Problem 96 - An Introduction to Mathematical Statistics and Its Applications

Question

If the random variable \(Y\) has the gamma pdf with integer parameter \(r\) and arbitrary \(\lambda>0\), show that $$ E\left(Y^{m}\right)=\frac{(m+r-1) !}{(r-1) ! \lambda^{m}} $$ (Hint: Use the fact that \(\int_{0}^{\infty} y^{r-1} e^{-y} d y=(r-1) !\) when \(r\) is a positive integer.)

Step-by-Step Solution

Answer

The expected value of \(Y^{m}\) for a random variable \(Y\) following a gamma distribution with parameters \(r\) and \(\lambda\) is \(E\left(Y^{m}\right)=\frac{(m+r-1)!}{(r-1)! \lambda^{m}}\)

Step 1: Define The Random Variable Y

The random variable \(Y\) follows a gamma distribution. The gamma pdf for \(Y\) with parameters \(r\) and \(\lambda\) is given by: \[ f(y)=\frac{e^{-y/\lambda} y^{r-1}}{\lambda^{r}(r-1)!}, y>0 \]

Step 2: Compute The Expected Value of Y^m

To compute the expected value of \(Y^{m}\) denoted \(E(Y^m)\), we need to evaluate the integral: \[ E[Y^{m}]=\int_{0}^{\infty} y^{m} f(y) dy \] Substitute \(f(y)\) in the above equation, we get:\[ E[Y^{m}]=\int_{0}^{\infty} y^{m} \frac{e^{-y/\lambda} y^{r-1}}{\lambda^{r}(r-1)!} dy \]

Step 3: Simplify The Integral

The integral in Step 2 simplifies to \[ E[Y^{m}]=\frac{1}{\lambda^{r}(r-1)!} \int_{0}^{\infty} y^{r+m-1} e^{-y/\lambda} dy \] Now change variables in the integral with \(z = y/\lambda\). The integral then becomes \[E[Y^{m}]=\frac{1}{(r-1)!} \int_{0}^{\infty} (\lambda z)^{r+m-1} e^{-z} \lambda dz \] Which simplifies to \[E[Y^{m}]= \frac{\lambda^m}{(r-1)!} \int_{0}^{\infty} z^{r+m-1} e^{-z} dz \]

Step 4: Apply The Gamma Function

The integral in the expression for \(E(Y^{m})\), \(\int_{0}^{\infty} z^{r+m-1} e^{-z} dz\), corresponds to the gamma function when \(r\) and \(m\) are positive integers. Using the hint provided in the exercise, this integral is equal to \((r+m-1)!\). Hence the expression for \(E(Y^{m})\) becomes \[E[Y^{m}]= \frac{\lambda^m (r+m-1)!}{(r-1)!}\]

Step 5: Final Result

The expression for \(E(Y^{m})\) simplifies to \[ E(Y^{m}) = \frac{(m + r - 1)!}{(r-1)! \lambda^m}\] as required. This is obtained by using the properties of the factorial function, and it aligns with the result we wanted to prove.

Key Concepts

Expected Value

The expected value of a random variable is a key concept in probability and statistics. It provides a measure of the center of a probability distribution and is often considered as the "average" value. For the gamma distribution, and especially for this exercise, we're interested in the expected value of powers of the random variable.

When you see the term "expected value," think of it as the theoretical mean of the random variable that describes the average result of an experiment after many repetitions. This is represented mathematically as:

• For a random variable \(Y\), the expected value \(E(Y)\) is calculated by integrating the product of \(Y\) with its probability density function over its entire range.
• The general form is \(E[g(Y)] = \int_{-fty}^{fty} g(y)f(y)dy\), where \(g(Y)\) is any function of \(Y\).

In the context of the exercise, the expected value of \(Y^m\) involves integrating \(Y^m\) times the gamma probability density function (pdf).

Gamma Function

The gamma function is an extension of the factorial function to complex numbers. For positive integers, it's quite similar to the factorial, as \(\Gamma(n) = (n-1)!\). However, the gamma function's true power is its ability to work for non-integer values as well.

The integral definition of the gamma function is:

• \(\Gamma(n) = \int_{0}^{\infty} t^{n-1} e^{-t} dt\) for \(n > 0\),
• For integers, it becomes \(\Gamma(n) = (n-1)!\).

Understanding the gamma function is critical when working with gamma distributions, which is evident in solving the given problem. It allows us to simplify integral expressions involving factorials of non-integers. In particular, it simplifies the integral of \(z^{r+m-1} e^{-z}\) to \((r+m-1)!\), aiding significantly in calculating the expected value.

Probability Density Function

A probability density function (pdf) describes the likelihood of a random variable to take on a particular value. For continuous random variables, the pdf is crucial as it helps determine probabilities for a range of values rather than specific outcomes.

For a gamma-distributed variable, the pdf is given by:

• \(f(y) = \frac{e^{-y/\lambda} y^{r-1}}{\lambda^{r}(r-1)!}\) for \(y > 0\).
• It incorporates an exponential term and a polynomial term \(y^{r-1}\), modulated by \(\lambda\) and \(r\).

This functional form of the pdf is central to calculating the expected values, as seen in the exercise, where the integral of this function with \(Y^m\) allows us to find \(E(Y^m)\). The pdf thus provides the foundation for further probabilistic calculations and analyses.

Random Variable

A random variable is a variable whose possible values are outcomes of a random phenomenon. In probability, we differentiate between continuous and discrete random variables. Gamma-distributed variables are continuous since they can take any value within a range.

When dealing with a random variable:

• It's denoted as \(Y\) or similar symbols in mathematics.
• Characterized by a probability distribution, which in this context is the gamma distribution.

Understanding what a random variable represents is essential in tackling the exercise. Here, \(Y\) is the random variable following a gamma distribution with parameters \(r\) and \(\lambda\). This influences everything from the form of the pdf to how we calculate expected values and other statistical measures.