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If \(X\) and \(Y\) have joint density function $$f_{X, Y}(x, y)=\left\\{\begin{array}{ll} 1 / y, & \text { if } 0

Short Answer

Expert verified
The short version of the answer is: (a) \(E[XY] = \frac{1}{6}\) (b) \(E[X] = \frac{1}{6}\) (c) \(E[Y] = \frac{1}{2}\)

Step by step solution

01

Calculate (a) E[XY]

To find \(E[XY]\), we need to compute the double integral: \[E[XY] = \int\int xyf_{X,Y}(x, y) \, dx \, dy\] Since \(f_{X, Y}(x, y) = 0\) outside the region where \(0<y<1, 0<x<y\), we will integrate over this region: \[E[XY] = \int_0^1\int_0^y (xy\frac{1}{y}) \,dx\,dy\] Now, simplify: \[E[XY] = \int_0^1\int_0^y x \, dx \, dy\]
02

Integrate with respect to x

Next, integrate with respect to x: \[E[XY] = \int_0^1 \left[\frac{1}{2}x^2\right]_0^y dy\] Now, substitute the limits of integration: \[E[XY] = \int_0^1 \frac{1}{2}y^2 dy\]
03

Integrate with respect to y

Now, integrate with respect to y: \[E[XY] = \left[\frac{1}{6}y^3\right]_0^1\] Substitute the limits of integration: \[E[XY] = \frac{1}{6}\] So, \(E[XY] = \frac{1}{6}\).
04

Calculate (b) E[X]

To find \(E[X]\), we need to compute the double integral: \[E[X] = \int\int xf_{X,Y}(x, y) \, dx \, dy\] Integrate over the region where \(0<y<1, 0<x<y\): \[E[X] = \int_0^1\int_0^y (x\frac{1}{y}) \,dx\,dy\] Now, simplify: \[E[X] = \int_0^1\int_0^y x\,dx\,dy\]
05

Integrate with respect to x

Next, integrate with respect to x: \[E[X] = \int_0^1 \left[\frac{1}{2}x^2\right]_0^y dy\] Now, substitute the limits of integration: \[E[X] = \int_0^1 \frac{1}{2}y^2 dy\]
06

Integrate with respect to y

Now, integrate with respect to y: \[E[X] = \left[\frac{1}{6}y^3\right]_0^1\] Substitute the limits of integration: \[E[X] = \frac{1}{6}\] So, \(E[X] = \frac{1}{6}\).
07

Calculate (c) E[Y]

To find \(E[Y]\), we need to compute the double integral: \[E[Y] = \int\int yf_{X,Y}(x, y)\, dx\, dy\] Integrate over the region where \(0<y<1, 0<x<y\): \[E[Y] = \int_0^1\int_0^y (y\frac{1}{y}) \,dx\,dy\] Now, simplify: \[E[Y] = \int_0^1\int_0^y 1 \, dx \, dy\]
08

Integrate with respect to x

Next, integrate with respect to x: \[E[Y] = \int_0^1 [x]_0^y dy\] Now, substitute the limits of integration: \[E[Y] = \int_0^1 y dy\]
09

Integrate with respect to y

Now, integrate with respect to y: \[E[Y] = \left[\frac{1}{2}y^2\right]_0^1\] Substitute the limits of integration: \[E[Y] = \frac{1}{2}\] So, \(E[Y] = \frac{1}{2}\). In conclusion: (a) \(E[XY] = \frac{1}{6}\) (b) \(E[X] = \frac{1}{6}\) (c) \(E[Y] = \frac{1}{2}\)

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Key Concepts

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

Expected Value
In the realm of probability and statistics, the expected value is a fundamental concept that represents the average outcome we expect from a random variable after many trials. In the case where we have two variables, such as X and Y, working together, as demonstrated in the exercise, we may want to calculate the expected value of their product, E[XY].

This calculation simply implies that we're interested in the center of mass, or the 'average' value, of their joint product. When calculating the expected value of a function g(X, Y) of the joint variables, such as XY, over their joint density function, which here is f_{X,Y}(x, y), it involves an integral - specifically a double integral if we have two continuous random variables.

To better understand, consider the expected value as the probability-weighted average of all possible values. For example, if we throw a six-sided die repeatedly and calculate the average of the results, that average would approach the expected value as the number of throws increases.
Double Integral
A double integral allows us to calculate the volume under the surface of a function of two variables over a specified area. It is an integral of an integral, which computes the accumulation of quantities across two dimensions. In the situation of our exercise, the double integral is used to find the expected value by multiplying the joint density function f_{X,Y}(x, y) by the expression xy, then integrating in terms of x and y over the appropriate domain.

Calculating a double integral, as seen in the exercise, involves taking two integrals in sequence. The inner integral, often with respect to x, computes the accumulation along the x-direction, holding y constant. The outer integral then adds up those values along the y-direction. This sequence is crucial in making sure we cover the entire domain specified by the limits of integration. In practical applications, double integrals have a plethora of uses ranging from physics to probability, to even economics and beyond.
Integration Limits
Understanding the limits of integration is crucial when performing a double integral. The integration limits define the region over which we evaluate the integral. In the context of joint probability density functions, these limits must respect the boundaries where the density function is non-zero.

For example, our problem provided a domain defined by the conditions 0
Setting up the correct limits of integration is a pivotal step. If these are incorrectly chosen, our calculations can bring inaccurate results. Hence, comprehending and correctly applying the integration limits for both x and y is essential for executing a double integral accurately in context of the given scenario.

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Most popular questions from this chapter

In Example \(6 \mathrm{c},\) suppose that \(X\) is uniformly distributed over (0,1) . If the discretized regions are determined by \(a_{0}=0, a_{1}=\frac{1}{2},\) and \(a_{2}=1\) calculate the optimal quantizer \(Y\) and compute \(E\left[(X-Y)^{2}\right]\)

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