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

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Explanation:

PDF = probability density function

The given joint PDF is

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

Let's focus on the [tex]1 \le x \le y \le 2[/tex]. Specifically the x term for now. Erasing out the y term, we have the inequality [tex]1 \le x \le 2[/tex] 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 [tex]1.5 \le y \le 2[/tex]. As another example, x = 1.7  leads to [tex]1.7 \le y \le 2[/tex]. In general, we would say [tex]x \le y \le 2[/tex] 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.

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

[tex]\displaystyle V = \int_{1}^{2}\int_{x}^{2}f(x,y)dydx\\\\[/tex]

The outer integral handles the x terms that range from 1 to 2, describing [tex]1 \le x \le 2[/tex]. 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 [tex]x \le y \le 2[/tex]

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.

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Let's simplify the double integral.

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

[tex]\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\\\\[/tex]

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

[tex]\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)\\\\[/tex]

[tex]\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\\\\[/tex]

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

Recall that V = 1. So,

[tex]\displaystyle V = \frac{2}{3}a\\\\\displaystyle \frac{2}{3}a = 1\\\\\displaystyle a = \frac{3}{2} = 1.5\\\\[/tex]

a = 3/2 is the final answer.

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Name:
Unit 4: Congruent Triangles
Date:
Per: :
Homework 1: Classifying Triangles
H
** This is a 2-page document!
Directions: Classify each triangle by its angles and sides.
1.
2.
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15.6 m
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2011
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60
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8.
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20
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