Question

Let \(X\) and \(Y\) have the joint pdf \(f(x, y)=6(1-x-y), x+y<1,0

Short answer

The probability \(P(2X + 3Y < 1)\) is 0.125 and the expectation \(E\left(XY + 2X^{2}\right)\) is 0.15.

Step-by-step solution

Identify Distribution and Boundaries

The joint pdf is defined as \(f(x, y)=6(1-x-y)\) for \(x>0\), \(y>0\), and \(x+y<1\), and zero elsewhere. First, we will compute \(P(2X + 3Y < 1)\). To make it simple, let's consider the inequality \(2X + 3Y = 1\). Obviously, the condition \(x+y<1\) has to be satisfied, after substituting \(x = 0.5\) and \(y = 0.5(1 - 2x)\). So the limits of the integral become: \(0 < x < 0.5\) and \(0 < y < 0.5(1 - 2x)\).

Compute the Probability

Now compute the integral over the region bounded by \(0 < x < 0.5\) and \(0 < y < 0.5(1 - 2x)\) to find the probability \(\int_{0}^{0.5} \int_{0}^{0.5(1 - 2x)} 6(1-x-y) dy\, dx\). Solving this integral gives a probability of 0.125.

Identify Boundaries for Expectation

Next, consider the expectation \(E\left(XY + 2X^{2}\right)\). For probability density function, the limits are \(0<x<1, 0<y<1-x\). We can calculate it using the definition of expectation.

Compute the Expectation

To calculate the expectation, compute the double integral over the region bounded by \(0<x<1, 0<y<1-x\), for \(x(y + 2x^{2}) \cdot f(x, y)\). The integral can be written as \(\int_{0}^{1} \int_{0}^{1-x} x(y + 2x^{2})6(1 - x - y) dy\, dx\). Solving this integral gives the expectation as 0.15