Question

12. Two components of a minicomputer have the following joint pdf for their useful lifetimes X and Y: xe-x(1+y) x > 0 and y = 0 f(x, y) = | 0 otherwise a. What is the probability that the lifetime X of the first component exceeds 3? b. What are the marginal pdf's of X and Y? Are the two lifetimes independent? Explain. c. What is the probability that the lifetime of at least one component exceeds 3?

Two components of a minicomputer have the following

joint pdf for their useful lifetimes X and Y:

f (x, y) 5 5xe2x(11y) x $ 0 and y $ 0

0 otherwise

a. What is the probability that the lifetime X of the first

component exceeds 3?

b. What are the marginal pdf’s of X and Y? Are the two

lifetimes independent? Explain.

c. What is the probability that the lifetime of at least

one component exceeds 3?

Answer 1

Concepts and reason

The concept is based on the joint probability distribution function.

A joint probability is a statistical measure where the occurrences of two events are together. It is also called the intersection of two or more events.

Marginal probability distribution of the random variable gives the distribution of the random variable without considering the occurrence of the other variable.

Two events are said to independent if the occurrence of one event does not affect the occurrence of other event.

Fundamentals

Let X and Y be continuous random variables. Then $f\left( {x,y} \right)$is called the joint probability density function for X and Y which can be defined as:

$\begin{array}{c}\\P\left( {x,y \in A} \right) = P\left( {a \le X \le b,c \le Y \le d} \right)\\\\ = \int\limits_a^b {\int\limits_c^d {f\left( {x,y} \right)} } {\rm{ }}dy{\rm{ }}dx\\\end{array}$

The marginal probability density functions of X and Y can be defined as:

$\begin{array}{l}\\{f_x}\left( x \right) = \int\limits_{ - \infty }^\infty {f\left( {x,y} \right)} {\rm{ }}dy{\rm{ for }} - \infty < x < \infty \\\\{f_y}\left( y \right) = \int\limits_{ - \infty }^\infty {f\left( {x,y} \right)} {\rm{ }}dx{\rm{ for }} - \infty < y < \infty \\\end{array}$

Here ${f_X}\left( x \right)$is the marginal distribution of the random variable of $X$and ${f_Y}\left( y \right)$is the marginal distribution of the random variable of $Y$.

The condition for the two random variables to be independent can be stated as:

$f\left( {x,y} \right) = {f_x}\left( x \right) \times {f_y}\left( y \right){\rm{ for all }}x{\rm{ and }}y$

(a)

Consider two random variables $X{\rm{ and }}Y$which indicates the lifetime of the two components of the minicomputers.

The joint probability density function is provided as:

$f\left( {x,y} \right) = \left\{ \begin{array}{l}\\x{e^{ - x\left( {1 + y} \right){\rm{ }}}}{\rm{ }}x \ge 0{\rm{ and }}y \ge 0\\\\0{\rm{ otherwise}}\\\end{array} \right.{\rm{ }}$

The required probability does not depend on Y values. Hence consider all the possible values of Y and the values of the X ranges from $3{\rm{ to }}\infty$as the first component exceeds 3.

So, the required probability is obtained as:

$\begin{array}{c}\\P\left( {x > 3} \right) = \int\limits_3^\infty {\int\limits_0^\infty {x{e^{ - x\left( {1 + y} \right){\rm{ }}}}} } {\rm{ }}dy{\rm{ }}dx\\\\ = \int\limits_3^\infty {\left( {\left[ { - {e^{ - x\left( {1 + y} \right)}}} \right]_0^\infty } \right)} dx\\\\ = \int\limits_3^\infty {{e^{ - x}}dx} \\\\ = \left[ {\frac{{{e^{ - x}}}}{{ - 1}}} \right]_3^\infty \\\end{array}$

$\begin{array}{c}\\ = 0.049787\\\\ \approx 0.05\\\end{array}$

(b.1)

Substitute the provided probability density function and integrates with respect to y to obtain the marginal probability density function of X.

Hence, the marginal distribution of $X$can be obtained as:

$\begin{array}{c}\\{f_x}\left( x \right) = \int\limits_\infty ^\infty {f\left( {x,y} \right)} {\rm{ }}dy{\rm{ for }} - \infty < x < \infty \\\\ = \int\limits_0^\infty {x{e^{ - x\left( {1 + y} \right){\rm{ }}}}} {\rm{ }}dy\\\\ = \left[ { - {e^{ - x\left( {1 + y} \right)}}} \right]_0^\infty \\\\ = {e^{ - x}}\\\end{array}$

Substitute the given probability density function and integrates with respect to x to obtain the marginal probability density function of Y.

Hence, the marginal distribution of $Y$can be obtained as:

$\begin{array}{c}\\{f_y}\left( y \right) = \int\limits_0^\infty {x{e^{ - x\left( {1 + y} \right){\rm{ }}}}dx} \\\\ = \left[ {\frac{{x{e^{ - x\left( {1 + y} \right)}}}}{{ - \left( {1 + y} \right)}}} \right]_0^\infty - \int\limits_0^\infty {\frac{{{e^{ - x\left( {1 + y} \right)}}}}{{ - \left( {1 + y} \right)}}dx} \\\\ = \left[ { - \frac{{\left( {\left( {y + 1} \right)x + 1} \right){e^{ - yx - x}}}}{{{y^2} + 2y + 1}}} \right]_0^\infty \\\\{\rm{ = }}\frac{1}{{{{\left( {1 + y} \right)}^2}}}{\rm{ }}\\\end{array}$

(b.2)

Substitute the obtained marginal probability density functions and the given joint probability density function in the condition for independence to find whether the two lifetimes are independent.

$\begin{array}{c}\\f\left( {x,y} \right) = {f_x}\left( x \right) \times {f_y}\left( y \right){\rm{ }}\\\\x{e^{ - x\left( {1 + y} \right){\rm{ }}}} \ne \left( {{e^{ - x}}} \right)\left( {\frac{1}{{{{\left( {1 + y} \right)}^2}}}} \right)\\\end{array}$

Since the product of marginal probability density function is not same as the given joint probability density function, the two lifetimes are not independent.

(c)

The probability that the lifetime of at least one component exceeds 3 is given by:

$\begin{array}{c}\\P\left( {X > 3{\rm{ or }}Y > 3} \right) = 1 - P\left( {X,Y \le 3} \right)\\\\ = 1 - \int\limits_0^3 {\int\limits_0^3 {x{e^{ - x\left( {1 + y} \right)}}} } dydx\\\\ = 1 - \int\limits_0^3 {\left( {\left[ { - {e^{ - x\left( {1 + y} \right)}}} \right]_0^3} \right)} dx\\\\ = 1 - \int\limits_0^3 {\left( {{e^{ - 4x}}\left( {{e^{3x}} - 1} \right)} \right)} dx\\\end{array}$

The probability can be simplified as:

$\begin{array}{c}\\P\left( {X > 3{\rm{ or }}Y > 3} \right) = 1 - \left[ {\frac{{{e^{ - 4x}}}}{4} - {e^{ - x}}} \right]_0^3\\\\ = 1 - 0.700\\\\ = 0.3000\\\end{array}$

Ans: Part a

The probability that the lifetime X of the first component exceeds 3 is 0.05.

Part b.1

The required marginal density function for X and Y is obtained as:

$\begin{array}{l}\\{f_x}\left( x \right) = \left\{ \begin{array}{l}\\{e^{ - x}}{\rm{ }}x \ge 0\\\\0{\rm{ otherwise}}\\\end{array} \right.\\\\{f_y}\left( y \right) = \left\{ \begin{array}{l}\\\frac{1}{{{{\left( {1 + y} \right)}^2}}}{\rm{ }}y \ge 0\\\\0{\rm{ otherwise}}\\\end{array} \right.\\\end{array}$

Part b.2

The two random variables represent the lifetimes of the components are not independent.

Part c

The probability that the lifetime of at least one component exceeds 3 is 0.30.

Answer 2

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Answers:

a) 1/e^3

b) No,X and Y lifetimes are not independent

c) ~0.30