Problem 197 - An Introduction to Mathematical Statistics and Its Applications

Question

Suppose that \(f_{X, Y}(x, y)=\lambda^{2} e^{-\lambda(x+y)}, 0 \leq x, 0 \leq y\). Find \(E(X+Y)\).

Step-by-Step Solution

Answer

The expected value of the sum \(X + Y\) is \(2/\lambda\).

Step 1: Identify the Marginal Distributions of X and Y

Since \(X\) and \(Y\) are independent, we can find their marginal distributions separately. The marginal probability density function of \(X\) is given by \(\int_{-\infty}^{\infty}f_{X,Y}(x,y) dy = f_X(x) = \lambda e^{-\lambda x}\), for \(x \geq 0\). Similarly, the marginal probability density function of \(Y\) is \(\int_{-\infty}^{\infty}f_{X,Y}(x,y) dx = f_Y(y) = \lambda e^{-\lambda y}\), for \(y \geq 0\). These are both exponential distributions with parameter \(\lambda\).

Step 2: Apply Expectation Rules

Since \(X\) and \(Y\) are independent, \(E(X+Y)=E(X)+E(Y)\). This property is known as the linearity of expectations.

Step 3: Calculate the Expected Values

The expected value of an exponential random variable with parameter \(\lambda\) is \(1/\lambda\). Therefore, we have \(E(X) = 1/\lambda\) and \(E(Y) = 1/\lambda\).

Step 4: Use Linearity of Expectation to Find the Final Value

Add the two expected values to find \(E(X+Y)\). Thus, \[E(X+Y) = E(X) + E(Y) = 1/\lambda + 1/\lambda = 2/\lambda\].

Key Concepts

Marginal Probability Density Function

Understanding the marginal probability density function (PDF) is critical when dealing with multiple random variables. It involves the process of finding the probability distribution for a subset of the variables within a multivariate distribution.

Consider a joint PDF, which gives the probability that two random variables take on specific values simultaneously. To find the marginal PDF of one variable, say X, we integrate the joint PDF over the range of the other variable, Y, essentially 'marginalizing' out Y. This gives us a function that only depends on X, which describes how X is distributed irrespective of Y.

In the exercise, the integration of the joint PDF of X and Y over all possible values of Y yields the marginal PDF of X. This function, \[\begin{equation}\int_{-fty}^{fty}f_{X,Y}(x,y) dy = f_X(x) = \lambda e^{-\lambda x}\end{equation}\],is for all x greater or equal to 0. It tells us how likely different values of X are, without considering Y. The same approach is used to find the marginal PDF of Y. When we calculate the marginal PDFs of X and Y, they both turn out to be exponential distributions.

Linearity of Expectations

One of the most powerful and convenient properties in the realm of expected values in statistics is the linearity of expectations. This principle allows us to say that the expected value of the sum of random variables is equal to the sum of their expected values, regardless of whether the variables are independent or not.

In formula terms, if X and Y are random variables, then\[\begin{equation}E(X + Y) = E(X) + E(Y)\end{equation}\].

This property greatly simplifies calculations, especially in cases where dealing with joint distributions directly would be cumbersome. In our exercise, since the random variables X and Y are independent, their joint distribution factors into the product of their marginal distributions, but we actually don't need to use this fact to utilize the linearity of expectations. We simply add up the separate expected values we find for X and Y.

Exponential Distribution

The exponential distribution is a continuous probability distribution often associated with the time between events in a Poisson point process. It's an important distribution in the field of reliability and life testing, as well as in describing the time until some specific event occurs, like the failure of a mechanical system.

Mathematically, if a random variable X follows an exponential distribution with rate parameter \(\lambda\), the PDF is given by\[\begin{equation}\lambda e^{-\lambda x}\end{equation}\],where x is greater or equal to 0, and \(\lambda\) is strictly positive. A key property of the exponential distribution is that it is memoryless, meaning that if X is the time until an event occurs, the probability that X is greater than some value does not depend on how much time has already elapsed without the event occurring.

The mean or expected value of an exponentially distributed random variable X with rate \(\lambda\) is \(1/\lambda\). This result is used in the exercise to quickly find the expected values of the random variables X and Y, since these are independently exponentially distributed with the same rate parameter \(\lambda\).