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Suppose that \(X\) and \(Y\) are continuous random variables with density functions \(f_{X}(x)\) and \(f_{Y}(y)\), respectively. Let \(f(x, y)\) denote the joint density function of \((X, Y)\). Show that $$ \int_{-\infty}^{\infty} f(x, y) d y=f_{X}(x) $$ and $$ \int_{-\infty}^{\infty} f(x, y) d x=f_{Y}(y) $$

Short Answer

Expert verified
Integrating the joint density over one variable gives the marginal density of the other variable.

Step by step solution

01

Understand the Concept of Marginal Density

When dealing with joint density functions, the marginal density of one variable can be found by integrating the joint density over the other variable. This allows us to isolate the probability distribution of one variable while accounting for all possible values of the other.
02

Set Up the Integral for Marginal Density of X

The marginal density function of a variable, say \(X\), is found by integrating the joint density function over all possible values of the other variable \(Y\). This is given by the expression \(\int_{-\infty}^{\infty} f(x, y) \, dy = f_{X}(x)\). This states that if you integrate the joint density over \(y\), you effectively 'sum' over all possible values of \(Y\), leaving the density function of \(X\) alone.
03

Set Up the Integral for Marginal Density of Y

Similarly, the marginal density function of \(Y\) can be found by integrating the joint density over all possible values of \(X\). This is represented as \(\int_{-\infty}^{\infty} f(x, y) \, dx = f_{Y}(y)\). Integrating over \(x\) sums over all possible values of \(X\), leaving just the density function of \(Y\).
04

Explanation and Conclusion

These integrals are based on the principle that the total probability distribution of one variable is the accumulation (integration) of probabilities over any other variables from the joint distribution. Thus, integrating the joint density function \(f(x, y)\) over \(y\) gives \(f_{X}(x)\), and over \(x\) gives \(f_{Y}(y)\). This approach validates that the integrals represent the marginal probability density functions.

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Key Concepts

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

Marginal Density
When considering joint probability density functions, it's useful to look at what we call **marginal densities**. Imagine you have two variables, say, a set of continuous random variables denoted by \( X \) and \( Y \). They are described together by a joint density function, \( f(x, y) \). But sometimes, we're interested in the behavior of just one variable at a time.

To find the marginal density of one of these variables, say \( X \), you need to 'sum out' the other variable, \( Y \). This isn't regular addition but a type of addition done through integration. We integrate the joint density over all values \(Y\) can take, from negative infinity to positive infinity. The result is a function that tells us about \( X \) alone. This is written mathematically as:
  • \( \int_{-\infty}^{\infty} f(x, y) \, dy = f_{X}(x) \)
This process is mirrored for \( Y \):
  • \( \int_{-\infty}^{\infty} f(x, y) \, dx = f_{Y}(y) \)
These computations prove that we can isolate variables and study their individual distributions from a joint one. This is the essence of marginal density.
Continuous Random Variables
Continuous random variables are a fundamental concept when working with probability density functions. These are the kind of variables that can take any value on a continuum. Unlike discrete random variables, which jump from one value to another, continuous variables flow.

For example, think of measuring the height of people. Heights don't jump; they flow continuously from one person to another throughout possible values. In probability, instead of assigning probabilities to specific individual outcomes (like you might with dice or cards), we describe continuous random variables in terms of ranges, often with density functions.

The joint probability density function (\( f(x, y) \)) is used to describe how two continuous random variables interact. It's like a map, telling us how likely different pairs of outcomes are. From this, the marginal densities are derived, depicting the individual behaviors of these flowing random variables. With continuous variables, much of probability revolves around this idea of describing likelihoods over intervals (or ranges), not specific points.
Integration
Integration is crucial when working with continuous random variables and their density functions. If you want to capture the behavior of a single variable from a joint density function, integration is your tool. It's like a mathematical way of summing everything up smoothly. Where with discrete variables, we can just count and sum, with continuous ones, we need integration.

Here's a simple take: think of integration as accumulating all the tiny probabilities across every possible value of your variable. To find the marginal density of \( X \), you integrate out \( Y \):
  • \( \int_{-\infty}^{\infty} f(x, y) \, dy = f_{X}(x) \)
And vice versa for \( Y \):
  • \( \int_{-\infty}^{\infty} f(x, y) \, dx = f_{Y}(y) \)
In simpler terms, integration serves as a spotlight focusing solely on one variable by collapsing all the others to their possible outcomes. By slicing through the structure imposed by the joint probability, you can pull out these single-variable layers (or marginal densities) effortlessly. This role of integration underscores its place as a powerful instrument in the world of continuous probability.

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Most popular questions from this chapter

(Chung \(^{9}\) ) In London, half of the days have some rain. The weather forecaster is correct \(2 / 3\) of the time, i.e., the probability that it rains, given that she has predicted rain, and the probability that it does not rain, given that she has predicted that it won't rain, are both equal to \(2 / 3 .\) When rain is forecast, Mr. Pickwick takes his umbrella. When rain is not forecast, he takes it with probability \(1 / 3\). Find (a) the probability that Pickwick has no umbrella, given that it rains. (b) the probability that he brings his umbrella, given that it doesn't rain.

A radioactive material emits \(\alpha\) -particles at a rate described by the density function \(f(t)=.1 e^{-.1 t}\) Find the probability that a particle is emitted in the first 10 seconds, given that (a) no particle is emitted in the first second. (b) no particle is emitted in the first 5 seconds. (c) a particle is emitted in the first 3 seconds. (d) a particle is emitted in the first 20 seconds.

Let \(x\) and \(y\) be chosen at random from the interval [0,1] . Which pairs of the following events are independent? (a) \(x>1 / 3\) (b) \(y>2 / 3\). (c) \(x>y\) (d) \(x+y<1\).

Suppose you toss a dart at a circular target of radius 10 inches. Given that the dart lands in the upper half of the target, find the probability that (a) it lands in the right half of the target. (b) its distance from the center is less than 5 inches. (c) its distance from the center is greater than 5 inches. (d) it lands within 5 inches of the point (0,5) .

Let \(\Omega\) be the sample space of an experiment. Let \(E\) be an event with \(P(E)>0\) and define \(m_{E}(\omega)\) by \(m_{E}(\omega)=m(\omega \mid E)\). Prove that \(m_{E}(\omega)\) is a distribution function on \(E,\) that is, that \(m_{E}(\omega) \geq 0\) and that \(\sum_{\omega \in \Omega} m_{E}(\omega)=1\). The function \(m_{E}\) is called the conditional distribution given \(E\).

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