Moment Generating Function of a Product of Independent Standard Normal Variables: A Comprehensive Guide
Moment Generating Function of a Product of Independent Standard Normal Variables: A Comprehensive Guide
In this article, we will delve into the moment generating function (MGF) of a specific type of random variable, which is a product of independent standard normal variables. Understanding the MGF is crucial for various applications in probability and statistics. Let us explore the problem in detail.
Introduction to the Problem
The problem statement can be framed as follows:
Exercise
Suppose (X_1, X_2, X_3) are independent and N(0,1)-distributed. What is the moment generating function (MGF) of (Y X_1X_2 - X_1X_3 - X_2X_3).
Solution Approach
To solve this problem, we will employ a few key mathematical concepts and tools, including matrix algebra and the properties of the standard normal distribution. Let us break down the steps involved in computing the MGF of (Y).
Step 1: Matrix Representation
We express (Y) as a quadratic form in terms of a matrix (A) and a vector (mathbf{X}):
[Y frac{1}{2} mathbf{X}^T A mathbf{X}]
Where (A begin{pmatrix} 0 1 1 1 0 1 1 1 0 end{pmatrix}) and (mathbf{X} begin{pmatrix} X_1 X_2 X_3 end{pmatrix}).
Step 2: PDF of Standard Normal Distribution
The probability density function (PDF) of a 3-dimensional standard normal distribution is given by:
[f_{mathbf{x}} frac{1}{(2pi)^{3/2}} e^{-frac{1}{2} mathbf{x}^T mathbf{x}}]
Step 3: Moment Generating Function (MGF) of (Y)
The MGF of (Y) is given by:
[m_Y(t) E[e^{tY}] int_{mathbb{R}^3} e^{frac{1}{2} tmathbf{x}^T A mathbf{x}} cdot frac{1}{(2pi)^{3/2}} e^{-frac{1}{2} mathbf{x}^T mathbf{x}} , dV]
Our objective is to simplify this integral by diagonalizing the matrix (I - tA).
Step 4: Diagonalization and Orthogonal Transformations
Since (A) is symmetric, (I - tA) is also symmetric. This allows us to find an orthogonal matrix (P) such that:
[I - tA P^TDP]
Where (D) is the diagonal matrix containing the eigenvalues of (I - tA).
Step 5: Simplification Using Diagonal Form
With the change of variables (mathbf{y} Pmathbf{x}), the integral becomes:
[m_Y(t) frac{1}{(2pi)^{3/2}} int_{mathbb{R}^3} e^{-frac{1}{2} mathbf{y}^T D mathbf{y}} , dV]
By rewriting the integral in terms of three independent integrals, we have:
[m_Y(t) frac{1}{(2pi)^{3/2}} left( int_{-infty}^{infty} e^{-frac{1}{2} lambda_1 y_1^2} , dy_1 right) left( int_{-infty}^{infty} e^{-frac{1}{2} lambda_2 y_2^2} , dy_2 right) left( int_{-infty}^{infty} e^{-frac{1}{2} lambda_3 y_3^2} , dy_3 right)]
Each of these integrals is a standard normal integral, which equals:
[frac{1}{sqrt{lambda_i}}]
Hence the MGF is:
[m_Y(t) frac{1}{sqrt{lambda_1 lambda_2 lambda_3}}]
Step 6: Determinant of Matrix ((I - tA))
The determinant of the matrix (I - tA) can be found by evaluating the determinant of the matrix:
[begin{vmatrix} 1 - t -t -t -t 1 - t -t -t -t 1 - t end{vmatrix}]
Expanding the determinant, we get:
[text{det}(I - tA) 1 - 3t^2 - 2t^3]
Therefore, the MGF simplifies to:
[m_Y(t) frac{1}{sqrt{|1 - 3t^2 - 2t^3|}}]
Conclusion
The moment generating function of (Y) is well-defined for values of (t) where the denominator is positive. This is given by:
[t in (-infty, -1) cup left(-1, -frac{1}{2}right)]
Thus, the final moment generating function is:
[boxed{frac{1}{sqrt{1 - 3t^2 - 2t^3}}}]