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Density and Distribution Functions : Cauchy Density Function and Joint Distribution

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4.17. X has the U(?pi/2, pi/2) distribution, and Y = tan(X). Show that V has density l/(pi(1 + y2)) for ?oo <y <oo . (This is the Cauchy density function.) What can be said about the mean and variance of Y? How could you simulate values from this distribution, given a supply of U(O, 1) values?

4.21. Let X and Y have joint density 2 exp(?x ? y) over 0 < x <y < oo. Find their marginal densities; the density of X, given Y = 4; and the density of Y, given X = 4. Show that X and Y are not independent.
Find the joint density of U = X + Y and V = X/Y. Are U and V independent?

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The Cauchy Density Function and Joint Distributions are investigated. The solution is detailed and well presented. The response received a rating of "5" from the student who originally posted the question.

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Solution to 4.17.

In order to find the density function of Y, we need to find its distribution function first. Since X follows uniformly over , we know that its density function is as follows.

Now , so

So, the density function of Y is given by


Then by definition of mean, we can compute the mean of Y as follows.

This is an improper ...

Solution provided by:
  • BSc , Wuhan Univ. China
  • MA, Shandong Univ.
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  • "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
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