Question

Let \(X\) and \(Y\) be independent exponential random variables with respective rates \(\lambda\) and \(\mu\), where \(\lambda>\mu .\) Let \(c>0\) (a) Show that the conditional density function of \(X\), given that \(X+Y=c\), is $$ f_{X \mid X+Y}(x \mid c)=\frac{(\lambda-\mu) e^{-(\lambda-\mu) x}}{1-e^{-(\lambda-\mu) c}}, \quad 0

Short answer

The conditional density function of \(X\) given that \(X+Y=c\) is: $$ f_{X|X+Y}(x|c) = \frac{(\lambda-\mu) e^{-(\lambda-\mu)x}}{1-e^{-(\lambda-\mu)c}}, \quad 0<x<c $$ Using this, we find the conditional expectations of \(X\) and \(Y\) given that \(X+Y=c\): $$ E[X|X+Y=c] = \frac{c}{\lambda+\mu} $$ $$ E[Y|X+Y=c] = \frac{c\mu}{\lambda+\mu} $$

Step-by-step solution

Write the individual PDFs for \(X\) and \(Y\)

Since \(X\) and \(Y\) are exponentially distributed, they have the following probability density functions: $$ f_X(x) = \lambda e^{-\lambda x}, \quad x > 0 \\ f_Y(y) = \mu e^{-\mu y}, \quad y > 0 $$

Write the joint PDF for \(X\) and \(Y\)

Since \(X\) and \(Y\) are independent, their joint probability density function is the product of their individual PDFs: $$ f_{X,Y}(x,y) = f_X(x)f_Y(y) = \lambda \mu e^{-\lambda x} e^{-\mu y}, \quad x > 0, y > 0 $$

Write the conditional density function for \(X\) given \(X+Y=c\)

To find the conditional density function \(f_{X|X+Y}(x|c)\), we need to find the joint probability density function of \(X\) and \(Y\) constrained to the condition \(X+Y=c\). We can do this by finding the Jacobian of the transformation and then integrating out the \(Y\) component. The transformation is: $$ u = x \\ v = x + y $$ Taking the inverse transformation, we get: $$ x = u \\ y = v - u $$ Computing the Jacobian: $$ J = \frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} = \begin{vmatrix} 1 & 0 \\ -1 & 1 \end{vmatrix} = 1 $$ Now, we can write the constrained joint density function: $$ f_{U,V}(u,v) = f_{X,Y}(u,v-u) |J| = \lambda \mu e^{-\lambda u} e^{-\mu (v-u)} $$ Now, to find the conditional density function \(f_{X|X+Y}(x|c)\), we need to integrate out the \(V\) component, setting \(v=c\): $$ f_{X|X+Y}(x|c) = \frac{f_{U,V}(x,c)}{\int_{0}^{c} f_{U,V}(x,v) dv} $$ Let's compute the integral and simplify the expression: $$ \int_{0}^{c} f_{U,V}(x,v) dv = \lambda \mu e^{-\lambda x} \int_{0}^{c} e^{-\mu (c-x)} dv = \frac{\lambda \mu e^{-(\lambda-\mu)x}(1-e^{-(\lambda-\mu)c})}{\lambda - \mu} $$ Now, we can substitute this back into the conditional density function: $$ f_{X|X+Y}(x|c) = \frac{(\lambda-\mu) e^{-(\lambda-\mu)x}}{1-e^{-(\lambda-\mu)c}}, \quad 0<x<c $$ #b) Find \(E[X|X+Y=c]\)#

Find the conditional mean

To find the conditional mean \(E[X|X+Y=c]\), we need to take the expectation with respect to the conditional density function \(f_{X|X+Y}(x|c)\): $$ E[X|X+Y=c] = \int_{0}^{c} x f_{X|X+Y}(x|c) dx $$

Evaluate the integral

Let's compute the integral: $$ E[X|X+Y=c] = \int_{0}^{c} x \frac{(\lambda-\mu) e^{-(\lambda-\mu)x}}{1-e^{-(\lambda-\mu)c}} dx $$ Substitute \(u = (\lambda - \mu)x\) and \(du = (\lambda - \mu)dx\), we get: $$ E[X|X+Y=c] = \frac{1}{(\lambda - \mu)(1-e^{-(\lambda-\mu)c})} \int_{0}^{\lambda c - \mu c} \frac{u}{\lambda-\mu} e^{-u} du $$ Integrating by parts, we have: $$ E[X|X+Y=c] = \frac{1}{(\lambda - \mu)(1-e^{-(\lambda-\mu)c})} \left[-\left(\frac{u}{\lambda-\mu}+1\right)e^{-u}\right]_0^{\lambda c - \mu c} = \frac{c}{\lambda + \mu} $$ Thus, \(E[X|X+Y=c] = \frac{c}{\lambda + \mu}\). #c) Find \(E[Y|X+Y=c]\)

Apply conditional expectation

From part (b) we found the conditional expectation of \(X\) given \(X + Y = c\): \(E[X|X+Y=c] = \frac{c}{\lambda + \mu}\). Now, we will use this result to find \(E[Y|X+Y=c]\): $$ E[Y|X+Y=c] = E[c-X|X+Y=c] = c - E[X|X+Y=c] $$

Substitute and simplify

Substitute the result from part (b) and simplify: $$ E[Y|X+Y=c] = c-\frac{c}{\lambda+\mu} = \frac{c\mu}{\lambda+\mu} $$ Thus, \(E[Y|X+Y=c] = \frac{c\mu}{\lambda+\mu}\).

Key concepts

Exponential Random Variables

Exponential random variables play a critical role in statistical analysis, particularly in modeling the time until an event occurs, such as the lifespan of an electronic component or the time between customer arrivals at a service center. An exponential random variable, denoted by the symbol 'X' or 'Y' in this context, is defined by a single parameter, often referred to as the rate parameter (denoted as \(\lambda\) or \(\mu\) respectively). This parameter captures the intensity or frequency of the event occurring.

The probability density function (PDF) for an exponential random variable is expressed mathematically as \(f(x) = \lambda e^{-\lambda x}\), with x being the time and \(\lambda > 0\). One characteristic feature of exponential random variables is their memoryless property, meaning the probability of an event occurring in the future is independent of how much time has already elapsed.

When dealing with scenarios involving two independent exponential random variables, such as 'X' and 'Y' in our problem, we consider the cumulative interaction of these two variables. For example, when summing two independent exponential random variables, we may be interested in the new random variable 'Z = X + Y', representing the total time for both events to occur. Understanding the behavior of 'Z', including its conditional density function given 'X + Y = c', where 'c' is constant, provides valuable insights into the combined process' dynamics.

Probability Density Function

The probability density function (PDF) is a fundamental concept used to specify the probability of a continuous random variable falling within a particular range of values. For an exponential random variable with rate parameter \(\lambda\), the PDF is given by \(f_X(x) = \lambda e^{-\lambda x}\) for \(x > 0\), which dictates the likelihood of the variable 'X' being at or near a specific value.

The key purpose of the PDF is to serve as a tool for finding probabilities for continuous variables. For example, to find the probability that 'X' is between two values 'a' and 'b', we would integrate the PDF from 'a' to 'b'. It's important to note that the integral of the PDF across the entire range of the random variable is always equal to 1, representing the certainty that the variable will take on a value within its domain.

In practical terms, when we have two independent random variables with their respective PDFs, as in the case of 'X' and 'Y', we can determine their joint PDF. Since 'X' and 'Y' are independent, the joint PDF is simply the product of their individual PDFs. This mathematical detail is crucial when deriving the conditional density function for 'X' given 'X + Y = c'.

Expectation Conditional

Conditional expectation extends the concept of expected value into the realm of conditioned probabilities. It represents the expected value (or mean) of a random variable given that a certain condition or event has occurred. In the context of our problem, we are interested in \(E[X | X+Y=c]\), which is the expected value of 'X' given that the sum of 'X' and 'Y' is equal to some constant 'c'.

Computing the conditional expectation involves integrating over the conditional density function, which provides a weighted average of all possible values that 'X' can take, given the conditioning event. The expression \(E[X | X+Y=c]\) can be calculated by integrating the product of 'x' with the conditional density function of 'X' over the interval from 0 to 'c'. The power of conditional expectations lies in their ability to provide insights into one variable's behavior when another variable's behavior is known.

As part of the solution improvement advice, a clear understanding of how to apply the process of integration by parts—an essential technique in calculating expectations—is pivotal. When properly applied, this method simplifies the computation of expected values for complex probability distributions, such as those resulting from conditioning one random variable on another.