Question

Let \(X\) and \(Y\) be independent exponential random variables with respective rates \(\lambda\) and \(\mu\). (a) Argue that, conditional on \(X>Y\), the random variables \(\min (X, Y)\) and \(X-Y\) are independent. (b) Use part (a) to conclude that for any positive constant \(c\) $$ \begin{aligned} E[\min (X, Y) \mid X>Y+c] &=E[\min (X, Y) \mid X>Y] \\ &=E[\min (X, Y)]=\frac{1}{\lambda+\mu} \end{aligned} $$ (c) Give a verbal explanation of why \(\min (X, Y)\) and \(X-Y\) are (unconditionally) independent.

Short answer

In this exercise, we have shown that given X>Y, the random variables min(X,Y) and X-Y are conditionally independent. We found the joint probability distribution of min(X,Y) and X-Y given X>Y, and showed that it is equal to the product of their marginal probability distributions. Using this conditional independence, we concluded that for any positive constant c, the conditional expectation E[min(X,Y) | X>Y+c] equals E[min(X,Y)] = 1/(λ+μ). Finally, we explained that the unconditional independence of min(X,Y) and X-Y is due to the fact that knowing the time until one event occurs in independent exponential random variables provides no information about the other, making their minimum and difference independent.

Step-by-step solution

Understand conditional independence

First, we need to understand what it means for two random variables to be conditionally independent. Given two random variables X and Y, they are conditionally independent given a third random variable Z if and only if the conditional probability distribution of X given Y and Z is independent of Y. In other words, P(X|Y,Z) = P(X|Z). Now, let's apply this concept to the given exercise.

Show conditional independence

Given that X>Y, we need to show that min(X,Y) and X-Y are independent random variables. To do this, we need to find the joint probability distribution of min(X,Y) and X-Y given X>Y, and show that it is equal to the product of their marginal probability distributions. Since X and Y are independent, their joint probability distribution is given by the product of their marginal probability distributions: \(f_{X,Y}(x,y) = f_X(x)f_Y(y) = \lambda e^{-\lambda x} \cdot \mu e^{-\mu y}\) for x>0 and y>0. Now, we need to find the joint probability distribution of min(X,Y) = Z and X-Y = W, given the condition X>Y. We can use the transformation technique for this: \(f_{Z,W|X>Y}(z,w) = \frac{f_{X,Y}(x,y)}{P(X>Y)}\Bigg \lvert \frac{\partial(x,y)}{\partial(z,w)} \Bigg \rvert \) where x(z,w) = z + w and y(z,w) = z. The Jacobian is given by: \(\Bigg \lvert \frac{\partial(x,y)}{\partial(z,w)} \Bigg \rvert = \lvert \det \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \rvert = 1\) As X and Y are independent, we know the joint distribution of X,Y: \(f_{X,Y}(x,y) = \lambda e^{-\lambda x} \mu e^{-\mu y}\) Therefore, \(f_{Z,W|X>Y}(z,w) = \frac{\lambda e^{-\lambda(z+w)} \mu e^{-\mu z}}{P(X>Y)}\)

Check for independence

Now, we need to check if the joint probability distribution is equal to the product of their marginal probability distributions. \(f_{Z|X>Y}(z) = \int_{w=0}^{\infty} f_{Z,W|X>Y}(z, w) dw = (\lambda+\mu)e^{-(\lambda+\mu)z}\) \(f_{W|X>Y}(w) = \int_{z=0}^{\infty} f_{Z,W|X>Y}(z, w) dz = \lambda e^{-\lambda w}\) \(f_{Z|X>Y}(z)f_{W|X>Y}(w) = (\lambda+\mu)e^{-(\lambda+\mu)z} \cdot \lambda e^{-\lambda w} = f_{Z,W|X>Y}(z, w)\) Since the joint probability distribution is equal to the product of their marginal probability distributions, min(X,Y) and X-Y are conditionally independent given X>Y.

Solve for conditional expectation

Using the derived conditional independence, we can solve for the given conditional expectation: \(E[\min (X, Y) \mid X>Y+c] = E[\min (X, Y) \mid X>Y] = E[\min (X, Y)]=\frac{1}{\lambda+\mu}\)

Verbal explanation

Verbally, the fact that min(X,Y) and X-Y are unconditionally independent can be explained as follows: The minimum of two independent exponential random variables represents the time until the first event occurs, while the difference between them represents the additional time until the second event occurs. Since both events are independent, knowing the time until one event occurs provides no information about the other, making their minimum and difference independent.