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Chapter 5: Problem 10

Let \(U, V\) be random numbers chosen independently from the interval [0,1] . Find the cumulative distribution and density for the random variables (a) \(Y=\max (U, V)\) (b) \(Y=\min (U, V)\)

Short Answer

Expert verified
(a) CDF: \( y^2 \), PDF: \( 2y \); (b) CDF: \( 2y - y^2 \), PDF: \( 2 - 2y \).

Step by step solution

01

Understand the Problem for Part (a)

We need to find the cumulative distribution function (CDF) of the random variable \( Y = \max(U, V) \), where \( U \) and \( V \) are independent random variables uniformly distributed over the interval [0,1].
02

Find the CDF for \( Y=\max(U,V) \)

The cumulative distribution function \( F_Y(y) = P(\max(U, V) \leq y) \) can be understood by considering that both \( U \leq y \) and \( V \leq y \) must hold. Thus, \( F_Y(y) = P(U \leq y)P(V \leq y) = y^2 \), for \( 0 \leq y \leq 1 \).
03

Differentiate to Find the PDF for \( Y=\max(U,V) \)

The probability density function (PDF) is the derivative of the CDF. Therefore, for \( Y = \max(U, V) \), the PDF is given by \( f_Y(y) = \frac{d}{dy}(y^2) = 2y \), for \( 0 \leq y \leq 1 \).
04

Understand the Problem for Part (b)

Now consider \( Y = \min(U, V) \), with the same condition that \( U \) and \( V \) are independent and uniformly distributed over [0,1].
05

Find the CDF for \( Y=\min(U,V) \)

The CDF \( F_Y(y) = P(\min(U, V) \leq y) \) is the probability that at least one of \( U \leq y \) or \( V \leq y \) holds. So, \( F_Y(y) = 1 - P(U > y)P(V > y) = 1 - (1-y)^2 \). Expanding gives \( F_Y(y) = 1 - (1 - 2y + y^2) = 2y - y^2 \) for \( 0 \leq y \leq 1 \).
06

Differentiate to Find the PDF for \( Y=\min(U,V) \)

The PDF is again the derivative of the CDF. Hence, for \( Y = \min(U, V) \), the PDF is \( f_Y(y) = \frac{d}{dy}(2y - y^2) = 2 - 2y \) for \( 0 \leq y \leq 1 \).

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Key Concepts

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

Cumulative Distribution Function (CDF)
The cumulative distribution function, or CDF, is a fundamental concept in probability that describes the probability that a random variable will take a value less than or equal to a particular number.
This is represented mathematically as \( F_Y(y) = P(Y \leq y) \).
The CDF begins at zero and increases to one as the variable reaches its maximum value within the given range. It helps to visualize the likelihood of various outcomes of a random variable.
  • For a random variable \( Y \), the CDF is a non-decreasing function.
  • The value of the CDF at any point \( y \) is the area under the PDF curve from the smallest possible value up to \( y \).
By understanding the CDF, you can gain insights into the distribution and probability of different values that a random variable may take.
Probability Density Function (PDF)
The probability density function, or PDF, provides a way to describe the likelihood of different outcomes for a continuous random variable.
While the CDF gives the cumulative probability, the PDF shows the rate of change, essentially indicating how likely a particular value is.
The PDF is connected to the CDF through differentiation:
  • The PDF is the derivative of the CDF, \( f_Y(y) = \frac{d}{dy} F_Y(y) \).
  • The area under the PDF curve, across its entire range, equals 1, representing the total probability.
In the context of our exercise, the PDF for \( Y = \max(U, V) \) and \( Y = \min(U, V) \) illustrates how these maximum or minimum values are distributed over the interval \([0, 1]\). Understanding the PDF is crucial because it provides more detailed information about where values are concentrated.
Uniform Distribution
A uniform distribution is characterized by having equal probabilities for all outcomes within a specified range, typically resulting in a flat and constant density function.
Random variables, such as \( U \) and \( V \), that are uniformly distributed over an interval, exemplify no bias towards any sub-interval within that range.
  • For random variables uniformly distributed on \( [0, 1] \), every subinterval has the same chance of containing the chosen random number.
  • The PDF of a uniform distribution is constant over its range.
The simplicity of the uniform distribution makes it a great starting point for understanding more complex random variables, as evidenced by its use in finding the maximum and minimum of two independent variables \( U \) and \( V \). Through uniform distribution, we can explore detailed computations involving their combinations.
Maximum and Minimum of Random Variables
Calculating the maximum or minimum of random variables introduces interesting scenarios in probability, allowing for analyses that involve seeing where extremal values occur.
These are particularly insightful with uniformly distributed independent random variables \( U \) and \( V \).
  • The distribution of the maximum, \( Y = \max(U, V) \), provides insight into how one or the other variable takes on larger values.
  • Conversely, the distribution of the minimum, \( Y = \min(U, V) \), shows insights into where smaller values are likely to occur.
For maximums, both \( U \) and \( V \) must be below \( y \) to consider a particular value, leading to calculations based on their joint probabilities. For minimums, at least one of the variables must be below \( y \), showing a different interaction pattern. Exploring these teaches us how distributions influence one another in different scenarios.

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Most popular questions from this chapter

The Poisson distribution with parameter \(\lambda=.3\) has been assigned for the outcome of an experiment. Let \(X\) be the outcome function. Find \(P(X=0)\), \(P(X=1),\) and \(P(X>1)\)

Let \(X_{1}\) and \(X_{2}\) be independent random variables and for \(i=1,2,\) let \(Y_{i}=\) \(\phi_{i}\left(X_{i}\right),\) where \(\phi_{i}\) is strictly increasing on the range of \(X_{i} .\) Show that \(Y_{1}\) and \(Y_{2}\) are independent. Note that the same result is true without the assumption that the \(\phi_{i}\) 's are strictly increasing, but the proof is more difficult.

Suppose we know a random variable \(Y\) as a function of the uniform random variable \(U: Y=\phi(U),\) and suppose we have calculated the cumulative distribution function \(F_{Y}(y)\) and thence the density \(f_{Y}(y)\). How can we check whether our answer is correct? An easy simulation provides the answer: Make a bar graph of \(Y=\phi(r n d)\) and compare the result with the graph of \(f_{Y}(y)\). These graphs should look similar. Check your answers to Exercises 1 and 2 by this method.

For which of the following random variables would it be appropriate to assign a uniform distribution? (a) Let \(X\) represent the roll of one die. (b) Let \(X\) represent the number of heads obtained in three tosses of a coin. (c) A roulette wheel has 38 possible outcomes: \(0,00,\) and 1 through \(36 .\) Let \(X\) represent the outcome when a roulette wheel is spun. (d) Let \(X\) represent the birthday of a randomly chosen person. (e) Let \(X\) represent the number of tosses of a coin necessary to achieve a head for the first time.

Let \(U\) be a uniformly distributed random variable on \([0,1] .\) What is the probability that the equation $$x^{2}+4 U x+1=0$$ has two distinct real roots \(x_{1}\) and \(x_{2} ?\)
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