Problem 195 - An Introduction to Mathematical Statistics and Its Applications

Question

Suppose that \(X\) and \(Y\) are two independent random variables, where \(f_{X}(x)=x e^{-x}, x \geq 0\), and \(f_{Y}(y)=\) \(e^{-y}, y \geq 0\). Find the pdf of \(Y / X\).

Step-by-Step Solution

Answer

The Probability Density Function (pdf) of \(Z = Y/X\) is given by \(f_{Z}(z) = 1 - e^{-z(1+z)}\) for \(z > 0\).

Step 1: Define New Random Variable

Define a new random variable \(Z= Y/X\). The job is to find \(f_{Z}(z)=\int_R f_{X,Y}(x,y)dx\), where \(R= \{(x,y): y/x = z\}\).

Step 2: Compute Marginal Joint PDF

Since the random variables \(X\) and \(Y\) are independent, the joint PDF is the product of their marginal PDFs, \(f_{X,Y}(x,y)=f_{X}(x)f_{Y}(y)\). Substituting the given values, \(f_{X,Y}(x,y) = xe^{-x} e^{-y} = xe^{-x-y}\). Now, R = \{(x,y) | 0< y/x (x>0)< z\}. This gives x < y < zx.

Step 3: Integrate Joint PDF Over R

We integrate the joint PDF over the region R to get the PDF of \(Z\). \(f_{Z}(z) = \int_{0}^{z} \int_{y/z}^{y}xe^{-x-y}dxdy\). Calculate the inner integral first.

Step 4: Inner Integration

Perform the inner integration with respect to 'x', \( = \int_{0}^{z} -e^{-y}(e^{-y/z}-e^{-y})dy\).

Step 5: Outer Integration

Perform the outer integration next. This gets us -e^{-z(1+z)}|_{0}^{z} = 1 - e^{-z(1+z)}.

Key Concepts

Random Variables

A random variable is a mathematical representation of a potential outcome of a random process. It's a variable whose values are determined by the outcomes of a random phenomenon. In many cases, these outcomes are quantity we can measure, like the result of rolling a die or the time it takes for a computer to boot.

• Random variables can be discrete or continuous. Discrete variables have countable outcomes, like integers from a die roll.
• Continuous variables can take on any value within a given range, such as height or temperature.

In the given problem, both \(X\) and \(Y\) are continuous random variables. They are independent, meaning the outcome of one doesn't affect the other. This independence simplifies calculations involving joint probabilities and distributions, as shown in the joint PDF calculation in the exercise solution.

Probability Density Function (PDF)

A Probability Density Function (PDF) is a function that describes the likelihood of a continuous random variable to take on a particular value. Although a PDF doesn't give probabilities directly, it helps calculate probabilities over intervals.

• The area under the PDF curve over an interval gives the probability that the random variable falls within that interval.
• P(f_{X}(x) = x \(e^{-x}\)) represents the PDF for the random variable \(X\), meaning \(X\) could take any value \(x\) from 0 to infinity with a density influenced by \(x e^{-x}\).
• For \(Y\), the PDF is \(f_{Y}(y) = e^{-y}\), corresponding to its behavior from 0 to infinity.

In the solution, these PDFs were used to find the joint PDF of \(X\) and \(Y\), \( f_{X,Y}(x,y)=xe^{-x-y} \), essential for finding the distribution of \(Z = Y / X\).

Independent Variables

In probability theory, when we say that two random variables are independent, we're asserting that knowing the outcome of one variable gives no information about the other. This characteristic greatly simplifies many statistical tasks, such as determining joint distributions.

• Independence is crucial when calculating joint PDFs, as the joint PDF of independent variables is simply the product of their separate PDFs.
• In our exercise, since \(X\) and \(Y\) are independent, we can easily write their joint PDF as \( f_{X,Y}(x,y) = f_{X}(x) \cdot f_{Y}(y) \), allowing for simpler integration.
• This concept is vital when transforming variables or conducting operations on multiple variables, like determining the distribution of \(Z = Y/X\).

Through this understanding, we efficiently solved for the PDF of \(Z\) in the exercise, using the independent nature of \(X\) and \(Y\) to simplify calculations.