. Since PDFs are integral to CDFs in continuous cases, mastery of integration techniques is necessary to transition from the density function to a cumulative probability description.

In summary, understanding how continuous random variables work and harnessing the tools like the PDF and CDF to describe them, helps statisticians and mathematicians quantify uncertainty and variation in continuous data, which is necessary for a variety of applications across different fields.

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Most popular questions from this chapter

Suppose a life insurance company sells a \(\$ 50,000\), five-year term policy to a twenty-five-year-old woman. At the beginning of each year the woman is alive, the company collects a premium of \(\$ P\). The probability that the woman dies and the company pays the \(\$ 50,000\) is given in the table below. So, for example, in Year 3 , the company loses \(\$ 50,000-\$ P\) with probability \(0.00054\) and gains \(\$ P\) with probability \(1-0.00054=0.99946\). If the company expects to make \(\$ 1000\) on this policy, what should \(P\) be? \begin{tabular}{cc} \hline Year & Probability of Payoff \\ \hline 1 & \(0.00051\) \\ 2 & \(0.00052\) \\ 3 & \(0.00054\) \\ 4 & \(0.00056\) \\ 5 & \(0.00059\) \\ \hline \end{tabular}

An urn contains four chips numbered 1 through 4 . Two are drawn without replacement. Let the random variable \(X\) denote the larger of the two. Find \(E(X)\).

Suppose a fair die is tossed three times. Let \(X\) be the largest of the three faces that appear. Find \(p_{X}(k)\).

As the owner of a chain of sporting goods stores, you have just been offered a "deal" on a shipment of one hundred robot table tennis machines. The price is right, but the prospect of picking up the merchandise at midnight from an unmarked van parked on the side of the New Jersey Turnpike is a bit disconcerting. Being of low repute yourself, you do not consider the legality of the transaction to be an issue, but you do have concerns about being cheated. If too many of the machines are in poor working order, the offer ceases to be a bargain. Suppose you decide to close the deal only if a sample of ten machines contains no more than one defective. Construct the corresponding operating characteristic curve. For approximately what incoming quality will you accept a shipment \(50 \%\) of the time?

Calculate \(E(Y)\) for the following pdfs: (a) \(f_{Y}(y)=3(1-y)^{2}, 0 \leq y \leq 1\) (b) \(f_{Y}(y)=4 y e^{-2 y}, y \geq 0\) (c) \(f_{Y}(y)= \begin{cases}\frac{3}{4}, & 0 \leq y \leq 1 \\ \frac{1}{4}, & 2 \leq y \leq 3 \\ 0, & \text { elsewhere }\end{cases}\) (d) \(f_{Y}(y)=\sin y, \quad 0 \leq y \leq \frac{\pi}{2}\)

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