Question

Let $X_1,X_2,X_3,X_4$ denote a random sample of size 4 from a distribution which is $N(0,{\sigma }^2).$ Let $Y=\sum _1^4{a}_i{X}_i$, where $a_1,a_2,a_3$, and $a_4$ are real constants. If $Y^2$ and $Q=X_1{X}_2-X_3{X}_4$ are independent, determine $a_1,a_2,a_3$, and $a_4$.

Short answer

The constants that make $Y^2$ and $Q$ independent are $a_1=a_2=-a_3=-a_4$.

Step-by-step solution

Define Variables

Let's define the given variables first. It is given that $X_1$, $X_2$, $X_3$, $X_4$ are random samples from a normal distribution that is $N(0,{\sigma }^2)$ . $Y=\sum _{i=1}^4{a}_i{X}_i$ and $Q=X_1{X}_2-X_3{X}_4$ are also defined.

Calculate Covariance

The first step is to calculate the covariance of $Y$ and $Q$. For the variables to be independent, their covariance should be zero. Using the equation for the calculation of covariance: COV($Y$,$Q$) = E[$YQ$] - E[$Y$]E[$Q$]. Because $X_1,X_2,X_3,X_4$ are distributed N(0, ${\sigma }^2$), then E[$X_n$] = 0 for any n. So E[$Y$] = E[$Q$] = 0. Then COV($Y$,$Q$) = E[$YQ$] becomes our new equation.

Expression for Expected Value of Y*Q

Now, let's find the expression for E[$YQ$]. By multiplying $Y$ and $Q$ and taking the expected value: E[$YQ$] = E[$(a_1{X}_1+a_2{X}_2+a_3{X}_3+a_4{X}_4)(X_1{X}_2-X_3{X}_4)$]. This simplifies to: E[$YQ$] = $a_1$E[$X_1^2{X}_2$] + $a_2$E[$X_1{X}_2^2$] - $a_3$E[$X_1{X}_2{X}_3$] - $a_4$E[$X_1{X}_2{X}_4$].

Simplify the Expression

The next step is to simplify the expected value expression obtained above. Because $X_1,X_2,X_3,X_4$ are independent, the expected value of their product is the product of their expected values. As E[$X_n$] = 0 for any n, E[$X_1{X}_2{X}_3$] = E[$X_1{X}_2{X}_4$] = 0. Now our E[$YQ$] becomes $a_1$E[$X_1^2{X}_2$] + $a_2$E[$X_1{X}_2^2$].

Independence of Variables

For $Y$ and $Q$ to be independent, the covariance of these two variables should be zero. So COV($Y$,$Q$) = E[$YQ$] = 0. Setting the expression $a_1$E[$X_1^2{X}_2$] + $a_2$E[$X_1{X}_2^2$] to be zero gives $a_1$ = -$a_2$. However, the choice of $X_1$ and $X_2$ is arbitary, thus, any pair of $a_i$ and $a_j$ should hold $a_i$ = -$a_j$. Hence, $a_1=a_2=-a_3=-a_4$.

Key concepts

Covariance

Covariance is a measure of how two random variables change together. In simpler terms, it shows the relationship between two variables, helping us understand if they tend to increase or decrease simultaneously. If the covariance is positive, it means that as one variable increases, the other tends to increase as well. Conversely, a negative covariance indicates that as one goes up, the other tends to go down.

To find the covariance between two variables, we use the formula:

• $\text{COV}(X,Y)=E[(X-{\mu }_X)(Y-{\mu }_Y)]$
• This can also be rephrased as $E[XY]-E[X]E[Y]$ for simplified calculation

In our exercise, determining the covariance between $Y$ and $Q$ being zero is key. It implies these two variables don't influence each other's fluctuations and suggests independence.

This is why it's crucial to simplify our calculations to find when $a_{1}E[X_1^2{X}_2]+a_{2}E[X_1{X}_2^2]=0$, helping us derive relations like $a_1=-a_2$ and extending this idea to find the solution for $a_1,a_2,a_3,$ and $a_4$.

Independence

Independence in probability and statistics refers to the idea that two random variables do not affect each other's outcomes. Simply put, whether one variable occurs or not has no impact on the probability of the other variable occurring.

In mathematical terms, two events $A$ and $B$ are independent if the probability of both events occurring together is the same as the product of their separate probabilities:

• $$P(A\cap B)=P(A)\times P(B)$$

For our exercise, the challenge is to demonstrate independence between $Y^2$ and $Q$ as this is given as a condition in the problem. Given that $COV(Y,Q)=0$, this statistical property helps affirm that the variables are independent if they follow normal distribution settings. The conclusion derived from the independence condition also conforms with calculating zero covariance, which reinforces their independent status under these conditions.

Thus, when you find that the covariance is zero, like in our task, it reinforces potential independence, especially in the context of normal distribution.

Normal Distribution

A normal distribution, also known as the Gaussian distribution, is one of the most important probability distributions in statistics. This distribution is characterized by its bell-shaped curve, where most of the data points cluster around a central mean, and the probabilities gradually decrease as you move towards the extremes or tails.

Key properties of a normal distribution include:

• The mean, median, and mode of the distribution are equal.
• It is symmetric about the mean.
• Approximately 68% of the data lies within one standard deviation ($\sigma$) of the mean, 95% within two, and 99.7% within three.

In this exercise, the variables $X_1,X_2,X_3,X_4$ are distributed as $N(0,{\sigma }^2)$. This implies they follow a normal distribution centered around zero with variance ${\sigma }^2$.

Understanding that $Y$ and $Q$ involve combinations of these normally distributed variables helps derive insights about the characteristics of $Y^2$ and $Q$. Knowing they are extracted from a normal distribution makes solving the problem more intuitive, leading to the independence and covariance concepts utilized in the solution.