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Let \(X\) have the uniform pdf \(f_{X}(x)=\frac{1}{\pi}\), for \(-\frac{\pi}{2}

Short Answer

Expert verified
The pdf of \(Y=tan(X)\) is \(f_{Y}(y)=\frac{1}{\pi |1+y^{2}|}\). This represents a Cauchy distribution.

Step by step solution

01

Identify the transformation

We are given that \(Y=\tan(X)\). Note that the transformation function is \(g(X)=\tan(X)\) which is a monotonic function between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\). This tells us that limits of \(Y\) will be -\(\infty\) and \(\infty\) because \(\tan(X)\) approaches \(-\infty\) when \(X\) approaches \(-\frac{\pi}{2}\) and \(\infty\) when \(X\) approaches \(\frac{\pi}{2}\).
02

Calculate Derivative

Now, let's calculate the derivative of the transformation function \(\frac{dg(X)}{dx}\). As \(g(X)= \tan(X)\), its derivative will be \( \frac{d}{dx} \tan(X) = \sec^{2}(X) = 1 + \tan^{2}(X) = 1 + Y^{2}\).
03

Apply Transformation Formula

We use the formula for the transformation of random variables which states that the pdf of \(Y\), say \(f_{Y}(y)\), is \(\frac{f_{X}(g^{-1}(Y))}{|\frac{dg^{-1}(Y)}{dy}|}\) where \(f_{X}(x)\) is the pdf of \(X\). We already have \(f_{X}(x)=\frac{1}{\pi}\) for \(-\frac{\pi}{2}<x<\frac{\pi}{2}\) and \(\frac{dg^{-1}(Y)}{dy}= \frac{1}{1+Y^{2}}\) from the previous step. Therefore, plugging these values into the formula will yield: \(f_{Y}(y)=\frac{f_{X}(g^{-1}(Y))}{|\frac{dg^{-1}(Y)}{dy}|}=\frac{1}{\pi |1+y^{2}|}\). By taking the absolute value, the range of \(Y\) is from \(-\infty\) to \(\infty\), which means the pdf \(f_{Y}(y)\) is non-zero for all real numbers.

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Key Concepts

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

Uniform Distribution
The uniform distribution is a probability distribution where all outcomes are equally likely within a certain range. In the given exercise, the variable X follows a uniform distribution within the interval \( -\frac{\text{\pi}}{2} < x < \frac{\pi}{2} \). Mathematically, the probability density function (pdf) for X is defined as \( f_X(x) = \frac{1}{\pi} \) for values of x within the specified range. This is because the width of the interval is \( \pi \) (from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \) ), and the height is \( \frac{1}{\pi} \) to ensure the area under the curve equals 1, satisfying the properties of a probability distribution.
Outside this interval, the pdf is zero, indicating that it is impossible for X to take on values beyond the given range in this context. The uniform distribution is often used in simulations and random sampling where each outcome in the range is as likely as any other.
Transformation of Random Variables
When dealing with random variables, we sometimes want to understand the behavior of a new variable that is a function of a known one, like Y = g(X). This process is called the transformation of random variables. In our exercise, we are transforming X, which has a uniform distribution, using the tangent function to find a new variable Y.
For continuous random variables, if the transformation is monotonic, as it is within our specified range \( (-\frac{\pi}{2}, \frac{\pi}{2}) \), we can use the formula \( f_Y(y) = \frac{f_X(g^{-1}(y))}{|\frac{dg^{-1}(y)}{dy}|} \) to find the pdf of Y. It's important to calculate the derivative \( \frac{dg(X)}{dx} \) correctly, as it gives us insight into how the interval for X gets mapped to the interval for Y. Specifically, in the exercise, \( \frac{dg(X)}{dx} \) informs us about how the density 'stretches' or 'compresses' during the transformation.
Cauchy Distribution
The Cauchy distribution is a type of probability distribution that has a particular bell shape but with fatter tails as compared to the normal distribution. This implies that it is more prone to producing outliers. In our exercise, the transformation of the uniform random variable X through the tangent function yields a random variable Y that follows a Cauchy distribution.
Its probability density function is given by \( f_Y(y) = \frac{1}{\pi(1+y^2)} \), which varies from \( -\infty \) to \( \infty \) for y. This pdf is characterized by its peak at zero and long tails, indicating a higher probability of extreme values. It's crucial to understand that despite its appearance, the area under the curve of a Cauchy pdf still integrates to 1, fulfilling the criterion of a probability distribution. The Cauchy distribution is often used in physics and finance, where systems can experience large fluctuations, and it is also notable for not having defined mean or variance, unlike many other distributions.

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Most popular questions from this chapter

Let \(\mathcal{C}=R\), where \(R\) is the set of all real numbers. Let \(\mathcal{I}\) be the set of all open intervals in \(R\). The Borel \(\sigma\) -field on the real line is given by $$ \mathcal{B}_{0}=\cap\\{\mathcal{E}: \mathcal{I} \subset \mathcal{E} \text { and } \mathcal{E} \text { is a } \sigma \text { -field }\\} $$ By definition, \(\mathcal{B}_{0}\) contains the open intervals. Because \([a, \infty)=(-\infty, a)^{c}\) and \(\mathcal{B}_{0}\) is closed under complements, it contains all intervals of the form \([a, \infty)\), for \(a \in R\). Continue in this way and show that \(\mathcal{B}_{0}\) contains all the closed and half- open intervals of real numbers.

Consider an urn that contains slips of paper each with one of the numbers \(1,2, \ldots, 100\) on it. Suppose there are \(i\) slips with the number \(i\) on it for \(i=1,2, \ldots, 100\). For example, there are 25 slips of paper with the number \(25 .\) Assume that the slips are identical except for the numbers. Suppose one slip is drawn at random. Let \(X\) be the number on the slip. (a) Show that \(X\) has the \(\operatorname{pmf} p(x)=x / 5050, x=1,2,3, \ldots, 100\), zero elsewhere. (b) Compute \(P(X \leq 50)\). (c) Show that the cdf of \(X\) is \(F(x)=[x]([x]+1) / 10100\), for \(1 \leq x \leq 100\), where \([x]\) is the greatest integer in \(x\).

A French nobleman, Chevalier de Méré, had asked a famous mathematician, Pascal, to explain why the following two probabilities were different (the difference had been noted from playing the game many times): (1) at least one six in four independent casts of a six-sided die; (2) at least a pair of sixes in 24 independent casts of a pair of dice. From proportions it seemed to de Méré that the probabilities should be the same. Compute the probabilities of (1) and (2).

The following game is played. The player randomly draws from the set of integers \(\\{1,2, \ldots, 20\\} .\) Let \(x\) denote the number drawn. Next the player draws at random from the set \(\\{x, \ldots, 25\\}\). If on this second draw, he draws a number greater than 21 he wins; otherwise, he loses. (a) Determine the sum that gives the probability that the player wins. (b) Write and run a line of \(\mathrm{R}\) code that computes the probability that the player wins. (c) Write an \(\mathrm{R}\) function that simulates the game and returns whether or not the player wins. (d) Do 10,000 simulations of your program in Part (c). Obtain the estimate and confidence interval, (1.4.7), for the probability that the player wins. Does your interval trap the true probability?

Let \(X\) have the \(\operatorname{pmf} p(x)=1 / 3, x=-1,0,1\). Find the pmf of \(Y=X^{2}\).

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