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Flashcards: Conditional Distributions and Independence

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Let $\uptext{X}$ , and $\uptext{Y}$ be the lifespans (in hours) of two electronic devices, and their joint probability mass function is given below.

$\operatorname{f_{X,Y}}{\left (x,y \right )} = \begin{cases} c e^{- x - 2 y} & 0<x<y<\infty \\ 0, & \text{Otherwise} \end{cases}$

Determine the value of $\uptext{c}$

$c = \frac{3}{2}$

c = 2

c = 1

c = 12

c = 6

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Flashcards: Conditional Distributions and Independence

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Let $\uptext{X}$ , and $\uptext{Y}$ be the lifespans (in hours) of two electronic devices, and their joint probability mass function is given below.

$\operatorname{f_{X,Y}}{\left (x,y \right )} = \begin{cases} c e^{- x - 2 y} & 0<x<y<\infty \\ 0, & \text{Otherwise} \end{cases}$

Determine the value of $\uptext{c}$

Your answer:

$c = \frac{3}{2}$

Your answer:

c = 2

Your answer:

c = 1

Your answer:

c = 12

Correct answer:

c = 6

Explanation:

In order to find the value of $\uptext{c}$ , we need to take find the double integral of the function

Let's find what the bounds are for both $\uptext{x}$ , and $\uptext{y}$

We look at the p.d.f to see that the bounds for $\uptext{x}$ are, 0<x<y , and for $\uptext{y}$ , $0<y<\infty$

Now let's set up the double integral

$\int_{0}^{\infty}\int_{0}^{y} c e^{- x - 2 y}\, dx\, dy$

Before we evaluate it, we need to remember to set the double integral equal to one, since we are essentially solving for the c.d.f.

$\int_{0}^{\infty}\int_{0}^{y} c e^{- x - 2 y}\, dx\, dy = 1$

Now evaluate the double integral

$\int_{0}^{\infty} c e^{- 2 y} - c e^{- 3 y}\, dy = 1$

To evaluate this, we need to use the limit definition

$\lim_{b \to \infty}\left(- \frac{c}{2} e^{- 2 y} + \frac{c}{3} e^{- 3 y}\right) \Big|_{ 0 }^{ b }$

$- \lim_{b \to \infty}\left(- \frac{c}{6}\right) + \lim_{b \to \infty}\left(- \frac{c}{2} e^{- 2 b} + \frac{c}{3} e^{- 3 b}\right)$

$\frac{c}{6} = 1$

Now we simply solve for $\uptext{c}$

c = 6

Flashcards: Conditional Distributions and Independence

Home Embed

Let $\uptext{X}$ , and $\uptext{Y}$ be the lifespans (in hours) of two electronic devices, and their joint probability mass function is given below.

$\operatorname{f_{X,Y}}{\left (x,y \right )} = \begin{cases} c e^{- x - 19 y} & 0<x<y<\infty \\ 0, & \text{Otherwise} \end{cases}$

Determine the value of $\uptext{c}$

c = 380

$c = \frac{1}{400}$

$c = \frac{1}{380}$

c = 400

c = 360

Report an Error Copyright Notice

Flashcards: Conditional Distributions and Independence

Home Embed

Let $\uptext{X}$ , and $\uptext{Y}$ be the lifespans (in hours) of two electronic devices, and their joint probability mass function is given below.

$\operatorname{f_{X,Y}}{\left (x,y \right )} = \begin{cases} c e^{- x - 19 y} & 0<x<y<\infty \\ 0, & \text{Otherwise} \end{cases}$

Determine the value of $\uptext{c}$

Your answer:

$c = \frac{1}{400}$

Your answer:

$c = \frac{1}{380}$

Your answer:

c = 400

Your answer:

c = 360

Correct answer:

c = 380

Explanation:

In order to find the value of $\uptext{c}$ , we need to take find the double integral of the function.

Let's find what the bounds are for both $\uptext{x}$ , and $\uptext{y}$ .

We look at the p.d.f to see that the bounds for $\uptext{x}$ are, 0<x<y , and for $\uptext{y}$ , $0<y<\infty$ .

Now let's set up the double integral.

$\int_{0}^{\infty}\int_{0}^{y} c e^{- x - 19 y}\, dx\, dy$

Before we evaluate it, we need to remember to set the double integral equal to one, since we are essentially solving for the c.d.f.

$\int_{0}^{\infty}\int_{0}^{y} c e^{- x - 19 y}\, dx\, dy = 1$

Now evaluate the double integral

$\int_{0}^{\infty} c e^{- 19 y} - c e^{- 20 y}\, dy = 1$
To evaluate this, we need to use the limit definition

$\lim_{b\to \infty}\left(- \frac{c}{19} e^{- 19 y} + \frac{c}{20} e^{- 20 y}\right) \Big|_{ 0 }^{ 19 }$

$- \lim_{b \to \infty}\left(- \frac{c}{380}\right) + \lim_{b\to \infty}\left(- \frac{c}{19 e^{361}} + \frac{c}{20 e^{380}}\right)$

$\frac{c}{380} = 1$

Now we simply solve for $\uptext{c}$

c = 380

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Flashcards: Conditional Distributions and Independence

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