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Magazine Chapter 4: Problem 115
Let \(X\) be uniformly distributed on \([0,1]\). Find the pdf of \(Y=\tan [\pi(X-.5)]\). The random variable \(Y\) has the Cauchy distribution after the famous mathematician.
Short Answer
Expert verified
The pdf of \(Y\) is \(f_Y(y) = \frac{1}{\pi(1 + y^2)}\), matching a standard Cauchy distribution.
Step by step solution
01
Identify the transformation
We are given that the random variable \(X\) is uniformly distributed on \([0,1]\), which means its probability density function (pdf) is constant over the interval. We need to find the pdf of \(Y = \tan(\pi(X - 0.5))\).
02
Calculate the inverse transformation
To find the pdf of \(Y\), we must first determine how \(X\) and \(Y\) are related through the transformation. We have \(Y = \tan(\pi(X - 0.5))\). Solving for \(X\), we get \(X = \frac{1}{\pi}\arctan(Y) + 0.5\).
03
Find the derivative of the inverse transformation
The next step is to find the derivative of \(X\) with respect to \(Y\). That is, calculate \(\frac{dX}{dY}\). Derivative: \(\frac{dX}{dY} = \frac{1}{\pi(1 + Y^2)}\). This gives the rate at which \(X\) changes with \(Y\).
04
Apply the change of variables formula
The change of variables in probability states that the pdf of \(Y\), \(f_Y(y)\), can be found as \(f_Y(y) = f_X(x) \left| \frac{dX}{dY} \right|\). Since \(X\) is uniformly distributed, \(f_X(x) = 1\) for \(x \in [0,1]\). Therefore, \(f_Y(y) = 1 \cdot \left| \frac{1}{\pi(1 + y^2)} \right| = \frac{1}{\pi(1 + y^2)}\).
05
Recognize the resulting distribution
The resulting pdf \(f_Y(y) = \frac{1}{\pi(1 + y^2)}\) is the pdf of a standard Cauchy distribution. Thus, \(Y\) follows the standard Cauchy distribution, confirming the statement about its name.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Uniform Distribution
When we say that a random variable, like \(X\), is uniformly distributed on \([0,1]\), we are dealing with a case of uniform distribution. This means that every outcome within this interval has an equal chance of occurring. It's as if you are picking a number at random between zero and one with no preference for any specific number.
A couple of key points about uniform distribution:
A couple of key points about uniform distribution:
- The probability density function (pdf) for a uniform distribution on \([0,1]\) is simply a constant value of 1 across the interval.
- Outside the interval, the pdf of \(X\) is zero, indicating no chance of selecting a number outside \([0,1]\).
- In terms of probabilities, if you choose any subinterval, the probability is just its length because the distribution is uniform.
Probability Density Function
The probability density function or pdf, is a core concept that describes the likelihood of a continuous random variable having specific values. It's the continuous counterpart to a probability mass function, which serves a similar purpose for discrete variables.
Here are a few essentials about pdfs:
Here are a few essentials about pdfs:
- The pdf provides the relative likelihood of the random variable being near a given value, but not the exact probability.
- Areas under the pdf curve represent probabilities. This means that the total area under the curve sums up to one, as it represents the entire probability space.
- For transformations, like converting \(X\) to \(Y\), we often need the original pdf to find the new pdf.
Transformation of Variables
Transformation of variables involves changing one random variable into another through a mathematical function. This process is critical in studying how distributions morph under different conditions.
Considerations during transformation:
Considerations during transformation:
- Firstly, identify the functional relationship between the two variables, such as \(Y = \tan(\pi(X - 0.5))\).
- This step involves manipulating the equation, solving for the original variable in terms of the new variable. In our example, solving for \(X\) gives \(X = \frac{1}{\pi}\arctan(Y) + 0.5\).
- Next, you need to determine how this change affects data spread, variance, and distribution characteristics. This includes determining how densities transform.
Inverse Transformation
Inverse transformation is about tracing the transformations backward, to express the initial variable in terms of the transformed one. This inverse relationship is crucial when deriving the new pdf for a transformed variable.
Here's how it works:
Here's how it works:
- The goal is to rewrite the equation, expressing the original variable as a function of the transformed one. For \(Y\), this expression was \(X = \frac{1}{\pi}\arctan(Y) + 0.5\).
- Once the inverse relationship is established, the next task involves finding its derivative. This derivative \(\frac{dX}{dY}\) is essential in applying the change of variables formula.
- The last part of the process is to use magnitude \(\left| \frac{dX}{dY} \right|\) when calculating the new pdf, ensuring that we account for the change in scale and orientation of transformation.
