Problem 75 - An Introduction to Mathematical Statistics and Its Applications

Question

If the joint pdf of the random variables \(X\) and \(Y\) is $$ f_{X, Y}(x, y)=k e^{-(2 / 3)\left[(1 / 4) x^{2}-(1 / 2) x y+y^{2}\right]} $$ find \(E(X), E(Y), \operatorname{Var}(X), \operatorname{Var}(Y), \rho(X, Y)\), and \(k\).

Step-by-Step Solution

Answer

The expected values are \(E(X) = E(Y) = 0\), the variances are \(\operatorname{Var}(X) = \operatorname{Var}(Y) = 4\), the correlation is \(\rho(X, Y) = 0.5\), and the constant \(k = \frac{1}{2\pi}\).

Step 1: Recognize the form and identify the parameters

The given pdf \(f_{X, Y}(x, y)\) is of the form of a bivariate normal distribution. We can rewrite it as \(f_{X, Y}(x, y)=k e^{-(\frac{1}{2 \sigma^2})(x-\mu_{x})^{2}+2 \rho \frac{(x-\mu_{x})(y-\mu_{y})}{\sigma_{x} \sigma_{y}}-(y-\mu_{y})^{2}/(2 \sigma_{y}^{2})}\). Now we need to identify \(\mu_{x}\), \(\mu_{y}\), \(\sigma_{x}\), \(\sigma_{y}\), and \(\rho\), which are the means, standard deviations and correlation of \(X\) and \(Y\), respectively. From the given distribution, we observe that \(\mu_{x} = \mu_{y} = 0\), \(\sigma_{x} = \sigma_{y} = 2\), and \(\rho = 0.5\).

Step 2: Find Expected Values

The expected value is simply the mean of the distribution, which is already identified as \(\mu_{x}\) and \(\mu_{y}\) for \(X\) and \(Y\) respectively. So their values are 0.

Step 3: Find Variances

The variance is simply the square of the standard deviation. Thus, \(\operatorname{Var}(X) = \sigma_{x}^{2} = 2^2 = 4\) and \(\operatorname{Var}(Y) = \sigma_{y}^{2} = 2^2 = 4\).

Step 4: Find Correlation

The correlation coefficient \(\rho(X, Y)\) is already given as 0.5.

Step 5: Find the constant k

To find the constant \(k\), we have to keep in mind that the integral of the function over the complete space must equal 1, based on the properties of a probability distribution function. Hence, \( \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} k e^{-(2 / 3)((1 / 4) x^{2}-(1 / 2) x y+y^{2})} dx dy = 1\). Solving this integral and setting equal to 1 will give us the value of \(k\). After solution, we find \(k = \frac{1}{2\pi}\).

Key Concepts

Joint Probability Density Function

The joint probability density function (joint pdf) for two random variables, in this case, \(X\) and \(Y\), is a mathematical function that describes the likelihood of different outcomes for the pair \((X, Y)\). This defines how the two random variables are related to each other as well as their individual probabilities.In a bivariate normal distribution, the joint pdf is expressed in an exponential form that includes several parameters: the means \(\mu_x\) and \(\mu_y\), the standard deviations \(\sigma_x\) and \(\sigma_y\), and the correlation coefficient \(\rho\). For our given function, we rewrite it as:\[f_{X, Y}(x, y) = k e^{-\left[\frac{1}{2 \sigma^2}(x - \mu_x)^2 + 2 \rho \frac{(x - \mu_{x})(y - \mu_{y})}{\sigma_{x} \sigma_{y}} - \frac{(y - \mu_y)^2}{2 \sigma_y^2}\right]}\]This shows how \(X\) and \(Y\) vary together within the distribution, and when these parameters are identified, they give us important information about the probability and characteristics of various outcomes.

Expected Value

The expected value, often denoted as \(E(X)\) for random variable \(X\), is essentially the average value you would expect to get if you repeated an experiment an infinite number of times. For a bivariate normal distribution, the expected values for \(X\) and \(Y\) are represented by their means \(\mu_x\) and \(\mu_y\), respectively.From the step-by-step solution and through observation of the joint pdf, we know:

• \(E(X) = \mu_x = 0\)
• \(E(Y) = \mu_y = 0\)

These values demonstrate that both variables are centered around zero in this distribution, signifying no general tendency to be larger or smaller.

Variance

Variance is a measure of how spread out the values of a random variable \(X\) are around the mean. For any normal distribution, it is calculated as the square of the standard deviation. Therefore, variance describes the dispersion or clustering of a set of values.In our bivariate normal distribution, we need to find the variances of \(X\) and \(Y\). Given:

• \(\sigma_x = \sigma_y = 2\)
• The variance formula: \(\operatorname{Var}(X) = \sigma_x^2\)

We determine the variances as:

• \(\operatorname{Var}(X) = 2^2 = 4\)
• \(\operatorname{Var}(Y) = 2^2 = 4\)

This uniform variance suggests that both variables have the same degree of spread around their means.

Correlation Coefficient

The correlation coefficient \(\rho\), specifically in the context of two random variables like \(X\) and \(Y\), measures the strength and direction of a linear relationship between them. It ranges from \(-1\) to \(1\), where \(1\) indicates a perfect positive linear relationship, \(-1\) indicates a perfect negative linear relationship, and \(0\) indicates no linear relationship.For our bivariate normal distribution, the correlation coefficient is identified as:

• \(\rho = 0.5\)

This value of \(0.5\) signifies a moderate positive correlation between \(X\) and \(Y\), meaning that as values of \(X\) increase, values of \(Y\) tend to also increase, but not perfectly or strongly.Understanding this coefficient helps in analyzing how changes in one variable might impact the other, which is crucial for predicting trends and making decisions based on linear relationships in data.