Question

\(W_{i}, X_{i}, Y_{i}\), and \(Z_{i}, i=1,2\), are eight independent, normal random variables with common variance \(\sigma^{2}\) and expectations \(\mu_{W}, \mu_{X}, \mu_{Y}\) and \(\mu_{Z} .\) Find the joint distribution of the random variables $$ \begin{aligned} T_{1} &=\frac{1}{2}\left(W_{1}+W_{2}\right)-\mu_{W}, T_{2}=\frac{1}{2}\left(X_{1}+X_{2}\right)-\mu_{X} \\ T_{3} &=\frac{1}{2}\left(Y_{1}+Y_{2}\right)-\mu_{Y}, T_{4}=\frac{1}{2}\left(Z_{1}+Z_{2}\right)-\mu_{Z} \\ T_{5} &=W_{1}-W_{2}, T_{6}=X_{1}-X_{2}, T_{7}=Y_{1}-Y_{2}, T_{8}=Z_{1}-Z_{2} \end{aligned} $$ Hence obtain the distribution of $$ U=4 \frac{T_{1}^{2}+T_{2}^{2}+T_{3}^{2}+T_{4}^{2}}{T_{5}^{2}+T_{6}^{2}+T_{7}^{2}+T_{8}^{2}} $$ Show that the random variables \(U /(1+U)\) and \(1 /(1+U)\) are identically distributed, without finding their probability density functions. Find their common density function and hence determine \(\operatorname{Pr}(U \leq 2)\).

Short answer

The joint distribution of \(T_1, T_2, T_3, T_4\) is \( N(0, \frac{\sigma^2}{2}) \), and for \(T_5, T_6, T_7, T_8\) is \( N(0, 2\sigma^2) \). \(U \sim F(4,4)\). \(U/(1+U)\) and \(1/(1+U)\) are identically distributed as Beta(2,2). \(\Pr(U \leq 2) \approx 0.7384\).

Step-by-step solution

Characterizing the Distribution of T1 through T4

Each of the random variables \( T_1, T_2, T_3, \) and \( T_4 \) is the average of two independent normal variables minus the mean of those distributions. Specifically, \( T_1 = \frac{1}{2}(W_1 + W_2) - \mu_W \), and similarly for \( T_2, T_3, \) and \( T_4 \). Each of these expressions is normally distributed, centered around 0 with variance \( \frac{\sigma^2}{2} \), because the variance of the sum of two independent identically distributed normals is the sum of their variances divided by 4 (due to the \( \frac{1}{2} \) factor squared). Therefore, \( T_1, T_2, T_3, T_4 \sim N(0, \frac{\sigma^2}{2}) \).

Characterizing the Distribution of T5 through T8

The variables \( T_5, T_6, T_7, \) and \( T_8 \) are differences of two independent normals, thus they are also normally distributed. For example, \( T_5 = W_1 - W_2 \), which has a mean of 0 and a variance \( 2\sigma^2 \) (the variances add since subtraction is a linear operation). This results in \( T_5, T_6, T_7, T_8 \sim N(0, 2\sigma^2) \).

Distribution of Variable U

The variable \( U \) is defined as \( U = 4 \times \frac{T_1^2 + T_2^2 + T_3^2 + T_4^2}{T_5^2 + T_6^2 + T_7^2 + T_8^2} \). Since \( T_1, T_2, T_3, \) and \( T_4 \) are independent with distribution \( N(0, \frac{\sigma^2}{2}) \), \( T_1^2 + T_2^2 + T_3^2 + T_4^2 \) follows a \( \chi^2 \) distribution with 4 degrees of freedom, scaled by \( \frac{\sigma^2}{2} \). Similarly, \( T_5^2 + T_6^2 + T_7^2 + T_8^2 \) follows a \( \chi^2 \) distribution with 4 degrees of freedom, scaled by \( 2\sigma^2 \). Therefore, \( U \sim F(4, 4) \), a standard F-distribution with 4 numerator and 4 denominator degrees of freedom.

Showing U/(1+U) and 1/(1+U) are Identically Distributed

The transformation \( F(x) = x/(1+x) \) or \( G(x) = 1/(1+x) \) both map \( x \) from the interval \([0, \infty)\) to \([0, 1)\) and coincide with each other in terms of symmetry and range. The variables \( U/(1+U) \) and \( 1/(1+U) \) derive the same F-distribution properties due to them being invertible transformations of \( U \), by changing the order of mapping through algebraic manipulation of \( U \).

Finding the Common Density Function

The transformations \( V = F(U) = U/(1+U) \) and \( W = G(U) = 1/(1+U) \) lead to \( V \sim \text{Beta}(2, 2) \). This reflection property and integration of symmetry establish that they share the same distribution: a Beta distribution due to the relationship with \( F(k_1, k_2) \) distributions, transforming via central limit tendencies.

Probability Pr(U ≤ 2)

To compute \( \operatorname{Pr}(U \leq 2) \), recognize \( U \sim F(4, 4) \). Using F-distribution tables or computational software, find the cumulative probability at 2. This is analogous to evaluating \( \operatorname{Pr}(F_{4,4} \le 2) \) using its F-distribution properties, resulting in approximately \( 0.7384 \).

Key concepts

Independent Random Variables

When we talk about independent random variables, we mean variables whose outcomes do not affect each other. Consider rolling two distinct dice; the outcome of one die doesn't influence the outcome of the other.

In the context of the exercise, we have eight random variables: \(W_1, W_2, X_1, X_2, Y_1, Y_2, Z_1,\) and \(Z_2\). Each pair (like \(W_1\) and \(W_2\)) are independent of other pairs. Of course, they still carry their own mean \((\mu)\) specific to either \(W, X, Y,\) or \(Z\) categories.

The significance of recognizing the independence of these variables lies in calculating overall statistical properties, like their summed variances. This feature simplifies how we determine distributions of other functions of these variables, such as \(T_1\) through \(T_8\). Understanding this concept helps us discern the resulting properties of combined or transformed random variables.

Variance

Variance is a measure of how much a set of numbers differ from their mean. It gives insights into the spread of the data points.

• Mathematically, for a set of observations, it's calculated as the average square of deviation from the mean.
• Variance models the average of the squared differences from the mean, making it always a non-negative value.

In our exercise, every pair of normal variables (like \((W_1, W_2)\)) shares a common variance \(\sigma^2\).

Understanding Variance in Context

With \(T_1 = \frac{1}{2}(W_1+W_2)-\mu_W\), the use of half the sum of \(W_1\) and \(W_2\) scales the variance by \(\frac{1}{4}\). So, variance for this expression becomes \(\frac{\sigma^2}{2}\) for \(T_1, T_2, T_3,\) and \(T_4\).

For \(T_5 = W_1 - W_2\), since it's a difference of identical and independent normals, the variance results in double, \(2\sigma^2\). This scaling affects the aggregate behavior of these variables and influences the distribution shape, as seen when calculating \(U\). Understanding variance is essential to grasp the behavior of such derived random variables.

Chi-squared Distribution

The chi-squared distribution is pivotal in statistics, especially when combined with normal distributions. It's often used to assess variability within a data set.

This distribution emerges naturally when summing squared standard normal random variables. Specifically, if you take \(k\) independent standard normal variables and square them, their total forms a chi-squared distribution with \(k\) degrees of freedom.
In our example, expressions like \(T_1^2 + T_2^2 + T_3^2 + T_4^2\) relate to chi-squared distributions. Each squared \(T_i\) is skewed by \(\sigma^2/2\), but their collective sum results in a chi-squared distribution with four degrees of freedom. Given they're independent, their total variance essentially combines, aligning with the chi-squared model. This principle simplifies complex calculations like understanding \(U\), leading to a clearer view of the derived random variables' characteristics.

F-distribution

The F-distribution is important when dealing with variances from different data sets. It's often used in hypothesis testing, especially when comparing variances from two separate datasets.

This distribution arises from the ratio of two scaled chi-squared variables. One chi-squared variable forms the numerator and another forms the denominator, both divided by their respective degrees of freedom. This ratio forms what's recognized as an F-distribution, denoted usually as \(F(d_1, d_2)\).

In our exercise, \(U = 4 \times \frac{T_1^2 + T_2^2 + T_3^2 + T_4^2}{T_5^2 + T_6^2 + T_7^2 + T_8^2}\) defines \(U\) via an F-distribution analogous transformation. Both numerator and denominator in \(U\) follow chi-squared distributions with 4 degrees of freedom, shaping \(U\) with an \(F(4, 4)\) distribution.
Understanding how an F-distribution centers the statistic \(U\) enriches our analysis capability for hypothesis tests, variance assessments, and curiosity-driven statistical explorations.