Question

Problem 4 [20 points]. The proportion of contaminant X and Y found among the con- taminants of a soil sample have the following probability density function: f(x, y)一亻0, otherwise a) (3 points) Sketch out the domain of the density function. Label all axes and boundaries b) (3 points) Using your sketch in (a), carefully set up the limits of your integration to find the value of c that would make this a valid probability density function

Answer 1

Solution:

The joint PDF is $f\left ( x,y \right )=cxy^2;0\leqslant x+y\leqslant 1$ .

a)The support of the PDF is shown shaded below.

x+y=1

b) The condition for PDF is

c) The marginal PDFs are

Since $f\left ( x,y \right )\neq f_X\left ( x \right )f_Y\left ( y \right )$ , are not independent.

d) The conditional PDF,

When $x=0.5$,

The conditional probability,

$P\left ( Y\geqslant 0.25|X=0.5 \right )=\int_{0.25}^{0.5}f_{Y|X=0.5}\left ( y|x \right )dy\\ P\left ( Y\geqslant 0.25|X=0.5 \right )=\int_{0.25}^{0.5}\frac{3y^2}{\left ( 1-0.5 \right )^3}dy\\ P\left ( Y\geqslant 0.25|X=0.5 \right )=\frac{\left [ y^3 \right ]_{0.25}^{0.5}}{\left ( 1-0.5 \right )^3}\\ P\left ( Y\geqslant 0.25|X=0.5 \right )=\frac{0.125-0.25^3}{0.125}\\ {\color{Blue} P\left ( Y\geqslant 0.25|X=0.5 \right )=0.875}$