To find the expected value of each variable $X$, $Y$, and $Z$ given the joint density function, we use the definition of the expected value for continuous random variables. The expected value of a random variable $X$ is given by:

$E[X] = \iiint x f(x, y, z) \, dx \, dy \, dz$

Similarly, the expected values for $Y$ and $Z$ are:

$E[Y] = \iiint y f(x, y, z) \, dx \, dy \, dz$ $E[Z] = \iiint z f(x, y, z) \, dx \, dy \, dz$

Given the joint density function $f(x, y, z)$:

12x^2yz & \text{if } 0 \leq x, y, z \leq 1 \\ 0 & \text{otherwise} \end{cases}$$ ### Expected Value of $$X$$ $$E[X] = \iiint_0^1 x \cdot 12x^2yz \, dx \, dy \, dz$$ Separate the integrals: $$E[X] = 12 \int_0^1 \int_0^1 \int_0^1 x^3 yz \, dx \, dy \, dz$$ $$E[X] = 12 \left( \int_0^1 x^3 \, dx \right) \left( \int_0^1 y \, dy \right) \left( \int_0^1 z \, dz \right)$$ Compute each integral separately: $$\int_0^1 x^3 \, dx = \left[ \frac{x^4}{4} \right]_0^1 = \frac{1}{4}$$ $$\int_0^1 y \, dy = \left[ \frac{y^2}{2} \right]_0^1 = \frac{1}{2}$$ $$\int_0^1 z \, dz = \left[ \frac{z^2}{2} \right]_0^1 = 1$$ Thus, $$E[X] = 12 \cdot \frac{1}{4} \cdot \frac{1}{2} \cdot 1 = \frac{12}{8} = \frac{3}{2} = 0.75$$ ### Expected Value of $$Y$$ $$E[Y] = \iiint_0^1 y \cdot 12x^2yz \, dx \, dy \, dz$$ Separate the integrals: $$E[Y] = 12 \int_0^1 \int_0^1 \int_0^1 x^2 y^2 z \, dx \, dy \, dz$$ $$E[Y] = 12 \left( \int_0^1 x^2 \, dx \right) \left( \int_0^1 y^2 \, dy \right) \left( \int_0^1 z \, dz \right)$$ Compute each integral separately: $$\int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3}$$ $$\int_0^1 y^2 \, dy = \left[ \frac{y^3}{3} \right]_0^1 = \frac{1}{3}$$ $$\int_0^1 z \, dz = \left[ \frac{z^2}{2} \right]_0^1 = 1$$ Thus, $$E[Y] = 12 \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot 1 = \frac{12}{9} = \frac{4}{3} = 0.5$$ ### Expected Value of $$Z$$ $$E[Z] = \iiint_0^1 z \cdot 12x^2yz \, dx \, dy \, dz$$ Separate the integrals: $$E[Z] = 12 \int_0^1 \int_0^1 \int_0^1 x^2 y z^2 \, dx \, dy \, dz$$ $$E[Z] = 12 \left( \int_0^1 x^2 \, dx \right) \left( \int_0^1 y \, dy \right) \left( \int_0^1 z^2 \, dz \right)$$ Compute each integral separately: $$\int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3}$$ $$\int_0^1 y \, dy = \left[ \frac{y^2}{2} \right]_0^1 = \frac{1}{2}$$ $$\int_0^1 z^2 \, dz = \left[ \frac{z^3}{3} \right]_0^1 = \frac{1}{3}$$ Thus, $$E[Z] = 12 \cdot \frac{1}{3} \cdot \frac{1}{2} \cdot \frac{1}{3} = \frac{12}{18} = \frac{2}{3} = 0.67$$ ### Summary The expected values are: - $$E[X] = 0.75$$ - $$E[Y] = 0.5$$ - $$E[Z] = 0.67$$ Do you want details or have any questions? Here are some more questions you might want to explore: 1. What is the variance of each variable $$X$$, $$Y$$, and $$Z$$? 2. What is the covariance between $$X$$ and $$Y$$? 3. What is the probability that $$X + Y + Z \leq 1$$? 4. How do the marginal density functions of $$X$$, $$Y$$, and $$Z$$ look like? 5. Can you find the moment generating function for $$X$$, $$Y$$, and $$Z$$? **Tip:** When calculating expected values, make sure to set up the correct limits of integration and factor the density function correctly for each variable.