Setup (in an appropriate coordinate system and JUSTIFY
your choice of the coordinate system) the integrals necessary to accomplish the
following tasks. You are not evaluating these integrals.

Problem Set 3: a joint density function for probability of three events is
given by the following equation:
f (x, y, z) ={(12x2yz [0, 1] × [0, 1] × [0, 1])
(0 otherwise)
Answer the following questions - set up any integrals necessary to do so.
– Explain (and set up any integrals necessary) how to verify this function
is in fact a probability density function
– Probability that a chosen point (x, y, z) is such that 2x + y ≤ 1
– Expected value of each variable

To address Problem Set 3, we'll use Cartesian coordinates, which are suitable for the given density function and the specified domain.

### Joint Density Function

The joint density function \(f(x, y, z)\) is given by:

\[ f(x, y, z) = \begin{cases} 
12x^2yz & \text{if } 0 \leq x \leq 1, \; 0 \leq y \leq 1, \; 0 \leq z \leq 1 \\
0 & \text{otherwise}
\end{cases} \]

### Verifying the Function is a Probability Density Function

To verify that \(f(x, y, z)\) is a probability density function, we need to show that:
1. \(f(x, y, z) \geq 0\) for all \((x, y, z)\)
2. \(\iiint\limits_{\mathbb{R}^3} f(x, y, z) \, dV = 1\)

Since \(f(x, y, z) \geq 0\) by definition, we need to check that the integral of \(f(x, y, z)\) over its domain is equal to 1.

Set up the integral for the domain \([0,1] \times [0,1] \times [0,1]\):

\[
\iiint\limits_{[0,1] \times [0,1] \times [0,1]} 12x^2yz \, dx \, dy \, dz
\]

\[
\int_0^1 \int_0^1 \int_0^1 12x^2yz \, dz \, dy \, dx
\]

### Probability that \(2x + y \leq 1\)

To find the probability that a chosen point \((x, y, z)\) is such that \(2x + y \leq 1\), we need to set up the integral of the joint density function over the region where \(2x + y \leq 1\):

\[
\iiint\limits_{0 \leq x \leq 1, \; 0 \leq y \leq 1, \; 0 \leq z \leq 1, \; 2x + y \leq 1} 12x^2yz \, dx \, dy \, dz
\]

The region where \(2x + y \leq 1\) can be rewritten in terms of \(y\):

\[
\int_0^1 \int_0^{1 - 2x} \int_0^1 12x^2yz \, dz \, dy \, dx
\]

### Expected Value of Each Variable

The expected value \(E[X]\) of \(x\) is given by:

\[
E[X] = \iiint\limits_{[0,1] \times [0,1] \times [0,1]} x \cdot f(x, y, z) \, dx \, dy \, dz
\]

\[
E[X] = \int_0^1 \int_0^1 \int_0^1 x \cdot 12x^2yz \, dz \, dy \, dx
\]

The expected value \(E[Y]\) of \(y\) is given by:

\[
E[Y] = \iiint\limits_{[0,1] \times [0,1] \times [0,1]} y \cdot f(x, y, z) \, dx \, dy \, dz
\]

\[
E[Y] = \int_0^1 \int_0^1 \int_0^1 y \cdot 12x^2yz \, dz \, dy \, dx
\]

The expected value \(E[Z]\) of \(z\) is given by:

\[
E[Z] = \iiint\limits_{[0,1] \times [0,1] \times [0,1]} z \cdot f(x, y, z) \, dx \, dy \, dz
\]

\[
E[Z] = \int_0^1 \int_0^1 \int_0^1 z \cdot 12x^2yz \, dz \, dy \, dx
\]

### Summary

The integrals set up for verifying the density function, finding the probability of the region \(2x + y \leq 1\), and calculating the expected values in Cartesian coordinates provide a structured approach to solving the problem. Although we are not evaluating these integrals, the setup forms the foundation for further computation.

If you have any questions or need further details, please let me know.

Here are some questions you might want to explore next:
1. How do you verify a joint density function over different types of regions?
2. What is the significance of the expected value in probability and statistics?
3. Can you provide an example of calculating probabilities for complex regions?
4. How does the joint density function affect the expected value

Explore the setup and verification of a joint density function for three events in probability theory. Calculate the probability of a specific region and expected values using integrals. Suitable for graduate-level students and researchers.

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