Please help me with this question #8

Suppose X and Y are random variables with joint density function. f(x, y) = 0.1e−(0.5x + 0.2y) if x ≥ 0, y ≥ 0 0 otherwise (a) I

Hatshy [7]

a. $f_{X,Y}$ is a joint density function if its integral over the given support is 1:

$\displaystyle\int_{-\infty}^\infty\int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=\frac1{10}\int_0^\infty\int_0^\infty e^{-x/2-y/5}\,\mathrm dx\,\mathrm dy$

$=\displaystyle\frac1{10}\left(\int_0^\infty e^{-x/2}\,\mathrm dx\right)\left(\int_0^\infty e^{-y/5}\,\mathrm dy\right)=\frac1{10}\cdot2\cdot5=1$

so the answer is yes.

b. We should first find the density of the marginal distribution, $f_Y(y)$ :

$f_Y(y)=\displaystyle\int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm dx=\frac1{10}\int_0^\infty e^{-x/2-y/5}\,\mathrm dy$

$f_Y(y)=\begin{cases}\dfrac15e^{-y/5}&\text{for }y\ge0\\\\0&\text{otherwise}\end{cases}$

Then

$P(Y\ge8)=\displaystyle\int_8^\infty f_Y(y)\,\mathrm dy=e^{-8/5}$

or about 0.2019.

For the other probability, we can use the joint PDF directly:

$P(X\le5,Y\le8)=\displaystyle\int_0^5\int_0^8f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=1+e^{-41/10}-e^{-5/2}-e^{-8/5}$

which is about 0.7326.

c. We already know the PDF for $Y$ , so we just integrate:

$E[Y]=\displaystyle\int_{-\infty}^\infty y\,f_Y(y)\,\mathrm dy=\frac15\int_0^\infty ye^{-y/5}\,\mathrm dy=\boxed5$

5 0

2 years ago

What's the difference between a independent variable and dependent variable

Travka [436]

Answer:

The dependent variable depends on the independent variable.

Step-by-step explanation:

The variable whose values are deliberately changed by the researcher in order to obtain the desired outcome is called independent variable...
The values of the independent variable can be changed as per requirement, by the researcher...
Manipulation can be done in the values of independent variable, but the researcher observes the value of a dependent variable during an experiment.
An independent variable is a presumed cause whereas dependent variable is a measured effect.
In a simple linear regression, ‘y’ denotes dependent variable while ‘x’ denotes independent variable, which means y depends on x

8 0

3 years ago

30 tens-3 tens 3tenths=

katovenus [111]

30 tens=300
3 ten=30
3 tenths=0.3
a number next to another number is multiplying
300-30=270
270 <em />×0.3=81

5 0

3 years ago

Which equations represent the line that is perpendicular to the line 5x − 2y = −6 and passes through the point (5, −4)? Select t

borishaifa [10]

Answer:

<em>Next time, please upload your option.</em>

The line 5x − 2y = −6 <=> 2y = 5x + 6 <=> y = (5/2)x + 3, has the slope a = 5/2

=> Another line (y = a'x + b) that is perpendicular with this line, must has slope a', that satisfies: a' x (5/2) = -1 => a' = -2/5

=> y = (-2/5)x + b

This line passes (5, -4), then we have:

-4 = (-2/5)*5 + b => -4 = -2 + b => b = -2

=> Solution is y = (-2/5)x -2

Hope this helps!

:)

8 0

3 years ago

The graph of a rational function f is shown below.

Mashutka [201]

We have y+2 = 0 and x - 2 = 0. The provided function has an x and y-intercept of -2 and +2, respectively. There is no vertical asymptote. Two is the horizontal asymptote.

<h3>What is a graph?</h3>

A diagram depicting the relationship between two or more variables, each measured along with one of a pair of axes at right angles.

The y-intercept of a function is determined by the intersection of its graph with the y-axis. The value of y on the y-axis at which the considered function crosses it is called the y-intercept.

Assume the following equation: y = f (x)

We have x =0- 2 and y+2 = 0,The x and y intercept of the given function is -2 and +2.

The vertical asymptote is none. The horizontal asymptote is 2.

Hence,we have y+2 = 0 and x - 2 = 0. The provided function has an x and y-intercept of -2 and +2, respectively. There is no vertical asymptote. Two is the horizontal asymptote.

To learn more about the graph, refer to the link;

brainly.com/question/14375099

#SPJ1

4 0

2 years ago