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Chapter 6: Problem 10

Let \(X\) and \(Y\) be independent random variables with uniform density functions on \([0,1] .\) Find (a) \(E(|X-Y|)\). (b) \(E(\max (X, Y))\) (c) \(E(\min (X, Y))\) (d) \(E\left(X^{2}+Y^{2}\right)\) (e) \(E\left((X+Y)^{2}\right)\)

Short Answer

Expert verified
(a) \(\frac{1}{3}\), (b) \(\frac{2}{3}\), (c) \(\frac{1}{3}\), (d) \(\frac{2}{3}\), (e) \(\frac{5}{6}\).

Step by step solution

01

Understanding Uniform Distribution

Both random variables \(X\) and \(Y\) are uniformly distributed over the interval \([0, 1]\). For a uniform distribution on \([a, b]\), the probability density function is \(f(x) = \frac{1}{b-a}\) within \([a, b]\). Thus, for \(X\) and \(Y\), \(f(x) = 1\) on \([0, 1]\).
02

Calculating \(E(|X-Y|)\)

The expected value \(E(|X-Y|)\) for independent uniform distributions is given by an integral of \(|x-y|\), calculated over the square \([0, 1] \times [0, 1]\). This can be split into two equal regions: above and below the line \(x = y\). The result is \(\int_0^1 \int_0^1 |x-y| \, dy \, dx = \frac{1}{3}\).
03

Calculating \(E(\max(X, Y))\)

The expected value of the maximum is the integral \( \int_0^1 x \cdot 2x(1-x) \, dx = \frac{2}{3}\), as the probability that \(X\) and \(Y\) are below \(x\) is \(x^2\), and \(2x(1-x)\) is derived as the density function for \(\max(X,Y)\) over \([0, 1]\).
04

Calculating \(E(\min(X, Y))\)

The expected value of the minimum is similar, \( \int_0^1 x \cdot (2x-x^2) \, dx = \frac{1}{3}\), using \(\min(X,Y)\) having a probability density of \(2(1-x)x\) since the probability that one is greater than \(x\) is \((1-x)^2\).
05

Calculating \(E(X^2 + Y^2)\)

Since \(X\) and \(Y\) are independent, \(E(X^2 + Y^2) = E(X^2) + E(Y^2) = \int_0^1 x^2 \, dx + \int_0^1 y^2 \, dy = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}\).
06

Calculating \(E((X + Y)^2)\)

Using linearity of expectation, \(E((X+Y)^2) = E(X^2 + 2XY + Y^2) = E(X^2) + 2E(XY) + E(Y^2)\). \(E(X^2) = E(Y^2) = \frac{1}{3}\) as before, and \(E(XY) = \int_0^1 \int_0^1 xy \, dy \, dx = \frac{1}{4}\). Thus, \(E((X+Y)^2) = \frac{1}{3} + \frac{1}{3} + 2 \times \frac{1}{4} = \frac{2}{3} + \frac{1}{2} = \frac{5}{6}\).

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Key Concepts

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

Uniform Distribution
Uniform distribution is a type of probability distribution in which all outcomes are equally likely. In the context of the exercise, both random variables \(X\) and \(Y\) follow a uniform distribution on the interval
  • The uniform distribution has a probability density function (PDF) defined as \(f(x) = \frac{1}{b-a}\) for \(a \leq x \leq b\).
  • Since \(X\) and \(Y\) are uniform on \([0,1]\), their PDF simplifies to \(f(x) = 1\) within this range.
This characteristic implies that the likelihood of \(X\) and \(Y\) taking any specific value within the interval is the same. This facilitates the calculation of expected values, making certain integrals simpler when computing properties like \(E(|X-Y|)\), \(E(\max(X,Y))\), and others.
Random Variables
Random variables are fundamental in probability theory, representing quantities with uncertain outcomes. Specifically, a random variable maps outcomes of a random phenomenon to numerical values. In our exercise:
  • We have two random variables, \(X\) and \(Y\), both of which are uniformly distributed over the interval \([0, 1]\).
  • They quantify the results of an experiment, where the exact outcome is unpredictable but governed by the underlying distribution.
Understanding them helps in computing their properties, such as expected values and variances, critical to solving problems involving probabilistic scenarios.
Independent Variables
In probability, independence between random variables indicates that the outcome of one variables does not affect the outcome of another. In the exercise:
  • \(X\) and \(Y\) are independent random variables.
  • This means that the joint probability of \(X\) and \(Y\) is simply the product of their individual probabilities: \(P(X = x, Y = y) = P(X = x) \times P(Y = y)\).
  • Independence simplifies calculations, as shown when finding the expected value of the sum \(E((X+Y)^2)\).
It allows the use of very useful relationships like \(E(XY) = E(X)E(Y)\) when \(X\) and \(Y\) are independent, essential for more complex calculations.
Probability Density Function
The Probability Density Function (PDF) is a function that describes the likelihood of a random variable to take on a particular value. For continuous random variables, the PDF is crucial for calculating probabilities within an interval:
  • The PDF must integrate to 1 over the interval for which the variable is defined, ensuring it represents a legitimate probability distribution.
  • For \(X\) and \(Y\) in the exercise, the PDF is constant at \(f(x) = 1\) over \([0, 1]\).
  • This uniformity simplifies integration when evaluating expectations, as seen when calculating \(E(X^2)\), \(E(|X-Y|)\), etc.
The PDF gives a visual representation of where values are more or less likely to occur, instrumental in determining characteristics like means and variances in probability distribution.

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Most popular questions from this chapter

A sequence of random numbers in [0,1) is generated until the sequence is no longer monotone increasing. The numbers are chosen according to the uniform distribution. What is the expected length of the sequence? (In calculating the length, the term that destroys monotonicity is included.) Hint: Let \(a_{1}, a_{2}, \ldots\) be the sequence and let \(X\) denote the length of the sequence. Then $$P(X>k)=P\left(a_{1}1)+P(X>2)+\cdots$$

A box contains two gold balls and three silver balls. You are allowed to choose successively balls from the box at random. You win 1 dollar each time you draw a gold ball and lose 1 dollar each time you draw a silver ball. After a draw, the ball is not replaced. Show that, if you draw until you are ahead by 1 dollar or until there are no more gold balls, this is a favorable game.

Let \(X\) be a continuous random variable with density function \(f_{X}(x) .\) Show that if $$\int_{-\infty}^{+\infty} x^{2} f_{X}(x) d x<\infty$$ then $$\int_{-\infty}^{+\infty}|x| f_{X}(x) d x<\infty$$

(Banach's Matchbox \(^{16}\) ) A man carries in each of his two front pockets a box of matches originally containing \(N\) matches. Whenever he needs a match, he chooses a pocket at random and removes one from that box. One day he reaches into a pocket and finds the box empty. (a) Let \(p_{r}\) denote the probability that the other pocket contains \(r\) matches. Define a sequence of counter random variables as follows: Let \(X_{i}=1\) if the \(i\) th draw is from the left pocket, and 0 if it is from the right pocket. Interpret \(p_{r}\) in terms of \(S_{n}=X_{1}+X_{2}+\cdots+X_{n} .\) Find a binomial expression for \(p_{r}\) (b) Write a computer program to compute the \(p_{r},\) as well as the probability that the other pocket contains at least \(r\) matches, for \(N=100\) and \(r\) from 0 to 50 . (c) Show that \((N-r) p_{r}=(1 / 2)(2 N+1) p_{r+1}-(1 / 2)(r+1) p_{r+1}\). (d) Evaluate \(\sum_{r} p_{r}\) (e) Use (c) and (d) to determine the expectation \(E\) of the distribution \(\left\\{p_{r}\right\\}\). (f) Use Stirling's formula to obtain an approximation for \(E .\) How many matches must each box contain to ensure a value of about 13 for the expectation \(E ?\) (Take \(\pi=22 / 7\).)

Let \(X\) be Poisson distributed with parameter \(\lambda\). Show that \(V(X)=\lambda\).
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