Chapter 6: Problem 29
Suppose that \(X\) is a random variable that has a uniform distribution on the interval \([0,1] .\) (See Problem 20.) The point \((1, X)\) is plotted in the plane. Let \(Y\) be the distance from \((1, X)\) to the origin. Find the CDF and the PDF of the random variable \(Y\). Hint : Find the CDF first.
Short Answer
Expert verified
CDF: \(F_Y(y) = \sqrt{y^2-1}\) for \(1 \leq y < \sqrt{2}\); PDF: \(f_Y(y) = \frac{y}{\sqrt{y^2-1}}\) for \(1 \leq y < \sqrt{2}\).
Step by step solution
01
Define the Random Variable Y
The point \((1, X)\) can be represented as \((x, y) = (1, X)\). The distance from this point to the origin \((0,0)\) is calculated using the distance formula. Thus, the random variable \(Y\) representing the distance is given by: \[ Y = \sqrt{(1 - 0)^2 + (X - 0)^2} = \sqrt{1^2 + X^2} = \sqrt{1 + X^2} \] where \(X\) is uniformly distributed over \([0, 1]\).
02
Derive the CDF of Y
To find the cumulative distribution function (CDF) of \(Y\), we express it as: \[ F_Y(y) = P(Y \leq y) = P(\sqrt{1 + X^2} \leq y) \] Squaring both sides gives: \[ 1 + X^2 \leq y^2 \Rightarrow X^2 \leq y^2 - 1 \] Given \(X\) is uniform on \([0, 1]\), we have \(0 \leq X \leq \sqrt{y^2 - 1}\). Thus, \(F_Y(y) = P(X \leq \sqrt{y^2 - 1})\). For \(y \geq \sqrt{1 + 1^2} = \sqrt{2}\), \(F_Y(y) = 1\). For \(y < \sqrt{1}\), \(F_Y(y) = 0\). For \(1 \leq y < \sqrt{2}\), \(F_Y(y) = \sqrt{y^2-1}\).
03
Calculate the PDF of Y
To find the probability density function (PDF), differentiate the CDF. From the previous step, for \(1 \leq y < \sqrt{2}\), \(F_Y(y) = \sqrt{y^2-1}\). Differentiating with respect to \(y\), we obtain: \[ f_Y(y) = \frac{d}{dy}F_Y(y) = \frac{d}{dy}\sqrt{y^2-1} = \frac{y}{\sqrt{y^2-1}} \] Therefore, the PDF of \(Y\) is \(f_Y(y) = 0\) for \(y < 1\) and \(f_Y(y) = \frac{y}{\sqrt{y^2 - 1}}\) for \(1 \leq y < \sqrt{2}\), and \(f_Y(y) = 0\) for \(y > \sqrt{2}\).
04
Verify the PDF Correctness
Ensure the integral of \(f_Y(y)\) over its range is 1: \[ \int_1^{\sqrt{2}} \frac{y}{\sqrt{y^2 - 1}} \, dy = 1 \] Solving this integral confirms the PDF is correctly normalized, ensuring \(f_Y(y)\) is a valid density function.
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 Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Cumulative Distribution Function
The Cumulative Distribution Function (CDF) is a fundamental concept in probability and statistics, describing the probability that a random variable takes a value less than or equal to a particular number. Given a random variable, the CDF provides a comprehensive view of the distribution, which aids in understanding the likelihood of various outcomes.
For a uniformly distributed random variable on the interval \([0, 1]\), the CDF is especially useful to determine probabilities over subregions of the interval. For the exercise, we defined \(Y\), which is a transformation of another uniform random variable \(X\). The CDF of \(Y\) is found by analyzing the probability that \(Y\) is less than or equal to a value \(y\), represented as \(F_Y(y) = P(Y \leq y)\).
Computing the CDF involves identifying the range of values \(X\) can take such that the expression \(\sqrt{1 + X^2}\) is within the bounds described by the chosen \(y\). By solving the inequality \(\sqrt{1 + X^2} \leq y\), we deduced that \(X^2 \leq y^2 - 1\), establishing bounds for \(X\). The CDF then provides a cumulative view of probabilities for these values.
For a uniformly distributed random variable on the interval \([0, 1]\), the CDF is especially useful to determine probabilities over subregions of the interval. For the exercise, we defined \(Y\), which is a transformation of another uniform random variable \(X\). The CDF of \(Y\) is found by analyzing the probability that \(Y\) is less than or equal to a value \(y\), represented as \(F_Y(y) = P(Y \leq y)\).
Computing the CDF involves identifying the range of values \(X\) can take such that the expression \(\sqrt{1 + X^2}\) is within the bounds described by the chosen \(y\). By solving the inequality \(\sqrt{1 + X^2} \leq y\), we deduced that \(X^2 \leq y^2 - 1\), establishing bounds for \(X\). The CDF then provides a cumulative view of probabilities for these values.
Probability Density Function
The Probability Density Function (PDF) provides insight into the distribution of a continuous random variable. It highlights where values are more or less likely to occur. Essentially, the PDF represents a relative likelihood for a random variable to take on a given value.
Unlike discrete random variables, where probabilities are straightforward, continuous random variables rely on the PDF to characterize distributions. For \(Y\), derived from \(X\), the PDF \(f_Y(y)\) helps us understand how the values of \(Y\) are spread across its range. This is found by differentiating the cumulative distribution function (CDF).
In our discussed problem, the PDF derivation involves differentiating the CDF, \(F_Y(y) = \sqrt{y^2-1}\), for the interval \(1 \leq y < \sqrt{2}\). This results in the expression \(f_Y(y) = \frac{y}{\sqrt{y^2 - 1}}\), which characterizes the density of \(Y\) in that interval. Outside this range, the PDF is zero, reflecting the absence of probability mass.
Unlike discrete random variables, where probabilities are straightforward, continuous random variables rely on the PDF to characterize distributions. For \(Y\), derived from \(X\), the PDF \(f_Y(y)\) helps us understand how the values of \(Y\) are spread across its range. This is found by differentiating the cumulative distribution function (CDF).
In our discussed problem, the PDF derivation involves differentiating the CDF, \(F_Y(y) = \sqrt{y^2-1}\), for the interval \(1 \leq y < \sqrt{2}\). This results in the expression \(f_Y(y) = \frac{y}{\sqrt{y^2 - 1}}\), which characterizes the density of \(Y\) in that interval. Outside this range, the PDF is zero, reflecting the absence of probability mass.
Random Variable
A random variable is a fundamental concept in probability, defined as a variable whose possible values are outcomes of a random phenomenon. In simpler terms, it is used to quantify the results of random processes and is central to performing and interpreting probability calculations.
The exercise involves a random variable \(X\) uniformly distributed over \([0, 1]\), meaning all values in this range are equally likely. From this, a new random variable \(Y = \sqrt{1 + X^2}\) is derived, representing the distance from a specific point to the origin. This transformation introduces a dependency on \(X\), where \(Y\) represents a specific outcome based on \(X\)'s behavior.
This setup exemplifies how transformations can create new outcomes from existing random structures, illustrating the continuous nature of values they may assume over different intervals. Understanding random variables in this context enhances the grasp of dependency relationships and the applications of transformations in probability.
The exercise involves a random variable \(X\) uniformly distributed over \([0, 1]\), meaning all values in this range are equally likely. From this, a new random variable \(Y = \sqrt{1 + X^2}\) is derived, representing the distance from a specific point to the origin. This transformation introduces a dependency on \(X\), where \(Y\) represents a specific outcome based on \(X\)'s behavior.
This setup exemplifies how transformations can create new outcomes from existing random structures, illustrating the continuous nature of values they may assume over different intervals. Understanding random variables in this context enhances the grasp of dependency relationships and the applications of transformations in probability.
Distance Formula
The Distance Formula is a tool used to determine the distance between two points in the Cartesian plane. It is especially useful in geometry, physics, and any field involving spatial calculations. It can compute distances from basic plots to complex structures.
In its simplest form, when given two points \((x_1, y_1)\) and \((x_2, y_2)\), the distance \(d\) is derived using the formula \[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\].
This exercise applies the Distance Formula to find the distance from the point \((1, X)\) to the origin \((0,0)\). Through simple algebraic substitution, we derive that \(Y = \sqrt{1 + X^2}\), which provides an exact measure of this distance. The simplicity of the formula belies its power, offering a direct method to analyze how variables relate spatially, as seen in calculating transformations like \(Y\) from \(X\). Understanding this formula is crucial for navigating both theoretical and practical problems involving distances.
In its simplest form, when given two points \((x_1, y_1)\) and \((x_2, y_2)\), the distance \(d\) is derived using the formula \[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\].
This exercise applies the Distance Formula to find the distance from the point \((1, X)\) to the origin \((0,0)\). Through simple algebraic substitution, we derive that \(Y = \sqrt{1 + X^2}\), which provides an exact measure of this distance. The simplicity of the formula belies its power, offering a direct method to analyze how variables relate spatially, as seen in calculating transformations like \(Y\) from \(X\). Understanding this formula is crucial for navigating both theoretical and practical problems involving distances.
