Problem 266 - An Introduction to Mathematical Statistics and Its Applications

Question

Find \(E\left(Y^{4}\right)\) if \(Y\) is an exponential random variable with \(f_{Y}(y)=\lambda e^{-\lambda y}, y>0\).

Step-by-Step Solution

Answer

The expected value of \(Y^{4}\) for the given exponential random variable \(Y\) is \(4!\lambda\).

Step 1: Understanding the Exercise

We're given that \(Y\) is an exponential random variable with a probability density function defined as \(f_{Y}(y)=\lambda e^{-\lambda y}, y>0\). We need to find \(E\left(Y^{4}\right)\), i.e., the expected value of \(Y^{4}\). To do this, we remind ourselves that the expected value is calculated as \[E[g(Y)] = \int_{-\infty}^{\infty} g(y) f_Y(y) dy\] where \(g(y) = y^{4}\) in this case.

Step 2: Substitution

Substitute \(g(y)\) and \(f_Y(y)\) in the definition of the expected value: \[E\left(Y^{4}\right) = \int_{0}^{\infty} y^{4} \lambda e^{-\lambda y} dy\] because \(y>0\) in the case of the exponential distribution.

Step 3: Calculating the Expectation

The integral might seem complicated at first glance but it can be calculated using the gamma function which is defined as \(\Gamma(z) = \int_{0}^{\infty} t^{z-1} e^{-t} dt\) for any \(z > 0\). The integral then becomes \(E\left(Y^{4}\right) = \lambda \Gamma(5) = 4! \lambda\) as the result of the gamma function is \(Gamma(5) = 4!\).

Step 4: Summary

After calculating, we thus have that the expected value \(E(Y^{4})\) of the random variable \(Y\) equals to \(4!\lambda\).

Key Concepts

Expected Value

The concept of expected value is fundamental in probability and statistics. It allows us to summarize a random variable by its average outcome. In the context of an exponential distribution, which is continuous and characterized by a probability density function (PDF), expected value provides insight into what we might expect from repeated observations of a random variable like \(Y\).

When we're asked to find the expected value of \(Y^4\), it means we need the average value of \(Y\) raised to the fourth power. This is expressed mathematically as \(E(Y^4)\). To find this, we employ the expected value formula for continuous random variables:

• \(E[g(Y)] = \int_{0}^{\infty} g(y) f_Y(y) \ dy\)

In our specific case, since \(g(y) = y^4\), and given the exponential PDF \(f_Y(y) = \lambda e^{-\lambda y}\), the formula becomes:

• \(E(Y^4) = \int_{0}^{\infty} y^4 \lambda e^{-\lambda y} dy\)

Understanding this allows us to compute expected values even for complex transformations of random variables. It's a powerful tool that provides a single number representing the mean outcome of those transformations.

Gamma Function

The Gamma function is a crucial mathematical concept when dealing with integrals, particularly in probability and statistics. It's especially useful when we encounter integrals that resemble its definition. The Gamma function is defined as:

• \(\Gamma(z) = \int_{0}^{\infty} t^{z-1} e^{-t} dt\)

for any \(z > 0\). This function generalizes the factorial function to non-integer values. In our exercise, it's used to simplify the computation of the expected value of \(Y^4\) from the exponential distribution.

The integral \(\int_{0}^{\infty} y^4 \lambda e^{-\lambda y} dy\) can be intricate, but by using a substitution method and relating it to the Gamma function, it becomes more manageable. Recognizing that the format resembles \(\Gamma(z)\), we rewrite the integral to make if fit. We find \(E(Y^4) = \lambda \Gamma(5)\). Here, \(\Gamma(5)\) equals \(4!\) because \(\Gamma(n) = (n-1)!\) when \(n\) is a positive integer. This transforms our integral into a much easier problem to evaluate and results in our solution.

Probability Density Function

A probability density function (PDF) is crucial for characterizing continuous random variables. It describes the likelihood of a random variable taking on a particular value. For an exponential distribution, the PDF is critical in understanding its properties.

The PDF for an exponential distribution is given by:

• \(f_Y(y) = \lambda e^{-\lambda y}, \quad y > 0\)

This function tells us that the probability of \(Y\) being in a small interval \((y, y+dy)\) is proportional to \(f_Y(y)\); however, since \(Y\) is continuous, we determine probabilities over intervals by integrating the PDF over those intervals.

The parameter \(\lambda\) in the PDF plays a double role; it is both the rate of the process and the inverse of the average or expected value \(E(Y)\). Understanding this function helps elucidate behaviors such as memorylessness where the probability of future events is independent of past events. This unique property of the exponential distribution is significant when modeling processes like radioactive decay or server response times.