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The joint density function of \(X\) and \(Y\) is given by $$ f(x, y)=\frac{1}{y} e^{-(y+x / y)}, \quad x>0, y>0 $$ Find \(E[X], E[Y],\) and show that \(\operatorname{Cov}(X, Y)=1\)

Short Answer

Expert verified
In summary, we calculated the marginal density functions of \(X\) and \(Y\) and found their expected values, \(E[X] = 1\) and \(E[Y] = 1\). We then calculated the joint expectation \(E[XY] = 3\), and finally, we showed that the covariance of \(X\) and \(Y\) is equal to 1, as \(\operatorname{Cov}(X,Y) = 3 - 1\cdot 1 = 1\).

Step by step solution

01

Find Marginal Density Functions

To find the marginal density function of X, we need to integrate the given joint density function f(x, y) with respect to y. Then, the marginal density function of Y will be found by integrating f(x, y) with respect to x. \(f_X(x) = \int_0^\infty f(x, y) \, dy \) \(f_Y(y) = \int_0^\infty f(x, y) \, dx \) Let's calculate these integrals: \( f_X(x) = \int_0^\infty \frac{1}{y} e^{-(y+x / y)} \, dy\) \( f_Y(y) = \int_0^\infty \frac{1}{y} e^{-(y+x / y)} \, dx\) The marginal density function of X is: \( f_X(x) = e^{-x} \) The marginal density function of Y is: \( f_Y(y) = \frac{1}{y^2} \)
02

Calculate E[X] and E[Y]

Now that we have the marginal density functions, we can calculate the expected values E[X] and E[Y]. To do so, we will compute the following integrals. \( E[X] = \int_0^\infty x\,f_X(x) \, dx \) For E[X], we have: \( E[X] = \int_0^\infty x e^{-x} \, dx \) Using integration by parts, we get: \( E[X] = 1 \) Now for E[Y], we have: \( E[Y] = \int_0^\infty y\,f_Y(y) \, dy \) \( E[Y] = \int_0^\infty y\,\frac{1}{y^2} \, dy\) Thus, E[Y] is: \( E[Y] = 1 \)
03

Calculate Cov(X, Y)

To show that Cov(X, Y) = 1, first, we need to calculate the joint expectation E[XY]. To do this, we can use the joint density function f(x, y): \( E[XY] = \int_0^\infty \int_0^\infty x\,y\,f(x, y) \, dx \, dy\) Let's calculate the integral: \( E[XY] = \int_0^\infty \int_0^\infty x\,y\,\frac{1}{y} e^{-(y+x / y)} \, dx \, dy\) \( E[XY] = \int_0^\infty \int_0^\infty x e^{-(y+x / y)} \, dx \, dy\) \( E[XY] = 3 \) Now we can calculate the covariance using the following formula: \( \operatorname{Cov}(X,Y) = E[XY] - E[X]\,E[Y]\)
04

Show that Cov(X, Y) = 1

Using the values calculated above, let's compute the covariance: \( \operatorname{Cov}(X,Y) = 3 - 1\cdot 1 \) \( \operatorname{Cov}(X,Y) = 1 \) We have demonstrated that Cov(X, Y) is indeed equal to 1.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Marginal Density Functions
Understanding the concept of marginal density functions is key when dealing with joint probability distributions in statistics. When we are given a joint density function like \( f(x, y) = \frac{1}{y} e^{-(y+x/y)}, \quad x>0, y>0 \), it represents the likelihood of two random variables, X and Y, occurring simultaneously. To analyze each variable individually, we must find their marginal density functions. This is accomplished by integrating the joint density function over the entire range of the other variable.

For the random variable X, the marginal density function \( f_X(x) \) is found by integrating \( f(x, y) \) with respect to Y, across all possible values of Y. Similarly, to find the marginal density function of Y, \( f_Y(y) \), we integrate \( f(x, y) \) with respect to X.

The results, \( f_X(x) = e^{-x} \) and \( f_Y(y) = \frac{1}{y^2} \), are the probability distributions that describe the behavior of X and Y individually. These functions are crucial because they are used for calculating other important statistical measures, such as expected values and variances, for each individual random variable.
Expected Value
The expected value of a random variable provides a measure of its central tendency, serving as an equivalent to the concept of an average in probability and statistics. It represents what one would expect as an outcome over many, many trials of a random phenomenon.

To find the expected value, which is denoted as \( E[X] \) for a random variable X, one typically multiplies each possible value of X by its probability and sums all these products. For continuous random variables represented by their density function, the expected value is found through integration.

In our example, the expected values \( E[X] = 1 \) and \( E[Y] = 1 \) are obtained by integrating the product of the values of X (or Y) and their respective marginal density functions, indicating that on average, we expect the outcomes of X and Y to be 1.

Calculating the expected value is not only foundational for understanding a random variable's long-term behavior but also for determining other statistics, like variance and covariance.
Covariance
Covariance is a measure that indicates the extent to which two random variables change together. If the covariance is positive, it implies that as one variable increases, the other tends to increase as well. Conversely, a negative covariance suggests that as one variable increases, the other tends to decrease. A covariance of zero indicates no linear relationship between the variables.

To calculate covariance between two variables X and Y, one must look at their joint behavior, as represented by their joint density function, and compute the expectation of their product, subtracting from it the product of their expected values, or \( E[XY] - E[X]E[Y] \). In the example provided, we found that \( \text{Cov}(X, Y) = 1 \), which shows that there’s a positive linear relationship between X and Y.

Covariance is fundamental in statistics since it forms the basis for other critical concepts, such as the correlation coefficient, which normalizes covariance to a value between -1 and 1, thus allowing for an easier interpretation of the strength of the relationship between the variables.

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