/*! 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 9 Let \(X\) be a continuous random... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(X\) be a continuous random variable. Express the distribution function and probability density of the random variable \(Y=-X\) in terms of those of \(X\).

Short Answer

Expert verified
The distribution function of \(Y = -X\) is \(F_Y(y) = 1 - F_X(-y)\) and the density function is \(f_Y(y) = f_X(-y)\).

Step by step solution

01

Understanding the Problem

We need to find the distribution function and probability density function of the random variable \(Y = -X\) based on the distribution function and probability density function of \(X\).
02

Define the Distribution Functions

The distribution function of \(X\), denoted as \(F_X(x)\), is defined as \(F_X(x) = P(X \leq x)\). Our goal is to express \(F_Y(y)\) in terms of \(F_X(x)\), where \(y = -x\).
03

Express the Distribution Function of Y

For the random variable \(Y\), the distribution function \(F_Y(y)\) is given by \(F_Y(y) = P(Y \leq y) = P(-X \leq y)\). This can be rewritten in terms of \(X\) as \(P(X \geq -y) = 1 - P(X < -y)\). Thus, \(F_Y(y) = 1 - F_X(-y)\).
04

Differentiate to Find the Probability Density Function

The probability density function (pdf) can be found by differentiating the cumulative distribution function (cdf). Thus, \(f_Y(y) = \frac{d}{dy} F_Y(y) = \frac{d}{dy} [1 - F_X(-y)] = f_X(-y)\). Note the use of the chain rule, since \(F_X(-y)\) is derived with respect to \(-y\).
05

Conclude the Result

We've determined that the distribution function of \(Y = -X\) is \(F_Y(y) = 1 - F_X(-y)\) and the probability density function is \(f_Y(y) = f_X(-y)\). This completes the expression of \(Y\) in terms of \(X\).

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Key Concepts

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

Random Variables
In probability theory, a random variable is a variable whose values are determined by the outcomes of a random phenomenon or experiment. There are two main types of random variables: discrete and continuous. Discrete random variables take on a countable number of distinct values, such as the result of rolling a die. Continuous random variables, on the other hand, can take on any value within a given range. These are represented on a continuous scale and include examples like the weight of an object or the temperature outside.
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A random variable acts as a bridge between a population and the mathematical functions that define probabilities. It is a way of assigning numbers to outcomes to simplify the computation and analysis of probability problems.
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For instance, consider a random variable \(X\) which takes on values based on flipping a fair coin. \(X\) might be set to 0 for heads and 1 for tails. The importance of random variables lies in their ability to simplify real-world randomness into a mathematical form, which can then be analyzed and interpreted using various probability functions such as distribution functions and probability density functions.
Distribution Functions
Distribution functions, specifically cumulative distribution functions (CDFs), are crucial in understanding and working with random variables. The CDF of a random variable \(X\) is denoted as \(F_X(x)\) and is defined as the probability that \(X\) will take a value less than or equal to \(x\). Mathematically, this can be expressed as \(F_X(x) = P(X \leq x)\).
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CDFs provide a comprehensive way to look at the probability distribution of a random variable by offering a complete view of the entire range of outcomes. With CDFs, you can determine the likelihood of a random variable falling within certain intervals. For example, to find the probability that \(X\) takes on values between \(a\) and \(b\), you would calculate \(F_X(b) - F_X(a)\).
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In our exercise, we use the concept of CDFs to express the relationship between two random variables \(X\) and \(Y = -X\). We find \(F_Y(y)\), the CDF of \(Y\), using \(F_X(x)\), the CDF of \(X\), giving \(F_Y(y)\) as \(1 - F_X(-y)\). This transformation reflects how changes in variables affect their probability distributions.
Probability Density Function
The probability density function (PDF) of a continuous random variable is a function that describes the likelihood of the variable taking a specific value. For a continuous random variable \(X\), the PDF is denoted as \(f_X(x)\). The primary property of the PDF is that the area under the curve within a given interval corresponds to the probability that the variable falls within that interval.
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Unlike the probability mass function for discrete variables, which assigns a probability to each possible outcome, the PDF is used for continuous variables and gives probabilities over intervals. This means while \(f_X(x)\) itself doesn't represent a probability, the integral of \(f_X(x)\) over an interval does.
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To find the PDF of the transformed random variable \(Y = -X\), we differentiate the cumulative distribution function \(F_Y(y)\) with respect to \(y\). This process, called taking the derivative of the CDF, yields the PDF \(f_Y(y) = f_X(-y)\). Essentially, if you know the PDF of \(X\), you can easily derive the PDF for \(Y = -X\) by substituting \(-y\) into \(f_X\). This insight into the relationship between PDFs of related variables can be very powerful in complex probability problems.

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Most popular questions from this chapter

Let \(X\) be a random variable with the following probability mass function: \begin{tabular}{ccccc} \(x\) & 0 & 1 & 100 & 10000 \\ \hline \(\mathrm{P}(X=x)\) & \(\frac{1}{4}\) & \(\frac{1}{4}\) & \(\frac{1}{4}\) & \(\frac{1}{4}\) \end{tabular} a. Determine the distribution of \(Y=\sqrt{X}\). b. Which is larger \(\mathrm{E}[\sqrt{X}]\) or \(\sqrt{\mathrm{E}[X]}\) ? c. Compute \(\sqrt{\mathrm{E}[X]}\) and \(\mathrm{E}[\sqrt{X}]\) to check your answer (and to see that it makes a big difference!).

Suppose \(X\) has a uniform distribution over the points \(\\{1,2,3,4,5,6\\}\) and that \(g(x)=\sin \left(\frac{\pi}{2} x\right)\). a. Determine the distribution of \(Y=g(X)=\sin \left(\frac{\pi}{2} X\right)\), that is, specify the values \(Y\) can take and give the corresponding probabilities. b. Let \(Z=\cos \left(\frac{\pi}{2} X\right)\). Determine the distribution of \(Z\). c. Determine the distribution of \(W=Y^{2}+Z^{2}\). Warning: in this example there is a very special dependency between \(Y\) and \(Z\), and in general it is much harder to determine the distribution of a random variable that is a function of two other random variables. This is the subject of Chapter 11 .

Let \(W\) have a \(U(\pi, 2 \pi)\) distribution. What is larger: \(\mathrm{E}[\sin (W)]\) or \(\sin (\mathrm{E}[W])\) ? Check your answer by computing these two numbers.

In this exercise we take a look at Jensen's inequality for the function \(g(x)=x^{3}(\) which is neither convex nor concave on \((-\infty, \infty)) .\) a. Can you find a (discrete) random variable \(X\) with \(\operatorname{Var}(X)>0\) such that $$ \mathrm{E}\left[X^{3}\right]=(\mathrm{E}[X])^{3} ? $$ b. Under what kind of conditions on a random variable \(X\) will the inequality \(\mathrm{E}\left[X^{3}\right]>(\mathrm{E}[X])^{3}\) certainly hold?

Often one is interested in the distribution of the deviation of a random variable \(X\) from its mean \(\mu=\mathrm{E}[X]\). Let \(X\) take the values \(80,90,100,110\), and 120, all with probability \(0.2 ;\) then \(\mathrm{E}[X]=\mu=100\). Determine the distribution of \(Y=|X-\mu|\). That is, specify the values \(Y\) can take and give the corresponding probabilities.

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