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Suppose the random vector \((\mathrm{X}, \mathrm{Y})\) is distributed with probability density, \(\mathrm{f}(\mathrm{x}, \mathrm{y})=\mathrm{x}+\mathrm{y}\) \(0<\mathrm{x}<1\) and \(=0 \quad 0<\mathrm{y}<1\) otherwise. Find \(E[X Y], E[X+Y]\) and \(E(X)\).

Short Answer

Expert verified
The short answer is: E[XY] = \(\frac{7}{12}\) E[X+Y] = \(\frac{5}{3}\) E[X] = \(\frac{5}{6}\)

Step by step solution

01

Compute the Marginal Probability Density Functions of X and Y

To find the marginal probability density functions of X and Y, we will integrate f(x,y) with respect to the other variable. For the marginal probability density function of X, we integrate f(x,y) with respect to y: \(f_X(x) = \int_{0}^{1} (x + y) dy\) For the marginal probability density function of Y, we integrate f(x,y) with respect to x: \(f_Y(y) = \int_{0}^{1} (x + y) dx\)
02

Evaluate the Integrals to Find the Marginal Density Functions of X and Y

Now let's compute both the integrals: \(f_X(x) = \int_{0}^{1} (x + y) dy = \left[ xy + \frac{1}{2}y^2 \right]_0^1 = x + \frac{1}{2}\) \(f_Y(y) = \int_{0}^{1} (x + y) dx = \left[ \frac{1}{2}x^2 + xy \right]_0^1 = \frac{1}{2} + y\)
03

Compute E[XY], E[X+Y], and E[X] Using the Marginal Density Functions

Now we will use the marginal density functions to compute the expected values: \(E[XY] = \int_{0}^{1} \int_{0}^{1} xy(x + y) dx dy\) \(E[X+Y] = \int_{0}^{1} \int_{0}^{1} (x+y)(x + y) dx dy\) \(E[X] = \int_{0}^{1} x f_X(x) dx\)
04

Evaluate the Integrals to Find the Expected Values

Finally, we compute the integrals to find the expected values: \(E[XY] = \int_{0}^{1} \int_{0}^{1} xy(x + y) dx dy = \frac{7}{12}\) \(E[X+Y] = \int_{0}^{1} \int_{0}^{1} (x+y)(x + y) dx dy = \frac{5}{3}\) \(E[X] = \int_{0}^{1} x (x + \frac{1}{2}) dx = \frac{5}{6}\) Hence, the expected values are: E[XY] = \(\frac{7}{12}\) E[X+Y] = \(\frac{5}{3}\) E[X] = \(\frac{5}{6}\)

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Key Concepts

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

Random Variables
Random variables are fundamental elements in probability theory, used to model numerical outcomes of random processes. They are often denoted by capital letters like \(X\) or \(Y\).
In this article, \(X\) and \(Y\) are random variables related through a joint probability density function (pdf) \(f(x, y)\). This function describes the likelihood of \(X\) and \(Y\) taking particular values within a certain range.
For our specific problem, \(f(x, y) = x + y\) within the range of \(0 < x, y < 1\), providing a structured way to determine how probable certain outcomes are.
It's crucial to understand that while random variables can take on a range of values, the pdf gives a snapshot of their behavior over a defined interval. It's a key tool for calculating probabilities and expected outcomes.
Marginal Probability Density Function
Marginal probability density functions (pdfs) allow us to focus on individual random variables from a joint distribution by integrating over the unrelated variable.
To find the marginal pdf of \(X\), denoted \(f_X(x)\), we integrate the joint pdf \(f(x, y)\) with respect to \(y\) over its entire range. Similarly, for the marginal pdf of \(Y\) (\(f_Y(y)\)), we integrate with respect to \(x\).
Here's how it's done:
  • \(f_X(x) = \int_{0}^{1} (x + y) \, dy\)
  • \(f_Y(y) = \int_{0}^{1} (x + y) \, dx\)
This process strips down the joint distribution to focus on a single variable, summarizing its behavior independently of the other. Marginal pdfs are essential for deriving other statistical measures like expected values.
Expected Value
The expected value, or mean, gives the average outcome of a random variable if an experiment is repeated many times.
For a continuous random variable, it's found using the integral of the variable multiplied by its pdf.
For our problem, expected values are found as follows:
  • \(E[XY]\) uses the joint pdf, evaluating \(E[XY] = \int_{0}^{1} \int_{0}^{1} xy(x + y) \, dx \, dy\)
  • \(E[X+Y]\) evaluates the sum of the variables: \(E[X+Y] = \int_{0}^{1} \int_{0}^{1} (x+y)(x + y) \, dx \, dy\)
  • \(E[X]\) uses the marginal pdf of \(X\): \(E[X] = \int_{0}^{1} x \, f_X(x) \, dx\)
The expected value helps us find the central tendency of our distribution, providing insights into what values the random variables are likely to take.
Integral Calculus
Integral calculus is a mathematical technique used to calculate areas under curves, which is critical in probability for finding probabilities and expected values.
When working with probability density functions, integration helps determine:
  • The total probability over a given interval
  • Marginal pdfs by integrating over the irrelevant variables
  • Expected values by integrating the product of the variable and its pdf
For example, finding \(f_X(x)\) and the expected value involves integrating the relevant functions:
\( \int_{0}^{1} (x+y) \, dy \) or \( \int_{0}^{1} x \, f_X(x) \, dx \).
Mastering integration allows us to transform complex relationships in probability theory into calculated probabilities and expected outcomes, making predictions about random processes.

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Most popular questions from this chapter

Let \(\mathrm{X}\) have the probability distribution defined by $$ \begin{array}{lcc} \mathrm{f}(\mathrm{x})=1-\mathrm{e}^{-\mathrm{x}} & \text { for } & \mathrm{x} \geq 0 \\ \text { and }=0 & \text { for } & \mathrm{x}<0 . \end{array} $$ Let \(\mathrm{Y}=\sqrt{\mathrm{X}}\) be a new random variable. Find \(\mathrm{G}(\mathrm{y})\), the distribution function of \(\mathrm{Y}\), using the cumulative distribution function technique.

Two independent reports on the value of a tincture for treating a disease in camels were available. The first report made on a small pilot series showed the new tincture to be probably superior to the old treatment with a Yates' \(\mathrm{X}^{2}\) of \(3.84, \mathrm{df}=1, \alpha=.05 .\) The second report with a larger trial gave a "not significant" result with a Yates \(\mathrm{X}^{2}=2.71, \mathrm{df}=1\), \(\alpha=.10 .\) Can the results of the 2 reports be combined to form a new conclusion?

Find the expected value of the random variable \(\mathrm{S}^{2}{*}=(1 / \mathrm{n})^{\mathrm{n}} \sum_{\mathrm{i}=1}\left(\mathrm{X}_{\mathrm{i}}-\underline{\mathrm{X}}\right)^{2}\), where \(\underline{\mathrm{X}}={ }^{\mathrm{n}} \sum_{\mathrm{i}=1} \mathrm{X}_{\mathrm{i}} / \mathrm{n}\) and the \(\mathrm{X}_{\mathrm{i}}\) are independent and identically distributed with \(\mathrm{E}\left(\mathrm{X}_{\mathrm{i}}\right)=\mu\), Var \(\mathrm{X}_{\mathrm{i}}=\sigma^{2}\) for \(\mathrm{i}=1,2 \ldots \ldots \mathrm{n}\)

The seven dwarfs challenged the Harlem Globetrotters to a basketball game. Besides the obvious difference in height, we are interested in constructing a \(95 \%\) confidence in ages between dwarfs and basketball players. The respective ages are, $$ \begin{array}{|l|l|l|l|} \hline \text { Dwardfs } & & \text { Globetrotters } & \\ \hline \text { Sneezy } & 20 & \text { Meadowlark } & 43 \\ \hline \text { Grumpy } & 39 & \text { Curley } & 37 \\ \hline \text { Dopey } & 23 & \text { Marques } & 45 \\ \hline \text { Doc } & 41 & \text { Bobby Joe } & 25 \\ \hline \text { Sleepy } & 35 & \text { Theodis } & 34 \\ \hline \text { Happy } & 29 & & \\ \hline \text { Bashful } & 31 & & \\ \hline \end{array} $$ Can you construct the interval? Assume the variance in age is the same for dwarfs and Globetrotters.

Let \(\mathrm{Y}=\) the Rockwell hardness of a particular alloy of steel. Assume that \(\mathrm{Y}\) is a continuous random variable that can take on any value between 50 and 70 with equal probability. Find the expected Rockwell hardness.

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