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Let \(Y_{1}\) and \(Y_{2}\) have the joint probability density function given by $$f\left(y_{1}, y_{2}\right)=\left\\{\begin{array}{ll} k y_{1} y_{2}, & 0 \leq y_{1} \leq 1,0 \leq y_{2} \leq 1, \\ 0, & \text { elsewhere } \end{array}\right.$$ a. Find the value of \(k\) that makes this a probability density function. b. Find the joint distribution function for \(Y_{1}\) and \(Y_{2}\). c. Find \(P\left(Y_{1} \leq 1 / 2, Y_{2} \leq 3 / 4\right)\).

Short Answer

Expert verified
The value of \( k \) is 4. The joint distribution function is \( y_1^2 y_2^2 \), and \( P(Y_1 \leq 1/2, Y_2 \leq 3/4) = 9/64 \).

Step by step solution

01

Determine the normalization constant

To find the constant \( k \) that makes this a valid probability density function, we need to integrate the joint density function over its entire range and set the integral equal to 1. The integration limits for \( y_1 \) and \( y_2 \) are both from 0 to 1. Thus, we calculate:\[\int_0^1 \int_0^1 k y_1 y_2 \, dy_1 \, dy_2 = 1.\]Evaluating the integrals:\[\int_0^1 y_1 \, dy_1 = \left[ \frac{y_1^2}{2} \right]_0^1 = \frac{1}{2}.\]\[\int_0^1 y_2 \, dy_2 = \left[ \frac{y_2^2}{2} \right]_0^1 = \frac{1}{2}.\]Therefore:\[k \cdot \frac{1}{2} \cdot \frac{1}{2} = 1 \Rightarrow k = 4.\]
02

Identify the joint distribution function

The joint distribution function \( F(y_1, y_2) \) is the double integral of the joint probability density function. Therefore,\[F(y_1, y_2) = \int_0^{y_2} \int_0^{y_1} k y_1' y_2' \, dy_1' \, dy_2',\]where \( y_1' \) and \( y_2' \) are dummy variables of integration.Substituting \( k = 4 \) and calculating, we get:\[= \int_0^{y_2} \left[ 4 \int_0^{y_1} y_1' y_2' \, dy_1' \right] \, dy_2'.\]Evaluating the inner integral:\[\int_0^{y_1} y_1' \, dy_1' = \left[ \frac{(y_1')^2}{2} \right]_0^{y_1} = \frac{y_1^2}{2}.\]Plug this result back into the outer integral:\[4 \int_0^{y_2} \left( \frac{y_1^2}{2} \cdot y_2' \right) \, dy_2' = 4 \cdot \frac{y_1^2}{2} \left[ \frac{(y_2')^2}{2} \right]_0^{y_2} = y_1^2 y_2^2.\]Consequently, the joint distribution function is:\[F(y_1, y_2) = y_1^2 y_2^2.\]
03

Calculate the specified probability

To find the probability \( P(Y_{1} \leq 1/2, Y_{2} \leq 3/4) \), we use the joint distribution function obtained in Step 2:\[F\left(\frac{1}{2}, \frac{3}{4}\right) = \left(\frac{1}{2}\right)^2 \left(\frac{3}{4}\right)^2.\]Calculating, we have:\[= \frac{1}{4} \cdot \frac{9}{16} = \frac{9}{64}.\]

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Key Concepts

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

Normalization Constant
In probability theory, a normalization constant ensures that a probability density function (PDF) correctly represents a valid probability distribution. For a function to be a PDF, the total probability across its entire range must equal 1. This is where the normalization constant, often represented as \( k \), comes into play.
To determine \( k \), we integrate the given joint probability density function (PDF) over its full range. For our given function \( f(y_1, y_2) = k y_1 y_2 \), where \( 0 \leq y_1, y_2 \leq 1 \), we find:
\[ \int_0^1 \int_0^1 k y_1 y_2 \, dy_1 \, dy_2 = 1. \]
This is because the integral of a probability density function (PDF) over its entire space must equal to 1. By solving this double integral, we discovered that \( k = 4 \).
This normalization constant adjusts the function ensuring the area under the curve equals 1, thus confirming it as a valid probability density function. So, without \( k \), we wouldn't correctly represent the distribution!
  • The total area under the curve of a PDF must always equal 1.
  • \( k \) ensures the function is properly scaled to meet this criterion.
Joint Distribution Function
The joint distribution function is a critical concept in understanding how two random variables are related. Unlike a single variable, the joint distribution considers the likelihood of both variables occurring together within a specified range.
In our case, we seek the joint distribution function \( F(y_1, y_2) \). This function is derived by integrating the joint probability density function (PDF) with respect to both variables.
The formula to achieve this is:
\[ F(y_1, y_2) = \int_0^{y_2} \int_0^{y_1} 4 y_1' y_2' \, dy_1' \, dy_2', \]
where \( y_1' \) and \( y_2' \) are dummy variables meant for integration simplicity. After performing the integration, we find:
\[ F(y_1, y_2) = y_1^2 y_2^2. \]
  • This joint distribution function now tells us the cumulative probability for events described by \( y_1 \) and \( y_2 \) together.
  • It enables probability calculations over a specific range of these two variables.
Understanding this joint distribution function helps in grasping how different values of \( y_1 \) and \( y_2 \) interact and the probability of their simultaneous occurrences.
Probability Calculation
Calculating probabilities for specific events using the joint distribution function is a practical application of the theoretical concepts we've learned.
Suppose we want to find the probability that \( Y_1 \leq 1/2 \) and \( Y_2 \leq 3/4 \). We simply substitute these values into our joint distribution function \( F(y_1, y_2) = y_1^2 y_2^2 \).
Here's the computation:
\[ F\left( \frac{1}{2}, \frac{3}{4} \right) = \left( \frac{1}{2} \right)^2 \left( \frac{3}{4} \right)^2. \]
This equals:
\[ = \frac{1}{4} \cdot \frac{9}{16} = \frac{9}{64}. \]
The result, \( \frac{9}{64} \), is the probability that \( Y_1 \) and \( Y_2 \) fall within those specified limits.
Such computations are invaluable when determining the likelihood of various outcomes within a joint probability framework.
  • Using the joint distribution function allows for precise probability calculations for specific conditions.
  • Simplifying these probabilistic scenarios supports data-driven decision-making and predictions.

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Most popular questions from this chapter

Let \(Y_{1}\) and \(Y_{2}\) have a bivariate normal distribution. Show that the conditional distribution of \(Y_{1}\) given that \(Y_{2}=y_{2}\) is a normal distribution with mean \(\mu_{1}+\rho \frac{\sigma_{1}}{\sigma_{2}}\left(y_{2}-\mu_{2}\right)\) and variance \(\sigma_{1}^{2}\left(1-\rho^{2}\right)\)

An electronic system has one each of two different types of components in joint operation. Let \(Y_{1}\) and \(Y_{2}\) denote the random lengths of life of the components of type 1 and type II, respectively. The joint density function is given by $$f\left(y_{1}, y_{2}\right)=\left\\{\begin{array}{ll} (1 / 8) y_{1} e^{-\left(y_{1}+y_{1}\right) / 2}, & y_{1}>0, y_{2}>0 \\ 0, & \text { elsewhere } \end{array}\right.$$ (Measurements are in hundreds of hours.) Find \(P\left(Y_{1}>1, Y_{2}>1\right)\).

Let \(Y_{1}\) denote the weight (in tons) of a bulk item stocked by a supplier at the beginning of a week and suppose that \(Y_{1}\) has a uniform distribution over the interval \(0 \leq y_{1} \leq 1\). Let \(Y_{2}\) denote the amount (by weight) of this item sold by the supplier during the week and suppose that \(Y_{2}\) has a uniform distribution over the interval \(0 \leq y_{2} \leq y_{1},\) where \(y_{1}\) is a specific value of \(Y_{1}\) a. Find the joint density function for \(Y_{1}\) and \(Y_{2}\) b. If the supplier stocks a half-ton of the item, what is the probability that she sells more than a quarter-ton? c. If it is known that the supplier sold a quarter-ton of the item, what is the probability that she had stocked more than a half-ton?

Refer to Exercises 5.6,5.24 , and \(5.50 .\) Suppose that a radioactive particle is randomly located in a square with sides of unit length. A reasonable model for the joint density function for \(Y_{1}\) and \(Y_{2}\) is $$f\left(y_{1}, y_{2}\right)=\left\\{\begin{array}{ll} 1, & 0 \leq y_{1} \leq 1,0 \leq y_{2} \leq 1 \\ 0, & \text { elsewhere } \end{array}\right.$$ a. What is \(E\left(Y_{1}-Y_{2}\right) ?\) b. What is \(E\left(Y_{1} Y_{2}\right) ?\) c. What is \(E\left(Y_{1}^{2}+Y_{2}^{2}\right) ?\) d. What is \(V\left(Y_{1} Y_{2}\right) ?\)

The joint density function of \(Y_{1}\) and \(Y_{2}\) is given by $$f\left(y_{1}, y_{2}\right)=\left\\{\begin{array}{ll} 30 y_{1} y_{2}^{2}, & y_{1}-1 \leq y_{2} \leq 1-y_{1}, 0 \leq y_{1} \leq 1 \\ 0, & \text { elsewhere } \end{array}\right.$$ a. Find \(F(1 / 2,1 / 2)\). b. Find \(F(1 / 2,2)\). c. Find \(P\left(Y_{1}>Y_{2}\right)\).

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