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Chapter 1: Problem 17

Let \(\psi(t)=\log M(t)\), where \(M(t)\) is the mgf of a distribution. Prove that \(\psi^{\prime}(0)=\mu\) and \(\psi^{\prime \prime}(0)=\sigma^{2} .\) The function \(\psi(t)\) is called the cumulant generating function.

Short Answer

Expert verified
The first derivative of the cumulant generating function at t=0 equals the mean (\(\mu\)) of the distribution, and the second derivative at t=0 equals the variance (\(\sigma^2\)).

Step by step solution

01

Understand the Requirement

We need to prove that the first derivative of the cumulant generating function, \(\psi(t) = \log{M(t)}\), evaluated at t=0, equals the mean (\(\mu\)) of a distribution, and the second derivative evaluated at t=0 equals the variance (\(\sigma^{2}\)).
02

Compute the First Derivative

We compute the first derivative of \(\psi(t)\), which we will denote as \(\psi^{\prime}(t)\). By using the chain rule, we obtain \(\psi^{\prime}(t) = \frac{M^{\prime}(t)}{M(t)}\). Then, we evaluate this at t=0, i.e. \(\psi^{\prime}(0) = \frac{M^{\prime}(0)}{M(0)}\).
03

Use Properties of Moment Generating Function

The moment generating function evaluated at t=0, i.e. M(0), is always 1 for any distribution. This gives us \(\psi^{\prime}(0) = M^{\prime}(0)\). It is also known that the first derivative of the moment generating function evaluated at t=0 is equal to the mean, \(\mu\), of the distribution. Hence \(\psi^{\prime}(0) = M^{\prime}(0) = \mu\).
04

Compute the Second Derivative

Next, we need to compute the second derivative of \(\psi(t)\), denoted as \(\psi^{\prime\prime}(t)\). We differentiate \(\psi^{\prime}(t)\) to get \(\psi^{\prime\prime}(t) = \frac{M^{\prime\prime}(t)}{M(t)} - \left(\frac{M^{\prime}(t)}{M(t)}\right)^2\). Then, we evaluate this at t=0, i.e. \(\psi^{\prime\prime}(0) = \frac{M^{\prime\prime}(0)}{M(0)} - \left(\frac{M^{\prime}(0)}{M(0)}\right)^2\).
05

Use Properties again of Moment Generating Function

Again, since M(0) is 1, we get \(\psi^{\prime\prime}(0) = M^{\prime\prime}(0) - \left(M^{\prime}(0)\right)^2\). We know that \(\psi^{\prime}(0) = M^{\prime}(0) = \mu\), and \(M^{\prime\prime}(0) = \mu^2 + \sigma^2\). Substituting these values into the formula, we have \(\psi^{\prime\prime}(0) = \mu^2 + \sigma^2 - \mu^2\), which simplifies to \(\psi^{\prime\prime}(0) = \sigma^2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Moment Generating Function
The Moment Generating Function (MGF) is a fundamental tool in probability theory used to characterize the distribution of a random variable. It is represented as \( M(t) \) and is defined for a random variable \( X \) as the expected value of \( e^{tX} \), which is \( M(t) = E[e^{tX}] \).

Key properties of the MGF include:
  • Existence: Not all random variables have an MGF because the expected value may not exist for all \( t \). However, when it exists, it uniquely determines the distribution.
  • Initial Value: At \( t = 0 \), the MGF is always 1, that is, \( M(0) = 1 \). This is because \( e^{tX} = 1 \) when \( t = 0 \).
  • Relationship with Moments: The derivatives of the MGF at \( t = 0 \) relate to the moments of the distribution. Specifically, the n-th derivative of \( M(t) \) at \( t = 0 \) gives the n-th moment of the distribution.
The MGF is particularly powerful because it simplifies the process of finding moments (such as the mean and variance) and is often used in proofs and theoretical analysis.
Mean and Variance
The mean and variance are crucial statistical measures used to describe the central tendency and dispersion of a probability distribution.

To understand these concepts, let's break them down:
  • Mean (\( \mu \)): This is the average or expected value of a random variable. In a probability distribution, the mean tells us the central point or the "center of mass" of the distribution.
  • Variance (\( \sigma^2 \)): This measures how spread out the distribution is around the mean. A larger variance indicates a wider spread, while a smaller variance indicates that the data points are closer to the mean.
  • Calculating Mean and Variance: The first derivative of the cumulant generating function \( \psi(t) = \log M(t) \) at \( t = 0 \) gives us the mean, \( \psi'(0) = \mu \).
    The second derivative at \( t = 0 \) gives us the variance, \( \psi''(0) = M''(0) - [M'(0)]^2 = \sigma^2 \).
Thus, calculating derivatives at specific points allows us to determine these key statistical measures, providing insights into the data鈥檚 distribution.
Derivatives in Statistics
Derivatives play a vital role in statistical calculations and are a key concept in understanding how changes in variables affect other variables.

In statistics, derivatives are used to:
  • Find Moments: The n-th derivative of the MGF at \( t = 0 \) helps to discover the n-th moment of a distribution, which can include finding the mean and variance as first and second moments, respectively.
  • Understand Growth Rates: Derivatives give insights into the rate of change, offering an understanding of how probability distributions shift with different variables.
  • Analyze Cumulant Generating Functions: The cumulant generating function \( \psi(t) \) uses derivatives to link the moments to cumulants, allowing for precise understanding of the skewness and kurtosis of a distribution.
Using derivatives in this way provides a deeper understanding of the underlying structure of statistical data, revealing complex relationships and supporting predictive modeling.

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