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 the given problem, let's proceed with the tasks step-by-step, using a Cartesian coordinate system, which is suitable here because the limits are straightforwardly defined on \([0, 1] \times [0, 1] \times [0, 1]\).

### 1. Verifying the Probability Density Function

To verify that \( f(x, y, z) = 12x^2yz \) is a probability density function, we need to ensure that the total integral over the domain equals 1:

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

Setting up the integral:

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

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

To find the probability that a chosen point \((x, y, z)\) satisfies \(2x + y \leq 1\), we need to integrate \( f(x, y, z) \) over the region where \(2x + y \leq 1\):

\[ \iiint_{D} 12x^2yz \, dx \, dy \, dz \]

where \(D\) is the region defined by \(0 \leq x \leq 1\), \(0 \leq y \leq 1 - 2x\), and \(0 \leq z \leq 1\).

Setting up the integral:

\[ \iiint_{0}^{1} \iiint_{0}^{1 - 2x} \iiint_{0}^{1} 12x^2yz \, dz \, dy \, dx \]

### 3. Expected Value of Each Variable

The expected value of a random variable \(X\) is given by:

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

Similarly, the expected values for \(Y\) and \(Z\) are:

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

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

Setting up the integrals:

- \(E[X]\):

\[ E[X] = \iiint_{0}^{1} \iiint_{0}^{1} \iiint_{0}^{1} x \cdot 12x^2yz \, dx \, dy \, dz \]

- \(E[Y]\):

\[ E[Y] = \iiint_{0}^{1} \iiint_{0}^{1} \iiint_{0}^{1} y \cdot 12x^2yz \, dx \, dy \, dz \]

- \(E[Z]\):

\[ E[Z] = \iiint_{0}^{1} \iiint_{0}^{1} \iiint_{0}^{1} z \cdot 12x^2yz \, dx \, dy \, dz \]

### Summary of Integrals

1. **Verifying the Probability Density Function:**

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

2. **Probability that \(2x + y \leq 1\):**

\[ \iiint_{0}^{1} \iiint_{0}^{1 - 2x} \iiint_{0}^{1} 12x^2yz \, dz \, dy \, dx \]

3. **Expected Values:**

- \(E[X]\):

\[ E[X] = \iiint_{0}^{1} \iiint_{0}^{1} \iiint_{0}^{1} 12x^3yz \, dx \, dy \, dz \]

- \(E[Y]\):

\[ E[Y] = \iiint_{0}^{1} \iiint_{0}^{1} \iiint_{0}^{1} 12x^2y^2z \, dx \, dy \, dz \]

- \(E[Z]\):

\[ E[Z] = \iiint_{0}^{1} \iiint_{0}^{1} \iiint_{0}^{1} 12x^2yz^2 \, dx \, dy \, dz \]

Do you want any details on the integration process or have any questions?

Here are five additional questions you might consider asking:
1. How do you perform these integrals step-by-step?
2. What are the interpretations of the expected values in probability?
3. How would the setup change if the limits of integration were different?
4. Can you explain how the region \(2x + y \leq 1\) is determined geometrically?
5. How can these integrals be evaluated using numerical methods?

**Tip:** When dealing with joint density functions, visualizing the region of integration can greatly help in setting up the correct limits.

Explore how to verify a joint density function as a probability density function and set up integrals for related tasks, including calculating probabilities and expected values. Includes detailed step-by-step solutions and explanations.

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