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The joint density of \(X\) and \(Y\) is given by $$ f(x, y)=\frac{e^{-y}}{y}, \quad 0

Short Answer

Expert verified
The short answer based on the given step-by-step solution is: \(E\left[X^{2} \mid Y=y\right] = \frac{y^2}{3}\).

Step by step solution

01

Find the conditional probability density function \(f(x \mid y)\)

First, we need to find the marginal probability density function for Y, which is defined as: $$ f_Y(y) = \int_{-\infty}^{\infty} f(x, y) \, dx. $$ But be aware of the bounds for x, which are \(0 < x < y\), so the integral becomes: $$ f_Y(y) = \int_0^y f(x, y) \, dx. $$ After finding the marginal probability density function for Y, we can find the conditional probability density function: $$ f(x \mid y) = \frac{f(x, y)}{f_Y(y)}. $$ Now, let's find the marginal probability density function for Y: $$ f_Y(y) = \int_0^y \frac{e^{-y}}{y} \, dx = \frac{e^{-y}}{y} \int_0^y dx. $$ This integral simplifies to: $$ \int_0^y dx = [x]_0^y = y, $$ So, $$ f_Y(y) = \frac{e^{-y}}{y} y = e^{-y}. $$ Now, we can find the conditional probability density function: $$ f(x \mid y) = \frac{f(x, y)}{f_Y(y)} = \frac{\frac{e^{-y}}{y}}{e^{-y}}. $$ This simplifies to: $$ f(x \mid y) = \frac{1}{y}. $$
02

Compute \(E\left[X^{2} \mid Y=y\right]\)

Now that we have the conditional probability density function, we can compute the conditional expectation: $$ E\left[X^{2} \mid Y=y\right] = \int_{-\infty}^{\infty} x^2 f(x \mid y) \, dx $$ But be aware of the bounds for x, which are \(0 < x < y\), so the integral becomes: $$ E\left[X^{2} \mid Y=y\right] = \int_0^y x^{2} \frac{1}{y} \, dx $$ Let's solve this integral: $$ \int_0^y x^{2} \frac{1}{y} \, dx = \frac{1}{y} \int_0^y x^2 \, dx = \frac{1}{y} \left[\frac{x^3}{3}\right]_0^y, $$ After evaluating the integral, we get: $$ E\left[X^{2} \mid Y=y\right] = \frac{1}{y} \left(\frac{y^3}{3}\right) $$ Finally, simplify the expression: $$ E\left[X^{2} \mid Y=y\right] = \frac{y^2}{3} $$

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Key Concepts

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

Joint Probability Density Function
Imagine we have two continuous random variables, let's call them X and Y. The joint probability density function (PDF), denoted as \( f(x, y) \), describes the likelihood of these variables taking on specific values simultaneously. To put it simply, it tells us how dense the probability is spread over different values of X and Y at the same time. Think of it like a 3D map showing how likely different outcomes are.

When dealing with a joint PDF, one important thing to note is the range over which the variables are defined. In our exercise, the range is \( 0 < x < y \) and \( 0 < y < \
The joint PDF is the foundation upon which we can build to find more specific probabilities using marginal and conditional density functions, which reveals more information about individual variables within the joint distribution.
Marginal Probability Density Function
Imagine you're focusing on just one of the variables, say Y, and you want to ignore X for a moment. This is where the marginal probability density function (PDF) comes into play. The marginal PDF, for our case \( f_Y(y) \), sums up the probabilities of all potential outcomes for Y across all values of X. It's like looking at the shadow of our 3D probability map on the Y-axis. We're no longer concerned with where X stands; we're summing up over X's entire range.

To find the marginal PDF, we need to integrate the joint PDF over the entire range of the other variable. Mathematically, if we integrate \( f(x, y) \) over all values of X, we obtain \( f_Y(y) \), which gives us a function that only depends on Y. This simplification allows us to work with one variable at a time, which is crucial in complex probability scenarios. The exercise guides us through the process, emphasizing the importance of paying close attention to the limits of integration, which are dictated by the definition of the variables and their relationship.
Conditional Probability Density Function
Now, what if we are interested in understanding the behavior of X, given that we already know the value of Y? This is where the conditional probability density function (PDF), \( f(x | y) \), steps in. It represents the probability distribution of X while considering that Y is fixed at a certain value. In essence, it tells us about the probability landscape of X on the condition or given the knowledge of Y’s value.

To obtain the conditional PDF from the joint PDF, we simply divide the joint PDF by the marginal PDF of the known variable. Mathematically, it's represented as \( f(x | y) = \frac{f(x, y)}{f_Y(y)} \). The exercise shows the application of this concept, which is akin to adjusting our viewpoint; we’re not looking at the entire hill anymore, but instead, we’re examining a single level of altitude (a given value of y) and seeing how the probability is distributed along that level for x.

The conditional PDF allows us to gain insights into the behavior of a subset of the random variables within a joint distribution, offering a crucial tool for making predictions in the presence of partial information.
Integration in Probability
Integration is a powerful mathematical tool that helps us sum up an infinite number of infinitesimally small quantities. In probability, it enables us to find marginal and conditional probabilities by adding up the probabilities over a continuum of outcomes. When we integrate a probability density function (PDF) over a certain range, we are essentially finding the total probability of observing a value within that range.

In our exercise, integration is used in two key places: First, to derive the marginal PDF of Y, we integrate the joint PDF of X and Y over the allowable range of X. Next, to calculate the expected value of \( X^2 \) given Y, we integrate the conditional PDF of X given Y, weighted by \( X^2 \), over X's allowable range. It's like calculating the average outcome weighted by probability, which for continuous variables, requires the integral as a summing mechanism.

Understanding integration within the context of probability is essential because it allows us to move from general joint distributions to specific marginal and conditional distributions, and ultimately to expected values, which are a cornerstone of probabilistic analysis.

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Most popular questions from this chapter

Let \(X_{i}, i \geqslant 0\) be independent and identically distributed random variables with probability mass function $$ p(j)=P\left[X_{i}=i\right\\}, \quad j=1, \ldots, m, \quad \sum_{j=1}^{m} P(j)=1 $$ Find \(E[N]\), where \(N=\min \left[n>0: X_{n}=X_{0}\right\\}\)

Use the conditional variance formula to find the variance of a geometric random variable.

The joint density of \(X\) and \(Y\) is given by $$ f(x, y)=\frac{e^{-x / y} e^{-y}}{y}, \quad 0

In a knockout tennis tournament of \(2^{n}\) contestants, the players are paired and play a match. The losers depart, the remaining \(2^{n-1}\) players are paired, and they play a match. This continues for \(n\) rounds, after which a single player remains unbeaten and is declared the winner. Suppose that the contestants are numbered 1 through \(2^{n}\), and that whenever two players contest a match, the lower numbered one wins with probability \(p\). Also suppose that the pairings of the remaining players are always done at random so that all possible pairings for that round are equally likely. (a) What is the probability that player 1 wins the tournament? (b) What is the probability that player 2 wins the tournament? Hint: Imagine that the random pairings are done in advance of the tournament. That is, the first-round pairings are randomly determined; the \(2^{n-1}\) first-round pairs are then themselves randomly paired, with the winners of each pair to play in round 2; these \(2^{n-2}\) groupings (of four players each) are then randomly paired, with the winners of each grouping to play in round 3, and so on. Say that players \(i\) and \(j\) are scheduled to meet in round \(k\) if, provided they both win their first \(k-1\) matches, they will meet in round \(k\). Now condition on the round in which players 1 and 2 are scheduled to meet.

Let \(X_{1}, \ldots, X_{n}\) be independent random variables having a common distribution function that is specified up to an unknown parameter \(\theta\). Let \(T=T(\mathrm{X})\) be a function of the data \(\mathrm{X}=\left(X_{1}, \ldots, X_{n}\right) .\) If the conditional distribution of \(X_{1}, \ldots, X_{n}\) given \(T(\mathrm{X})\) does not depend on \(\theta\) then \(T(\mathrm{X})\) is said to be a sufficient statistic for \(\theta .\) In the following cases, show that \(T(\mathbf{X})=\sum_{i=1}^{n} X_{i}\) is a sufficient statistic for \(\theta\). (a) The \(X_{i}\) are normal with mean \(\theta\) and variance \(1 .\) (b) The density of \(X_{i}\) is \(f(x)=\theta e^{-\theta x}, x>0\). (c) The mass function of \(X_{i}\) is \(p(x)=\theta^{x}(1-\theta)^{1-x}, x=0,1,0<\theta<1\). (d) The \(X_{i}\) are Poisson random variables with mean \(\theta\).

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