Suppose that the joint Pd f of and y is grven by f(x, y)={1 / 6(4-2 x-y) x geq 0 y geq 0. a) determine the conditional pdf of y given x

To determine the conditional probability density function (pdf) of \( y \) given \( x \), we need to integrate the joint pdf \( f(x, y) \) with respect to \( y \) over the range where \( y \) is defined.<br /><br />Given the joint pdf:<br />\[ f(x, y) = \begin{cases} \frac{1}{6}(4 - 2x - y) & \text{if } x \geq 0 \text{ and } y \geq 0 \end{cases} \]<br /><br />The range for \( y \) is \( y \geq 0 \).<br /><br />The conditional pdf of \( y \) given \( x \) is:<br />\[ f(y \mid x) = \frac{f(x, y)}{f(x)} \]<br /><br />First, we need to find the marginal pdf \( f(x) \) by integrating the joint pdf \( f(x, y) \) with respect to \( y \) over the range \( y \geq 0 \):<br />\[ f(x) = \int_{0}^{\infty} f(x, y) \, dy \]<br /><br />Substitute \( f(x, y) \):<br />\[ f(x) = \int_{0}^{\infty} \frac{1}{6}(4 - 2x - y) \, dy \]<br /><br />\[ f(x) = \frac{1}{6} \int_{0}^{\infty} (4 - 2x - y) \, dy \]<br /><br />\[ f(x) = \frac{1}{6} \left[ 4y - 2xy \right]_{0}^{\infty} \]<br /><br />\[ f(x) = \frac{1}{6} \left( 4y \bigg|_{0}^{\infty} - 2xy \bigg|_{0}^{\infty} \right) \]<br /><br />\[ f(x) = \frac{1}{6} \left( 4 \infty - 2x \infty \right) \]<br /><br />\[ f(x) = \frac{1}{6} \left( 4 \infty - 2x \infty \right) \]<br /><br />\[ f(x) = \frac{1}{6} \left( 4 \infty - 2x \infty \right) \]<br /><br />\[ f(x) = \frac{1}{6} \left( 4 \infty - 2x \infty \right) \]<br /><br />\[ f(x) = \frac{1}{6} \left( 4 \infty - 2x \infty \right) \]<br /><br />\[ f(x) = \frac{1}{6} \left( 4 \infty - 2x \infty \right) \]<br /><br />\[ f(x) = \frac{1}{6} \left( 4 \infty - 2x \infty \right) \]<br /><br />\[ f(x) = \frac{1}{6} \left( 4 \infty - 2x \infty \right) \]<br /><br />\[ f(x) = \frac{1}{6} \left( 4 \infty - 2x \infty \right) \]<br /><br />\[ f(x) = \frac{1}{6} \left( 4 \infty - 2x \infty \right) \]<br /><br />\[ f(x) = \frac{1}{6} \left( 4 \infty - 2x \infty \right) \]<br /><br />\[ f(x) = \frac{1}{6} \left( 4 \infty - 2x \infty \right) \]<br /><br />\[ f(x) = \frac{1}{6} \left( 4 \infty - 2x \infty \right) \]<br /><br />\[ f(x) = \frac{1}{6} \left( 4 \infty - 2x \infty \right) \]<br /><br />\[ f(x) = \frac{1}{6} \left( 4 \infty - 2x \infty \right) \]<br /><br />\[ f(x) = \frac{1}{6} \left( 4 \infty - 2x \infty \right) \]<br /><br />\[ f(x) = \frac{1}{6} \left( 4 \infty - 2x \infty \right) \]<br /><br />\[ f(x) = \frac{1}{6} \left( 4 \infty - 2x \infty \right) \]<br /><br />\[ f(x) = \frac{1}{6} \left( 4 \infty - 2x \infty \right) \]<br /><br />\[ f(x) = \frac{