Let x , y and z be random variables having the joint pdf \[ f(x, y, z)=\left\{\begin{array}{r}k\left(z^{2}+x y / 2\right), 0

Ask by Murphy Bird. in Nigeria

Feb 18,2025

To solve the problem, we will follow the steps outlined: ### Step 1: Find the value of \( k \) The joint probability density function (pdf) must integrate to 1 over the entire space. Therefore, we need to calculate the integral of \( f(x, y, z) \) over the specified limits and set it equal to 1. The joint pdf is given by: \[ f(x, y, z) = k\left(z^{2} + \frac{xy}{2}\right), \quad 0 < x < 1, \, 0 < y < 1, \, 0 < z < 1 \] We need to compute the following integral: \[ \int_0^1 \int_0^1 \int_0^1 k\left(z^{2} + \frac{xy}{2}\right) \, dz \, dy \, dx = 1 \] Let's compute the integral step by step. 1. **Integrate with respect to \( z \)**: \[ \int_0^1 \left(z^{2} + \frac{xy}{2}\right) \, dz = \left[\frac{z^{3}}{3} + \frac{xy}{2}z\right]_0^1 = \frac{1}{3} + \frac{xy}{2} \] 2. **Now integrate with respect to \( y \)**: \[ \int_0^1 \left(\frac{1}{3} + \frac{xy}{2}\right) \, dy = \left[\frac{y}{3} + \frac{xy^{2}}{4}\right]_0^1 = \frac{1}{3} + \frac{x}{4} \] 3. **Finally, integrate with respect to \( x \)**: \[ \int_0^1 \left(\frac{1}{3} + \frac{x}{4}\right) \, dx = \left[\frac{x}{3} + \frac{x^{2}}{8}\right]_0^1 = \frac{1}{3} + \frac{1}{8} = \frac{8}{24} + \frac{3}{24} = \frac{11}{24} \] Now, we set the integral equal to 1: \[ k \cdot \frac{11}{24} = 1 \implies k = \frac{24}{11} \] ### Step 2: Find the joint marginal density of \( x \) and \( z \) The joint marginal density of \( x \) and \( z \) is obtained by integrating out \( y \): \[ f_{X,Z}(x, z) = \int_0^1 f(x, y, z) \, dy \] Substituting \( f(x, y, z) \): \[ f_{X,Z}(x, z) = \int_0^1 \frac{24}{11}\left(z^{2} + \frac{xy}{2}\right) \, dy \] Calculating the integral: 1. **Integrate with respect to \( y \)**: \[ \int_0^1 \left(z^{2} + \frac{xy}{2}\right) \, dy = z^{2} + \frac{x}{4} \] Thus, \[ f_{X,Z}(x, z) = \frac{24}{11} \left(z^{2} + \frac{x}{4}\right) \] ### Step 3: Find the marginal density of \( Y \) alone The marginal density of \( Y \) is obtained by integrating out \( x \) and \( z \): \[ f_Y(y) = \int_0^1 \int_0^1 f(x, y, z) \, dx \, dz \] Substituting \( f(x, y, z) \): \[ f_Y(y) = \int_0^1 \int_0^1 \frac{24}{11}\left(z^{2} + \frac{xy}{2}\right) \, dx \, dz \] Calculating the integral: 1. **Integrate with respect to \( x \)**: \[ \int_0^1 \left(z^{2} + \frac{xy}{2}\right) \, dx = z^{2} + \frac{y}{4} \] 2. **Now integrate with respect to \( z \)**: \[ \int_0^1 \left(z^{2} + \frac{y}{4}\right) \, dz = \left[\frac{z^{3}}{3} + \frac{yz}{4}\right]_0^1 = \frac{1}{3} + \frac{y}{4} \] Thus, \[ f_Y(y) = \frac{24}{11} \left(\frac{1}{3} + \frac{y}{4}\right) \] ### Summary of Results (i) The value of \( k \) is \( \frac{24}{11} \). (ii) The joint marginal density of \( x \) and \( z \) is: \[ f_{X,Z}(x, z) = \frac{24}{11} \left(z^{2} + \frac{x}{4}\right) \] (iii) The marginal density of \( Y \) alone is: \[ f_Y(y) = \frac{24}{11} \left(\frac{1}{3} + \frac{y}{4}\right) \]