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

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Step-by-step 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.

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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.

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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 - (1)/(3)ax^3\right)\Bigg|_(1)^(2)\\\\\displaystyle V = \left(a(2)^2 - (1)/(3)a(2)^3\right) - \left(a(1)^2 - (1)/(3)a(1)^3\right)\\\\\displaystyle V = \left(4a - (8)/(3)a\right)-\left(a - (1)/(3)a\right)\\\\`

`\displaystyle V = 4a - (8)/(3)a-a + (1)/(3)a\\\\\displaystyle V = 3a - (8)/(3)a + (1)/(3)a\\\\\displaystyle V = (9)/(3)a - (8)/(3)a + (1)/(3)a\\\\\displaystyle V = (9-8+1)/(3)a\\\\\displaystyle V = (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 = (2)/(3)a\\\\\displaystyle (2)/(3)a = 1\\\\\displaystyle a = (3)/(2) = 1.5\\\\`

a = 3/2 is the final answer.