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If \(X\) and \(Y\) are independent continuous positive random variables, express the density function of (a) \(Z=X / Y\) and (b) \(Z=X Y\) in terms of the density functions of \(X\) and \(Y\). Evaluate these expressions in the special case where \(X\) and \(Y\) are both exponential random variables.

Short Answer

Expert verified
For \(Z = \frac{X}{Y}\), the density function is: \(f_Z(z) = \int_0^\infty \lambda_x e^{-\lambda_x uz} \lambda_y e^{-\lambda_y u} \cdot \frac{1}{u^2} du.\) For \(Z = XY\), the density function is: \(f_Z(z) = \frac{1}{z} \int_0^{\ln(z)} \lambda_x e^{-\lambda_x e^v} \lambda_y e^{-\lambda_y \frac{z}{e^v}} e^v dv.\) These expressions are valid when X and Y are both exponential random variables with parameters \(\lambda_x\) and \(\lambda_y\), respectively.

Step by step solution

01

Define the probability density functions

Let f_X(x) and f_Y(y) denote the probability density functions of X and Y, respectively.
02

Find the density function of Z=X/Y

Define a new random variable \(Z = \frac{X}{Y}\) by the transformation \(X = UZ\), \(Y = U\). We can easily find the Jacobian of the transformation: \[J = \begin{vmatrix}\frac{\partial z}{\partial x} & \frac{\partial z}{\partial y} \\ \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y}\end{vmatrix} = \begin{vmatrix}\frac{1}{y} & -\frac{x}{y^2} \\ 0 & 1\end{vmatrix} = \frac{1}{y^2}.\] Then, the joint density function of U and Z is given by: \[f_{U, Z}(u, z) = f_{X, Y}(x, y) \cdot |J|.\] Since X and Y are independent, we have \(f_{X, Y}(x, y) = f_X(x) f_Y(y)\). Hence, we can rewrite the joint density function as: \[f_{U, Z}(u, z) = f_X(uz) f_Y(u) \cdot \frac{1}{u^2}.\] Now finding the marginal density function for Z by integrating over U: \[f_Z(z) = \int_0^\infty f_{U, Z}(u, z) du = \int_0^\infty f_X(uz) f_Y(u) \cdot \frac{1}{u^2} du.\]
03

Find the density function of Z=XY

To find the density function for Z=XY, we use the convolution theorem. The convolution theorem states that if Z = X + Y, then the density function of Z is given by the convolution of the density functions of X and Y: \[f_Z(z) = \int_0^\infty f_X(u) f_Y\left(\frac{z}{u}\right) \cdot \frac{1}{u} du.\] Since we want to find the density function for Z = XY, we make the change of variables W = ln(Z) and V = ln(X). Then ln(Y) = ln(Z) - ln(X) = W - V and we are back to working with the sum of two random variables, i.e., W = V + (W - V). Applying the convolution theorem for variables V and (W - V) we get: \[f_W(w) = \int_{-\infty}^w f_V(v) f_{W - V}(w - v) dv.\] Now change back to the original variables: W = ln(Z) and V = ln(X). We obtain: \[f_Z(z) = \frac{1}{z} \int_0^{\ln(z)} f_X(e^v) f_Y\left(\frac{z}{e^v}\right) e^v dv.\]
04

Evaluate the density functions for exponential random variables

If X and Y are exponential random variables with parameters \(\lambda_x\) and \(\lambda_y\) respectively, their density functions are given by: \[f_X(x) = \lambda_x e^{-\lambda_x x}\] for \(x > 0\), and \[f_Y(y) = \lambda_y e^{-\lambda_y y}\] for \(y > 0\). Now we substitute these density functions into the above solutions for outcome (a) and (b): (a) Density function of Z=X/Y: \[f_Z(z) = \int_0^\infty \lambda_x e^{-\lambda_x uz} \lambda_y e^{-\lambda_y u} \cdot \frac{1}{u^2} du.\] (b) Density function of Z=XY: \[f_Z(z) = \frac{1}{z} \int_0^{\ln(z)} \lambda_x e^{-\lambda_x e^v} \lambda_y e^{-\lambda_y \frac{z}{e^v}} e^v dv.\] These are the expressions for the density functions of Z=X/Y and Z=XY when X and Y are exponential random variables.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Density Function
The concept of a probability density function (PDF) is fundamental in understanding continuous random variables. It describes the likelihood of a random variable taking on a particular value. In mathematical terms, the PDF of a continuous random variable is a function that assigns probabilities to intervals of outcomes rather than discrete outcomes.

The PDF is crucial because it allows us to calculate the probability that a random variable falls within a certain range by integrating the function over that range. For example, the probability that a random variable X is between a and b is given by the integral of its PDF, denoted as f_X(x), over the interval [a, b]:\[\begin{equation} P(a \leq X \leq b) = \int_a^b f_X(x) dx.\end{equation}\]In the provided exercise, we are asked to determine the PDF of a new random variable Z, which is a function of two other independent random variables X and Y. This involves transforming the PDFs of X and Y into that of Z using mathematical methods such as Jacobian transformation or the convolution theorem.
Independent Random Variables
When we work with multiple random variables, an important concept is their interdependence or independence. Two random variables, X and Y, are termed independent if the occurrence of one does not affect the probability distribution of the other. Mathematically, X and Y are independent if for every pair of intervals A and B, the probability that X is in A and Y is in B is the product of their individual probabilities:

\[\begin{equation} P(X \in A \text{ and } Y \in B) = P(X \in A) \times P(Y \in B).\end{equation}\]Thus, for independent random variables, their joint PDF can be expressed as the product of their individual PDFs:\[\begin{equation} f_{X, Y}(x, y) = f_X(x) f_Y(y).\end{equation}\]In our exercise, X and Y are given as independent, enabling us to use their individual PDFs to find the PDF of the new variables Z=X/Y and Z=XY through appropriate transformations and integration strategies.
Exponential Random Variables
Exponential random variables are a special case of continuous random variables that model the time between events in a Poisson process, which is a type of statistical process where events occur continuously and independently at a constant average rate. The PDF of an exponential random variable is defined for x > 0 as:\[\begin{equation} f_X(x) = \lambda e^{-\lambda x},\end{equation}\]where \(\lambda\) is the rate parameter. The exponential distribution is characterized by the 'memoryless' property, meaning the probability of an event occurring in the next t units of time is independent of how much time has already elapsed.In the exercise, the fact that X and Y are both exponential random variables simplifies the calculations. We can directly plug in the exponential PDFs into the derived formulas for the PDF of Z=X/Y and Z=XY. This gives us explicit expressions for these density functions in terms of the rate parameters \(\lambda_x\) and \(\lambda_y\), and allows the calculation of probabilities and other statistical measures related to the new random variable Z.

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