Problem 27 - A First Course in Probability

Question

If \(X_{1}\) and \(X_{2}\) are independent exponential random variables with respective parameters \(\lambda_{1}\) and \(\lambda_{2},\) find the distribution of \(Z=X_{1} / X_{2} .\) Also compute \(P\left\\{X_{1}

Step-by-Step Solution

Answer

The distribution of Z = X₁ / X₂ is: \(f_{Z}(z) = \frac{\lambda_{1} \lambda_{2}}{(\lambda_{1}z + \lambda_{2})^{2}}\) for \(z \geq 0\), and the probability P{X₁ < X₂} is: \(\frac{\lambda_{1}}{\lambda_{1} + \lambda_{2}}\).

Step 1: Write down the PDFs of X₁ and X₂

Since X₁ and X₂ are independent exponential random variables, their PDFs are given by: PDF of X₁: \(f_{1}(x_{1}) = \lambda_{1} e^{-\lambda_{1} x_{1}}\) for \(x_{1} \geq 0\) PDF of X₂: \(f_{2}(x_{2}) = \lambda_{2} e^{-\lambda_{2} x_{2}}\) for \(x_{2} \geq 0\)

Step 2: Find the JPDF of X₁ and X₂

Since X₁ and X₂ are independent, their joint probability density function (JPDF) is the product of their individual PDFs: JPDF: \(f(x_{1}, x_{2}) = f_{1}(x_{1})f_{2}(x_{2}) = \lambda_{1} \lambda_{2} e^{-(\lambda_{1} x_{1} + \lambda_{2} x_{2})}\) for \(x_{1}, x_{2} \geq 0\)

Step 3: Apply the transformation technique

We want to find the distribution of Z = X₁ / X₂. Define another variable Y = X₂: Z = X₁ / X₂ Y = X₂ Now we'll find the JPDF of (Z, Y) by transforming the variables. First, we need to find the Jacobian: \( x_{1} = zy \) and \( x_{2} = y \) The Jacobian matrix is: \[ J = \begin{bmatrix} \frac{\partial x_{1}}{\partial z} & \frac{\partial x_{1}}{\partial y} \\ \frac{\partial x_{2}}{\partial z} & \frac{\partial x_{2}}{\partial y} \end{bmatrix} = \begin{bmatrix} y & z \\ 0 & 1 \end{bmatrix} \] The Jacobian determinant is |J| = y. Now, we find the JPDF of (Z, Y) using the transformation: \( f_{Z,Y}(z, y) = f_{X_{1},X_{2}}(zy, y)|J| = (\lambda_{1} \lambda_{2} e^{-(\lambda_{1} zy + \lambda_{2} y)})y \) for \(z \geq 0\) and \(y \geq 0\)

Step 4: Find the PDF of Z

To find the PDF of Z, we need to marginalize Y from the joint distribution of (Z, Y): PDF of Z: \(f_{Z}(z) = \int_{0}^{\infty} f_{Z,Y}(z, y)dy \) \( f_{Z}(z) = \int_{0}^{\infty} (\lambda_{1} \lambda_{2} e^{-(\lambda_{1} zy + \lambda_{2} y)})y dy \) After the integration, we get: PDF of Z: \( f_{Z}(z) = \frac{\lambda_{1} \lambda_{2}}{(\lambda_{1}z + \lambda_{2})^{2}} \) for \(z \geq 0\)

Step 5: Compute P{X₁ < X₂}

Now, we want to compute the probability P{X₁ < X₂}. Using the joint density function of X₁ and X₂, the probability can be found as follows: P{X₁ < X₂} = \(\int_{0}^{\infty} \int_{0}^{x_{2}} \lambda_{1} \lambda_{2} e^{-(\lambda_{1} x_{1} + \lambda_{2} x_{2})} dx_{1} dx_{2}\) After the integration, we get: P{X₁ < X₂} = \(\frac{\lambda_{1}}{\lambda_{1} + \lambda_{2}}\) Now we have found both the distribution of Z and the probability P{X₁ < X₂}.

Key Concepts

Exponential Random Variables

Exponential random variables are widely used in probability and statistics due to their simplicity and powerful applications. They model waiting times between events in a Poisson process. For example, if events happen at a constant average rate, then the time until the next event will follow an exponential distribution.

Each exponential random variable is characterized by a rate parameter, \(\lambda\), which is the inverse of the mean. The probability density function (PDF) of an exponential random variable \(X\) is given by:

• \(f(x) = \lambda e^{-\lambda x}\) for \(x \geq 0\)

This formula indicates that the probability decreases exponentially as \(x\) increases. Understanding these variables is crucial for analyzing processes where events occur continuously and independently over time.

Independent Variables

Understanding independent variables is key to solving probability problems, especially when dealing with joint distributions. Two random variables \(X_1\) and \(X_2\) are considered independent if the occurrence of one does not affect the probability of occurrence of the other.

The joint probability density function (JPDF) for two independent random variables is the product of their individual PDFs:

• \(f(x_1, x_2) = f_1(x_1)f_2(x_2)\)

In our example, \(X_1\) and \(X_2\) are independent exponential random variables. This independence allows us to easily find their joint distribution and further manipulate it to find the distribution of a transformed variable.
It's an essential concept that simplifies complex distributions.

Transformation Technique

The transformation technique is a powerful tool used to find the distribution of a new variable derived from existing random variables. When we need to transform a pair of random variables to another pair, understanding the Jacobian is crucial.

This exercise involves finding the distribution of \(Z = X_1 / X_2\). By setting up new variables \(Z = X_1 / X_2\) and \(Y = X_2\), we employ the Jacobian matrix to aid in this transformation.

The Jacobian matrix, derived from partial derivatives, helps adjust the probability densities for changes in scale when transforming variables:

• \( x_1 = zy \)
• \( x_2 = y \)
• Jacobian: \[ J = \begin{bmatrix} y & z \ 0 & 1 \end{bmatrix} \]
• Determinant: \(|J| = y\)

Understanding how to use these calculations ensures accurate estimation of new distribution.

Probability Calculation

Probability calculation is fundamental in determining the likelihood of events in different scenarios. In this context, calculating \(P\{X_1 < X_2\}\) is crucial, as it provides insights into the behavior of our random variables.

Using the joint density function, the probability is found by integrating over the desired region:

• \(P\{X_1 < X_2\} = \int_{0}^{\infty} \int_{0}^{x_2} \lambda_1 \lambda_2 e^{-(\lambda_1 x_1 + \lambda_2 x_2)} dx_1 dx_2 \)

After integration, we obtain:

• \(P\{X_1 < X_2\} = \frac{\lambda_1}{\lambda_1 + \lambda_2}\)

This solution requires understanding integration techniques and properties of exponential random variables. By mastering these concepts, you can determine probabilities effectively in various complex situations.