Question

Let \(X_{n}\) and \(Y_{n}\) be \(p\) dimensional random vectors such that \(X_{n}\) and \(\mathbf{Y}_{n}\) are independent for each \(n\) and their mgfs exist. Show that if $$ \mathbf{X}_{n} \stackrel{D}{\rightarrow} \mathbf{X} \text { and } \mathbf{Y}_{n} \stackrel{D}{\rightarrow} \mathbf{Y} $$

Short answer

If two random vectors \(X_{n}\) and \(Y_{n}\) converge in distribution to \(X\) and \(Y\) respectively, and they are independent, then their product \(X_{n}Y_{n}\) also converges in distribution to \(XY\), as the moment generating functions (mgfs) of \(X_{n}Y_{n}\) converge to the product of mgfs of \(X\) and \(Y\).

Step-by-step solution

Define moment generating functions (mgfs)

The moment generating function (mgf) of a random vector \(X\), denoted as \(M_{X}(t)\), is defined as the expected value of \(e^{tX}\). Similarly, the mgf of \(Y\) can be defined. Note that \(X_{n}\) and \(Y_{n}\) are independent, thus \(M_{X_{n}Y_{n}}(t)=M_{X_{n}}(t)M_{Y_{n}}(t)\) due to this property.

Applying conditions of convergence

When \(X_{n}\) and \(Y_{n}\) converge in distribution to \(X\) and \(Y\), their mgfs also converge, i.e., \(M_{X_{n}}(t)\rightarrow M_{X}(t)\) and \(M_{Y_{n}}(t)\rightarrow M_{Y}(t)\), as \(n\rightarrow \infty\).

Proving convergence in distribution of their product

We need to prove that the mgf of \(X_{n}Y_{n}\) converges to the product of mgfs of \(X\) and \(Y\). From step 2, mgfs of \(X_{n}\) and \(Y_{n}\) converge to mgfs of \(X\) and \(Y\) respectively. So, \(M_{X_{n}Y_{n}}(t)=M_{X_{n}}(t)M_{Y_{n}}(t) \rightarrow M_{X}(t)M_{Y}(t)=M_{XY}(t)\). This implies that \(X_{n}Y_{n}\) converges in distribution to \(XY\).

Key concepts

Moment Generating Function

A moment generating function, often abbreviated as mgf, is an incredibly useful tool in probability and statistics. It helps in studying the distribution of a random variable and finding properties like the mean and variance. The mgf of a random vector \(X\) is denoted by \(M_X(t)\) and is calculated as the expected value \(E[e^{tX}]\). Simply put, you take the average of the expression \(e^{tX}\) over the probability distribution of \(X\).

One fascinating property of mgfs is that they can uniquely determine the distribution of a random variable. This means, if two random variables have the same mgf, they have the same distribution. This property is what makes mgfs so powerful in proving convergence properties.

• MGF helps in finding moments (mean, variance) of a distribution.
• It is instrumental in proving theorems related to the convergence of distributions.
• Key advantage: mgfs can be used to determine the distribution uniquely.

Independent Random Vectors

When dealing with random vectors such as \(X_n\) and \(Y_n\), the term 'independence' has a special significance. Two random vectors are independent if the occurrence of one does not affect the occurrence of the other. This is crucial in proving many results in probability.

In mathematical terms, if \(X_n\) and \(Y_n\) are independent, then the probability distribution of their joint behavior can be expressed as the product of their individual distributions. For moment generating functions, this implies \(M_{X_nY_n}(t) = M_{X_n}(t)M_{Y_n}(t)\). Independence leads to a simplified analysis of joint distributions and is often a key assumption in statistical modeling.

• Independence simplifies the calculation of joint probabilities.
• Independent random vectors have separate distributions that do not influence each other.
• In mgfs, independence allows the product of the mgfs to express the joint distribution.

Convergence of Moment Generating Functions

Convergence in distribution is an important concept when studying sequences of random vectors. A random vector sequence \(X_n\) converges in distribution to \(X\) if the distribution of \(X_n\) gets closer to the distribution of \(X\) as \(n\) increases. Moment generating functions are highly effective in proving this type of convergence: if the mgfs of \(X_n\) converge to the mgf of \(X\), then \(X_n\) converges in distribution to \(X\).

For two independent random vectors \(X_n\) and \(Y_n\), if their respective mgfs converge to those of \(X\) and \(Y\), then their product \(X_nY_n\) also converges to \(XY\). This is because independence allows the product to reflect simple multiplication of individual mgfs, making the result straightforward.

• Convergence in distribution: essential for statistical inferences.
• MGF convergence serves as a tool to prove distributional convergence.
• Independence simplifies convergence of product distributions using mgfs.