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Let \(X\) and \(Y\) be two independent uniformly distributed random variables over the intervals \((0,1)\) and \((0,2)\), respectively. Find the probability density function of \(X / Y\).

Short Answer

Expert verified
The PDF of the variable Z = X/Y, given X and Y are independent uniformly distributed random variables over intervals (0,1) and (0,2) respectively, is \(f_Z(z) = 2z\) for \(0<z\leq1\), \(2-z\) for \(1<z<2\), and 0 otherwise.

Step by step solution

01

Define the probability density functions of \(X\) and \(Y\)

Since both \(X\) and \(Y\) are uniformly distributed, their pdf's over the intervals \((0,1)\) and \((0,2)\) respectively are easy to define. For \(X\), the pdf is \(f_X(x) = 1\) for \(0 \leq x \leq 1\) and 0 otherwise. Similarly, for \(Y\), the pdf is \(f_Y(y) = 0.5\) for \(0 \leq y \leq 2\) and 0 otherwise.
02

Define the new random variable \(Z = X/Y\)

Let \(Z = X/Y\). To find the pdf of \(Z\), we need to establish the cumulative distribution function (CDF) of \(Z\) which is \(F_Z(z) = P(Z \leq z) = P(X/Y \leq z)\), and then differentiate the CDF to get the pdf.
03

Find the cumulative distribution function (CDF) of Z

We are looking at the event that \(Z\), or \(X/Y\), is less than or equal to \(z\). This is equivalent to the event that \(X \leq zY\). We must integrate the joint pdf of \(X\) and \(Y\) (which is the product of their independent pdfs) over the region defined by the above inequality. Note here that it matters whether \(z\) is less than or greater than 1, because this determines the bounds of our region. If \(0 2\), the probability is 0.
04

Find the probability density function (pdf) of Z

We obtain \(f_Z(z)\), the pdf of \(Z\), by taking derivative of the CDF \(F_Z(z)\). That yields: if \(0 2\), the probability is 0.

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Key Concepts

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

Uniformly Distributed Random Variables
When we talk about uniformly distributed random variables, we're describing a situation where all outcomes within a certain range are equally likely to occur. Imagine a perfectly fair ruler, one inch long, held above a very long, perfectly straight line. If you drop the ruler and it could land with its starting point anywhere along the line, the landing point of the starting edge of the ruler is uniformly distributed between 0 and 1 inch. In probabilistic terms, it means that the probability density function (pdf) for a uniformly distributed random variable is constant across the interval for which it is defined.

In our exercise, the random variable X being uniformly distributed over the interval (0,1) resembles a perfectly even distribution of values – anywhere you 'drop' in this interval is equally plausible. Similarly, another variable Y is uniformly distributed over (0,2), meaning any landing point in this range has an even chance. When these variables are independent, knowing the outcome of one gives us no information about the other, which simplifies the calculations of their joint behavior.

Key Characteristics of Uniform Distribution:

  • The pdf is constant over the specified interval.
  • Every outcome in the interval is equally likely.
  • The integral of the pdf over the interval is equal to 1, ensuring it's a valid probability.
Cumulative Distribution Function
The cumulative distribution function (CDF), represented as F(x), is crucial in understanding any random variable. It tells us the probability that the variable takes on a value less than or equal to a certain point. Essentially, it bundles up the probabilities for all outcomes below a threshold, adding them together.

For uniformly distributed variables, the CDF is simply the integral of the pdf up to the point of interest. It gives us a quick way to measure 'how much' of the probability has passed by as we move along the range of possible outcomes. In the given exercise, we used the CDF to find the distribution function of a new variable Z = X/Y, which necessitated integration over specified bounds – an operation that sums up the probabilities up to that point.

Understanding the CDF:

  • The CDF accumulates probabilities up to a certain value.
  • For a uniform pdf, the CDF is linear and increases steadily over the interval.
  • The CDF is useful in finding probabilities for ranges of values and for obtaining the pdf by differentiation.
Independent Random Variables
The concept of independent random variables is pivotal in probability theory. Two random variables are independent if the occurrence of one event has no influence whatsoever on the probability of the occurrence of another event. For independent variables, the joint probability density function is simply the product of their individual pdfs.

In our exercise scenario, X and Y do not affect each other – the value of X offers no hint about Y, and vice versa. This property allowed us to calculate the probability of their ratio by multiplying their pdfs directly. Independence is a powerful assumption that simplifies our calculations significantly.

Benefits of Variable Independence:

  • Allows easy calculation of joint probabilities.
  • Reduces complexity in stochastic processes.
  • Facilitates the assumption of a product joint pdf.
Integration in Probability
Finally, let's talk about integration in probability. Integration serves as a tool for accumulating probabilities over a continuous interval. It's the mathematical equivalent of adding up infinitely small probabilities across a range to find the probability of a multi-faceted event.

We use integration to determine the cumulative distribution function (CDF) and, subsequently, the probability density function (pdf). In the exercise, we integrated the product of the pdfs of X and Y over a certain region bounded by the ratio involved in the variable Z. The integral gives us the total probability that falls within that region. It's the continuous counterpart to summing probabilities in discrete distributions.

Integral Uses in Probability:

  • Calculates cumulative probabilities across intervals.
  • Finds expected values for random variables.
  • Derives probability density functions from cumulative distribution functions.

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Most popular questions from this chapter

A point is selected at random from the bounded region between the curves \(y=x^{2}-1\) and \(y=1-x^{2}\). Let \(X\) be the the \(x\)-coordinate, and let \(Y\) be the \(y\)-coordinate of the point selected. Determine if \(X\) and \(Y\) are independent.

Let \(X\) and \(Y\) be continuous random variables with the joint probability density function $$ f(x, y)= \begin{cases}e^{-y} & \text { if } y>0, \quad 0

(The Wallet Paradox) Consider the following "paradox" given by Martin Gardner in his book "Aha! Gotcha," W. H. Freeman and Company, New York, \(1981 .\) Each of two persons places his wallet on the table. Whoever has the smallest amount of money in his wallet, wins all the money in the other wallet. Each of the players reason as follows: "I may lose what I have but I may also win more than I have. So the game is to my advantage." As Kent G. Merryfield, Ngo Viet, and Saleem Watson have observed in their paper "The Wallet Paradox" in the August-September 1997 issue of the American Mathematical Monthly: Paradoxically, it seems that the game is to the advantage of both players. ... However, the inference that "the game is to my advantage" is the source of the apparent paradox, because it does not take into account the probabilities of winning or losing. In other words, if the game is played many times, how often does a player win? How often does he lose? And by how much? Following the analysis of Kent G. Merryfield, Ngo Viet, and Saleen Watson, let \(X\) and \(Y\) be the amount of money in the wallets of players A and \(\mathrm{B}\), respectively. Let \(W_{A}\) and \(W_{B}\) be the amount of money that player A and \(\mathrm{B}\) will win, respectively. \(W_{A}(X, Y)=-W_{B}(X, Y)\) and $$ W_{A}(X, Y)= \begin{cases}-X & \text { if } X>Y \\ Y & \text { if } X

The joint probability density function of random variables \(X, Y\), and \(Z\) is given by $$ f(x, y, z)= \begin{cases}c(x+y+2 z) & \text { if } 0 \leq x, y, z \leq 1 \\ 0 & \text { otherwise } .\end{cases} $$ (a) Determine the value of \(c\). (b) Find \(P(X<1 / 3 \mid Y<1 / 2, Z<1 / 4)\).

Let the joint probability density function of \(X\) and \(Y\) be given by $$ f(x, y)= \begin{cases}c / x & \text { if } 0

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