Question

A random variables X and Y are distributed according to the joint PDF. The value of constant a = _______.f(x,y)=ax if 1<=1x<=y<=2 and 0 otherwise

Answer 1

Answer:    3/2

======================================================

Explanation:

PDF = probability density function

The given joint PDF is

$f(x,y) = \begin{cases}ax \ \ \ \text{ if } 1 \le x \le y \le 2\\0 \ \ \ \ \ \text{ otherwise}\end{cases}$

Let's focus on the $1 \le x \le y \le 2$. Specifically the x term for now. Erasing out the y term, we have the inequality $1 \le x \le 2$ which says x is between 1 and 2, inclusive. We have almost the same story for y, but there's another condition attached to it: y must also be equal to or larger than x.

So let's say x = 1.5. This would mean $1.5 \le y \le 2$. As another example, x = 1.7  leads to $1.7 \le y \le 2$. In general, we would say $x \le y \le 2$ where x is between 1 and 2.

As x gets bigger, the range of possible y values gets smaller. If x = 2, then y has no choice but to be 2 as well.

-----------------

Based on that, we'll have a double integral that looks like this:

$\displaystyle V = \int_{1}^{2}\int_{x}^{2}f(x,y)dydx$

The outer integral handles the x terms that range from 1 to 2, describing $1 \le x \le 2$. Note the dx on the outside. The order of the dy and dx matters.

On the inside, we have the integral for dy ranging from x to 2 to describe the interval $x \le y \le 2$

To have f(x,y) be a PDF, the volume under the f(x,y) surface must be 1, where the volume is based on the bounds set up. So we must have V = 1. We'll use this later.

-----------------

Let's simplify the double integral.

We'll start by computing the inner integral with respect to y.

$\displaystyle V = \int_{1}^{2}\int_{x}^{2}f(x,y)dydx\\\\\displaystyle V = \int_{1}^{2}\int_{x}^{2}\left(ax\right)dydx\\\\\displaystyle V = \int_{1}^{2}\left(axy\Bigg|_{x}^{2}\right)dx\\\\\displaystyle V = \int_{1}^{2}\left(ax(2) - ax(x)\right)dx\\\\\displaystyle V = \int_{1}^{2}\left(2ax - ax^2\right)dx$

Then we'll finish it off by integrating with respect to x.

$\displaystyle V = \int_{1}^{2}\left(2ax - ax^2\right)dx\\\\\displaystyle V = \left(ax^2 - \frac{1}{3}ax^3\right)\Bigg|_{1}^{2}\\\\\displaystyle V = \left(a(2)^2 - \frac{1}{3}a(2)^3\right) - \left(a(1)^2 - \frac{1}{3}a(1)^3\right)\\\\\displaystyle V = \left(4a - \frac{8}{3}a\right)-\left(a - \frac{1}{3}a\right)$

$\displaystyle V = 4a - \frac{8}{3}a-a + \frac{1}{3}a\\\\\displaystyle V = 3a - \frac{8}{3}a + \frac{1}{3}a\\\\\displaystyle V = \frac{9}{3}a - \frac{8}{3}a + \frac{1}{3}a\\\\\displaystyle V = \frac{9-8+1}{3}a\\\\\displaystyle V = \frac{2}{3}a$

Side note: We don't have to worry about the "plus C" integration constant when working with definite integrals.

Recall that V = 1. So,

$\displaystyle V = \frac{2}{3}a\\\\\displaystyle \frac{2}{3}a = 1\\\\\displaystyle a = \frac{3}{2} = 1.5$

a = 3/2 is the final answer.