Chapter 6: Q 6.22 (page 272)

The joint density function of $X$ and $Y$ is
$f (x, y) = \{\begin{cases} x + y 0 < x < 1, 0 < y < 1 \\ 0 otherwise \end{cases}$
(a) Are $X$ and $Y$ independent?
(b) Find the density function of $X$ .
(c) Find $P {X + Y < 1}$ .

Short Answer

Expert verified

The variables X and Y are dependent.
The density function of random variable X is:

$f_{X} (x) = x + \frac{1}{2}; 0 < x < 1$

$c . P (X + Y < 1) = \frac{1}{3}$

Step by step solution

Step 1. Given information:

$f (x, y) = \{\begin{matrix} x + y; 0 < x < 1, 0 < y < 1 \\ 0;otherwise \end{matrix}\}$

Step 2. Prove the independence of X and Y.

In order to test the independence of X and Y, we need to find the marginal distribution function of X and Y. Therefore,

$\begin{matrix} f_{X} (x) = \int_{y} f (x, y) d y \\ = \int_{0}^{1} (x + y) d y \\ = {(x y + \frac{y^{2}}{2})}_{0}^{1} \\ = x + \frac{1}{2} \\ f_{X} (x) = x + \frac{1}{2}; 0 < x < 1 \end{matrix}$

Since the variables $X$ and $Y$ are interchangeable/symmetric, therefore,

$f_{Y} (y) = y + \frac{1}{2}; 0 < y < 1$

Since $f (x, y) \neq f_{X} (x) f_{Y} (y)$

Therefore, $X$ and $Y$ are dependent random variables.

Density function of X

From the above step, we get-

$f_{X} (x) = x + \frac{1}{2}; 0 < x < 1$

Calculation of P(X+Y<1)

$\begin{array}{l} P (X + Y < 1) = \int_{x} \int_{y} f (x, y) d x d y \\ = \int_{x = 0}^{1} \int_{y = 0}^{1 - x} (x + y) d x d y \\ = \int_{0}^{1} {|x y + \frac{y^{2}}{2}|}_{0}^{1 - x} d x \\ = \int_{0}^{1} [x (1 - x) + \frac{{(1 - x)}^{2}}{2}] d x \\ = \int_{0}^{1} (\frac{1 - x^{2}}{2}) d x \\ = {|\frac{3 x - x^{3}}{6}|}_{0}^{1} \\ \Rightarrow P (X + Y < 1) = \frac{1}{3} \end{array}$

which is the required solution.

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The joint density function of $X$ and $Y$ is
$f (x, y) = \{\begin{cases} x + y 0 < x < 1, 0 < y < 1 \\ 0 otherwise \end{cases}$
(a) Are $X$ and $Y$ independent?
(b) Find the density function of $X$ .
(c) Find $P {X + Y < 1}$ .

Short Answer

Step by step solution

Step 1. Given information:

Step 2. Prove the independence of X and Y.

Density function of X

Calculation of P(X+Y<1)

One App. One Place for Learning.

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The joint density function of Xand Yisf(x,y)=x+y 0<x<1,0<y<10 otherwise(a) Are Xand Yindependent?(b) Find the density function of X.(c) FindP{X+Y<1}.

Short Answer

Step by step solution

Step 1. Given information:

Step 2. Prove the independence of X and Y.

Density function of X

Calculation of P(X+Y<1)

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Mechanics Maths

Logic and Functions

Discrete Mathematics

Statistics

Geometry

Decision Maths

Study anywhere. Anytime. Across all devices.

The joint density function of $X$ and $Y$ is
$f (x, y) = \{\begin{cases} x + y 0 < x < 1, 0 < y < 1 \\ 0 otherwise \end{cases}$
(a) Are $X$ and $Y$ independent?
(b) Find the density function of $X$ .
(c) Find $P {X + Y < 1}$ .