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Chapter 1: Problem 22

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

Short Answer

Expert verified
The probability density function of \(Y=\tan X\) when \(X\) has uniform distribution in \(-\pi/2 < x < \pi/2\) is \(f_Y(y) = \frac{1}{\pi(1+y^2)}\) for \(-\infty < y < \infty\), which is the formula for a Cauchy distribution.

Step by step solution

01

Find inverse function

First, identify the inverse function, \(g^{-1}(y)\), which is the function that corresponds to \(x\) when given \(y\). Since \(y = \tan x\), so \(g^{-1}(y) = x = \arctan y\).
02

Find derivative

The next step is to find the derivative of the inverse function with respect to \(y\), \(\frac{d}{dy} g^{-1}(y)\). Taking the derivative of \(\arctan y\) yields \(\frac{1}{1 + y^2}\).
03

Substitute into the formula

Now we substitute these values into the formula for the transformation of random variables. The range of \(y\) will be from \(-\infty\) to \(\infty\) as \(\tan x\) approaches \(-\infty\) as \(x\) approaches \(-\frac{\pi}{2}\) from the right and \(\infty\) as \(x\) approaches \(\frac{\pi}{2}\) from the left. The result is \[f_Y(y) = \frac{1}{\pi} \cdot \left| \frac{1}{1+y^2} \right| = \frac{1}{\pi(1+y^2)} \] for \(-\infty < y < \infty\). This is the probability density function of the Cauchy distribution.
04

Simplify the solution

Finally, we don't need absolute value bars around the derivative as \(1/(1+y^2)\) is always positive. Thus the final form of the pdf of \(Y\) is \[f_Y(y) = \frac{1}{\pi(1+y^2)}, -\infty < y < \infty.\]

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Most popular questions from this chapter

Each bag in a large box contains 25 tulip bulbs. It is known that \(60 \%\) of the bags contain bulbs for 5 red and 20 yellow tulips while the remaining \(40 \%\) of the bags contain bulbs for 15 red and 10 yellow tulips. A bag is selected at random and a bulb taken at random from this bag is planted. (a) What is the probability that it will be a yellow tulip? (b) Given that it is yellow, what is the conditional probability it comes from a bag that contained 5 red and 20 yellow bulbs?

Let \(X\) be a random variable with mean \(\mu\) and let \(E\left[(X-\mu)^{2 k}\right]\) exist. Show, with \(d>0\), that \(P(|X-\mu| \geq d) \leq E\left[(X-\mu)^{2 k}\right] / d^{2 k}\). This is essentially Chebyshev's inequality when \(k=1\). The fact that this holds for all \(k=1,2,3, \ldots\), when those \((2 k)\) th moments exist, usually provides a much smaller upper bound for \(P(|X-\mu| \geq d)\) than does Chebyshev's result.

Cast a die a number of independent times until a six appears on the up side of the die. (a) Find the pmf \(p(x)\) of \(X\), the number of casts needed to obtain that first six. (b) Show that \(\sum_{x=1}^{\infty} p(x)=1\). (c) Determine \(P(X=1,3,5,7, \ldots)\). (d) Find the cdf \(F(x)=P(X \leq x)\).

Let \(X\) be a number selected at random from a set of numbers \(\\{51,52, \ldots, 100\\}\). Approximate \(E(1 / X)\) Hint: Find reasonable upper and lower bounds by finding integrals bounding \(E(1 / X)\).

Let \(X\) have the cdf \(F(x)\) that is a mixture of the continuous and discrete types, namely $$F(x)=\left\\{\begin{array}{ll}0 & x<0 \\\\\frac{x+1}{4} & 0 \leq x<1 \\ 1 & 1 \leq x\end{array}\right.$$ Determine reasonable definitions of \(\mu=E(X)\) and \(\sigma^{2}=\operatorname{var}(X)\) and compute each.
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