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An exam consists of a problem section and a shortanswer section. Let \(X_{1}\) denote the amount of time (hr) that a student spends on the problem section and \(X_{2}\) represent the amount of time the same student spends on the short- answer section. Suppose the joint pdf of these two times is $$ f\left(x_{1}, x_{2}\right)=\left\\{\begin{array}{cl} c x_{1} x_{2} & \frac{x_{1}}{3}

Short Answer

Expert verified
a. c = 72. b. Probability is 0. c. Probability is 0. d. Use transformation to find joint and marginal distributions.

Step by step solution

01

Determine the limits for integration

To find the constant \(c\), we integrate \(f(x_1, x_2)\) over its support where \(0 < x_1 < 1\) and \(\frac{x_1}{3} < x_2 < \frac{x_1}{2}\). This gives the limits for \(x_2\) as a function of \(x_1\).
02

Compute the integral

Integrate the function \(f(x_1, x_2) = c x_1 x_2\) over its defined region:\[\int_0^1 \int_{x_1/3}^{x_1/2} c x_1 x_2 \, dx_2 \, dx_1 = 1\]. Compute the inner integral over \(x_2\), resulting in:\[\int_{x_1/3}^{x_1/2} x_2 \, dx_2 = \left[\frac{x_2^2}{2}\right]_{x_1/3}^{x_1/2} = \frac{1}{2}\left(\frac{x_1^2}{4} - \frac{x_1^2}{9}\right) = \frac{x_1^2}{18}\].
03

Solve for c

Substitute the inner integral back and integrate over \(x_1\):\[\int_0^1 c x_1 \frac{x_1^2}{18} \, dx_1 = \int_0^1 \frac{c x_1^3}{18} \, dx_1 = \frac{c}{18} \times \frac{1}{4} = 1\]. Solve for \(c\):\[\frac{c}{72} = 1 \implies c = 72\].
04

Setup for conditional probability

Given \(X_2 = 0.25\), find the conditional probability \(P(X_1 \leq 0.60 | X_2 = 0.25)\). Determine the interval for \(X_1\) using the condition \(\frac{x_1}{3} < 0.25 < \frac{x_1}{2}\), which implies \(0.75 < x_1 < 0.5\). The valid range is empty, thus \(P(X_1 \leq 0.60 | X_2 = 0.25) = 0\).
05

Compute probability for exceedance condition

Evaluate \(P(X_1 - X_2 \geq 0.5)\). This occurs when \(x_2 = x_1 - 0.5\). Since \(x_2\) is bounded by \(\frac{x_1}{3} < x_2 < \frac{x_1}{2}\), this situation is impossible. Thus, the probability is \(0\).
06

Joint distribution transformation

To find the joint distribution of \(Y_1 = X_2/X_1\) and \(Y_2 = X_2\), use transformation techniques. Convert variables \((X_1, X_2)\) to \((Y_1, Y_2)\) and find the Jacobian determinant:\[J = \begin{vmatrix} \frac{\partial x_1}{\partial y_1} & \frac{\partial x_1}{\partial y_2} \ \frac{\partial x_2}{\partial y_1} & \frac{\partial x_2}{\partial y_2} \end{vmatrix} = \begin{vmatrix} x_2^{-1} & 0 \ -x_2/x_1^2 & 1/x_1 \end{vmatrix} = \frac{1}{x_1}\].
07

Marginal distribution of the ratio

From the joint distribution, derive the marginal distribution of \(Y_1 = X_2/X_1\) by integrating over \(Y_2\). Determine the transformed support and perform integration:\[\int_{0}^{1} f(y_1, y_2) \, dy_2\]. Due to complex transformations, solve numerically or analytically based on simplified assumptions.

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Key Concepts

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

Conditional Probability
Conditional probability helps us figure out the likelihood of one event occurring given that another event has already taken place. In the context of the exam times, if we want to know the probability of a student completing the problem section in less than 0.60 hours, given they take exactly 0.25 hours on the short-answer section, we use conditional probability. This involves finding the relevant conditional distribution of times.

First, identify the valid range for the problem time (denoted by \(X_1\)) based on the given short-answer time (\(X_2 = 0.25\)). We set up inequalities: \(\frac{x_1}{3} < 0.25 < \frac{x_1}{2}\).

This simplifies to determine that \(0.75 < x_1 < 0.5\), but since this range is not practically possible (as one number cannot be both greater than and less than another in this way), the conditional probability ends up being zero. This illustrates how conditions on probabilities define feasible ranges, and how mathematical constraints can lead to definitive outcomes.
Probability Density Function
A Probability Density Function (pdf) is a function that describes the likelihood of a continuous random variable to take on a particular value. It gives a complete description of the possible outcomes for the variable and their probabilities.

In the exercise, the joint pdf is presented as \(f(x_1, x_2) = cx_1x_2\) in a specific region defined by \(\frac{x_1}{3} < x_2 < \frac{x_1}{2}\) and \(0 < x_1 < 1\). The pdf is zero outside this region, reflecting that outcomes outside these constraints have no probability of occurrence.

Calculating the value of \(c\), the constant in the pdf, is crucial. It ensures that the total probability of all potential outcomes equals one, thereby maintaining the function's integrity. This involves integrating over the specified bounds and setting the result to 1, which standardizes the distribution.
Transformation of Variables
Transforming variables is a powerful technique used in probability to simplify complex distributions. In our exercise, we transform the original variables \((X_1, X_2)\) into new variables \((Y_1, Y_2)\), where \(Y_1 = X_2/X_1\) and \(Y_2 = X_2\).

The transformation can be beneficial for tackling the problem, as it can simplify boundaries or lead to more straightforward integrations. A common step in this process is calculating the Jacobian, a determinant that measures how the area or volume in the new system relates to that in the original system. For this exercise, the Jacobian is found to be \(\frac{1}{x_1}\), which helps in the transition from one set of variables to another.

This method is part of the change of variables technique, used widely in multivariable calculus to handle transformations and find new distributions that charm with analysis.
Marginal Distribution
A marginal distribution provides the probabilities of one of the variables in a multivariate distribution, independently from the others. To find a marginal distribution, we integrate the joint distribution over the unwanted variables.

In the given problem, once we have the joint distribution of the transformed variables \((Y_1, Y_2)\), we focus on deriving the marginal distribution of \(Y_1 = X_2/X_1\). This involves integrating out \(Y_2\), effectively looking at only the behavior of the ratio independent of the second variable.

This process abstracts the problem further, allowing us to understand the distribution of each variable individually. Marginal distributions are crucial for isolating the effects and likelihoods of singular aspects within a joint framework and are integral to comprehending complex statistical data distributions.

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Most popular questions from this chapter

According to the Mars Candy Company, the longrun percentages of various colors of M\&M milk chocolate candies are as follows: \(\begin{array}{llllll}\text { Blue: } & \text { Orange: } & \text { Green: } & \text { Yellow: } & \text { Red: } & \text { Brown: } \\ 24 \% & 20 \% & 16 \% & 14 \% & 13 \% & 13 \%\end{array}\) a. In a random sample of 12 candies, what is the probability that there are exactly two of each color? b. In a random sample of 6 candies, what is the probability that at least one color is not included? c. In a random sample of 10 candies, what is the probability that there are exactly 3 blue candies and exactly 2 orange candies? d. In a random sample of 10 candies, what is the probability that there are at most 3 orange candies? [Hint: Think of an orange candy as a success and any other color as a failure.] e. In a random sample of 10 candies, what is the probability that at least 7 are either blue, orange, or green?

Conjecture the form of the joint pdf of three order statistics \(Y_{i}, Y_{j}, Y_{k}\) in a random sample of size \(n\).

This week the number \(X\) of claims coming into an insurance office is Poisson with mean 100 . The probability that any particular claim relates to automobile insurance is \(.6\), independent of any other claim. If \(Y\) is the number of automobile claims, then \(Y\) is binomial with \(X\) trials, each with "success" probability .6. a. Determine \(E(Y \mid X=x)\) and \(V(Y \mid X=x)\). b. Use part (a) to find \(E(Y)\). c. Use part (a) to find \(V(Y)\).

Show that if \(X, Y\), and \(Z\) are rv's and \(a\) and \(b\) are constants, then \(\operatorname{Cov}(a X+b Y, Z)=a \operatorname{Cov}(X, Z)+\) \(b \operatorname{Cov}(Y, Z)\)

A restaurant serves three fixed-price dinners costing $$\$ 20, \$ 25$$, and $$\$ 30$$. For a randomly selected couple dining at this restaurant, let \(X=\) the cost of the man's dinner and \(Y=\) the cost of the woman's dinner. The joint pmf of \(X\) and \(Y\) is given in the following table: $$ \begin{array}{ll|ccc} & & {l}{y} \\ {p(x, y)} & & 20 & 25 & 30 \\ \hline {x} & 20 & .05 & .05 & .10 \\ & 25 & .05 & .10 & .35 \\ & 30 & 0 & .20 & .10 \end{array} $$ a. Compute the marginal pmf's of \(X\) and \(Y\). b. What is the probability that the man's and the woman's dinner cost at most \(\$ 25\) each? c. Are \(X\) and \(Y\) independent? Justify your answer. d. What is the expected total cost of the dinner for the two people? e. Suppose that when a couple opens fortune cookies at the conclusion of the meal, they find the message "You will receive as a refund the difference between the cost of the more expensive and the less expensive meal that you have chosen." How much does the restaurant expect to refund?

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