Conditional Distributions and Independence

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Probability Theory › Conditional Distributions and Independence

Questions 1 - 10

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

$\operatorname{f_{X,Y}}{\left (x,y \right )} = \begin{cases} c e^{- 9 x - 11 y} & 0<x<y<\infty \ 0, & \text{Otherwise} \end{cases}$

Determine the value of $\uptext{c}$ .

$c = \frac{220}{3}$

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, , and for $\uptext{y}$ , $0<y<\infty$

Now let's set up the double integral

$\int_{0}^{\infty}\int_{0}^{y} c e^{- 9 x - 11 y}, dx, dy$

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.

$\int_{0}^{\infty}\int_{0}^{y} c e^{- 9 x - 11 y}, dx, dy = 1$

Now evaluate the double integral

$\int_{0}^{\infty} \frac{c}{9} e^{- 11 y} - \frac{c}{9} e^{- 20 y}, dy = 1$

To evaluate this, we need to use the limit definition

$\lim_{b \to \infty}\left(- \frac{c}{99} e^{- 11 y} + \frac{c}{180} e^{- 20 y}\right) \Big|_{ 0 }^{ b }$

$- \lim_{b \to \infty}\left(- \frac{c}{220}\right) + \lim_{b \to \infty}\left(- \frac{c}{99} e^{- 11 b} + \frac{c}{180} e^{- 20 b}\right)$

$\frac{c}{220} = 1$

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

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

$\operatorname{f_{X,Y}}{\left (x,y \right )} = \begin{cases} c e^{- 6 x - 10 y} & 0<x<y<\infty \ 0, & \text{Otherwise} \end{cases}$

Determine the value of $\uptext{c}$ .

$c = \frac{160}{3}$

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, , and for $\uptext{y}$ , $0<y<\infty$

Now let's set up the double integral

$\int_{0}^{\infty}\int_{0}^{y} c e^{- 6 x - 10 y}, dx, dy$

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.

$\int_{0}^{\infty}\int_{0}^{y} c e^{- 6 x - 10 y}, dx, dy = 1$

Now evaluate the double integral

$\int_{0}^{\infty} \frac{c}{6} e^{- 10 y} - \frac{c}{6} e^{- 16 y}, dy = 1$

To evaluate this, we need to use the limit definition

$\lim_{b \to \infty}\left(- \frac{c}{60} e^{- 10 y} + \frac{c}{96} e^{- 16 y}\right) \Big|_{ 0 }^{ b }$

$- \lim_{b \to \infty}\left(- \frac{c}{160}\right) + \lim_{b \to \infty}\left(- \frac{c}{60} e^{- 10 b} + \frac{c}{96} e^{- 16 b}\right)$

$\frac{c}{160} = 1$

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

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

$\operatorname{f_{X,Y}}{\left (x,y \right )} = \begin{cases} c e^{- 10 x - 10 y} & 0<x<y<\infty \ 0, & \text{Otherwise} \end{cases}$

Determine the value of $\uptext{c}$

$c = \frac{200}{3}$

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, , and for $\uptext{y}$ , $0<y<\infty$

Now let's set up the double integral

$\int_{0}^{\infty}\int_{0}^{y} c e^{- 10 x - 10 y}, dx, dy$

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.

$\int_{0}^{\infty}\int_{0}^{y} c e^{- 10 x - 10 y}, dx, dy = 1$

Now evaluate the double integral

$\int_{0}^{\infty} \frac{c}{10} e^{- 10 y} - \frac{c}{10} e^{- 20 y}, dy = 1$

To evaluate this, we need to use the limit definition

$\lim_{b \to \infty}\left(- \frac{c}{100} e^{- 10 y} + \frac{c}{200} e^{- 20 y}\right) \Big|_{ 0 }^{ b }$

$- \lim_{b \to \infty}\left(- \frac{c}{200}\right) + \lim_{b \to \infty}\left(- \frac{c}{100} e^{- 10 b} + \frac{c}{200} e^{- 20 b}\right)$

$\frac{c}{200} = 1$

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

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

$\operatorname{f_{X,Y}}{\left (x,y \right )} = \begin{cases} c e^{- 10 x - 13 y} & 0<x<y<\infty \ 0, & \text{Otherwise} \end{cases}$

Determine the value of $\uptext{c}$

$c = \frac{299}{3}$

$c = \frac{299}{4}$

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, , and for $\uptext{y}$ , $0<y<\infty$

Now let's set up the double integral

$\int_{0}^{\infty}\int_{0}^{y} c e^{- 10 x - 13 y}, dx, dy$

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.

$\int_{0}^{\infty}\int_{0}^{y} c e^{- 10 x - 13 y}, dx, dy = 1$

Now evaluate the double integral

$\int_{0}^{\infty} \frac{c}{10} e^{- 13 y} - \frac{c}{10} e^{- 23 y}, dy = 1$

To evaluate this, we need to use the limit definition

$\lim_{b \to \infty}\left(- \frac{c}{130} e^{- 13 y} + \frac{c}{230} e^{- 23 y}\right) \Big|_{ 0 }^{ b }$

$- \lim_{b \to \infty}\left(- \frac{c}{299}\right) + \lim_{b \to \infty}\left(- \frac{c}{130} e^{- 13 b} + \frac{c}{230} e^{- 23 b}\right)$

$\frac{c}{299} = 1$

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

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

$\operatorname{f_{X,Y}}{\left (x,y \right )} = \begin{cases} c e^{- 7 x - 14 y} & 0<x<y<\infty \ 0, & \text{Otherwise} \end{cases}$

Determine the value of $\uptext{c}$ .

$c = \frac{147}{2}$

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, , and for $\uptext{y}$ , $0<y<\infty$

Now let's set up the double integral

$\int_{0}^{\infty}\int_{0}^{y} c e^{- 7 x - 14 y}, dx, dy$
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.

$\int_{0}^{\infty}\int_{0}^{y} c e^{- 7 x - 14 y}, dx, dy = 1$

Now evaluate the double integral

$\int_{0}^{\infty} \frac{c}{7} e^{- 14 y} - \frac{c}{7} e^{- 21 y}, dy = 1$

To evaluate this, we need to use the limit definition
$\lim_{b \to \infty}\left(- \frac{c}{98} e^{- 14 y} + \frac{c}{147} e^{- 21 y}\right) \Big|_{ 0 }^{ b }$

$- \lim_{b \to \infty}\left(- \frac{c}{294}\right) + \lim_{b \to \infty}\left(- \frac{c}{98} e^{- 14 b} + \frac{c}{147} e^{- 21 b}\right)$

$\frac{c}{294} = 1$

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

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

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

Determine the value of $\uptext{c}$ .

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

$c = \frac{1}{2}$

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, , and for $\uptext{y}$ , $0<y<\infty$

Now let's set up the double integral

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

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.

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

Now evaluate the double integral

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

To evaluate this, we need to use the limit definition

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

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

$\frac{c}{2} = 1$

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

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 $\uptext{c}$

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

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

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, , 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 $\uptext{c}$

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

$\operatorname{f_{X,Y}}{\left (x,y \right )} = \begin{cases} c e^{- 5 x - 6 y} & 0<x<y<\infty \ 0, & \text{Otherwise} \end{cases}$

Determine the value of $\uptext{c}$ .

$c = \frac{33}{2}$

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, , and for $\uptext{y}$ , $0<y<\infty$

Now let's set up the double integral

$\int_{0}^{\infty}\int_{0}^{y} c e^{- 5 x - 6 y}, dx, dy$
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.

$\int_{0}^{\infty}\int_{0}^{y} c e^{- 5 x - 6 y}, dx, dy = 1$

Now evaluate the double integral

$\int_{0}^{\infty} \frac{c}{5} e^{- 6 y} - \frac{c}{5} e^{- 11 y}, dy = 1$
To evaluate this, we need to use the limit definition

$\lim_{b \to \infty}\left(- \frac{c}{30} e^{- 6 y} + \frac{c}{55} e^{- 11 y}\right) \Big|_{ 0 }^{ b }$

$- \lim_{b \to \infty}\left(- \frac{c}{66}\right) + \lim_{b \to \infty}\left(- \frac{c}{30} e^{- 6 b} + \frac{c}{55} e^{- 11 b}\right)$

$\frac{c}{66} = 1$

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

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

$\operatorname{f_{X,Y}}{\left (x,y \right )} = \begin{cases} c e^{- 9 x - 12 y} & 0<x<y<\infty \ 0, & \text{Otherwise} \end{cases}$

Determine the value of $\uptext{c}$

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, , and for $\uptext{y}$ , $0<y<\infty$

Now let's set up the double integral

$\int_{0}^{\infty}\int_{0}^{y} c e^{- 9 x - 12 y}, dx, dy$

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.

$\int_{0}^{\infty}\int_{0}^{y} c e^{- 9 x - 12 y}, dx, dy = 1$

Now evaluate the double integral

$\int_{0}^{\infty} \frac{c}{9} e^{- 12 y} - \frac{c}{9} e^{- 21 y}, dy = 1$

To evaluate this, we need to use the limit definition

$\lim_{b \to \infty}\left(- \frac{c}{108} e^{- 12 y} + \frac{c}{189} e^{- 21 y}\right) \Big|_{ 0 }^{ b }$

$- \lim_{b \to \infty}\left(- \frac{c}{252}\right) + \lim_{b \to \infty}\left(- \frac{c}{108} e^{- 12 b} + \frac{c}{189} e^{- 21 b}\right)$

$\frac{c}{252} = 1$

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

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 $\uptext{c}$

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

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, , 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 $\uptext{c}$

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