Question

Find the moments of the distribution that has mgf \(M(t)=(1-t)^{-3}, t<1\). Hint: Find the MacLaurin's series for \(M(t)\).

Short answer

The moments of the distribution that has MGF \(M(t)=(1-t)^{-3}, t<1\) are in the form of an alternating sequence starting from -3 and increasing by two every time.

Step-by-step solution

Identify MacLaurin series

The Maclaurin series for the function \(1/(1-t)^3\) can be obtained from the binomial series, which, in general, is \( (1-x)^n = \sum_{k=0}^\infty \binom{n}{k} (-x)^k \) where \(-x\) is replaced with \(t\).

Calculate coefficients

Using n = -3 in the binomial series will give the Maclaurin expansion for \(M(t)\), which is \(M(t)= \sum_{k=0}^\infty \binom{-3}{k} t^k\). The binomial coefficient \(\binom{n}{k}\) can be computed as \((n)(n-1)...(n-k+1)/k!\). Using this formula with n = -3 gives coefficients for the first few terms as: \(k=0: 1; k=1: -3, k=2: 6, k=3: -10, ...\) each multiplied by \(t^k\).

Identify moments

The coefficients of the Maclaurin series of \(M(t)\) represent the moments of the distribution. Hence, based on the computations in step 2, the first moment of the distribution is -3, and the second moment is 6 and so on.

Key concepts

Maclaurin Series

The Maclaurin series is essentially a Taylor series expansion of a function around zero. It's a powerful tool for expressing complex functions as an infinite sum of terms calculated from the values of their derivatives at a single point, typically 0.
The general formula for the Maclaurin series of a function \(f(x)\) is:

• \( f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \ldots \)

In this context, knowing the Maclaurin series allows us to convert a more complicated function into something easier to handle. For the exercise, the function \((1-t)^{-3}\) is expanded using this series.
The coefficients of \(t^k\) in this series represent the moments of a distribution when dealing with moment generating functions.

Binomial Series

The binomial series provides a way to express powers of the form \((1 - x)^n\) as an infinite sum when \(n\) is any real number, not just an integer.
It's given by:

• \((1-x)^n = \sum_{k=0}^{\infty} \binom{n}{k} (-x)^k\)

Here, \(\binom{n}{k}\) represents the binomial coefficient, which tells us the number of ways to choose \(k\) elements from \(n\) possibilities, and can be calculated using the formula:

• \(\binom{n}{k} = \frac{n(n-1)(n-2)\ldots (n-k+1)}{k!}\)

In our problem, using \(n = -3\) gives the series expansion for \((1-t)^{-3}\) as an infinite sum. This helps us find the Maclaurin series by substituting \(t\) for \(-x\). The coefficients obtained by this expansion are crucial in deriving the moments of the distribution.

Moments of a Distribution

The moments of a distribution provide key insights into its shape and properties. They quantify aspects such as the mean, variance, and skewness.
The \(n\)-th moment of a probability distribution is the expected value of the \(n\)-th power of deviations from a fixed point, typically the mean. However, when dealing with a moment generating function (mgf) like in this problem, the moments can be extracted directly from the series expansion.
Thus, for the function \(M(t) = (1-t)^{-3}\), its Maclaurin series expansion gives us the coefficients directly as the moments:

• First moment (Mean): coefficient of \(t\), which is \(-3\).
• Second moment: coefficient of \(t^2\), which is \(6\).

Recognizing these coefficients helps in understanding the behavior and characteristics of the underlying distribution.