What is the area of the shaded part

2) X and Y are jointly continuous with joint pdf

Nady [450]

From what I gather from your latest comments, the PDF is given to be

$f_{X,Y}(x,y)=\begin{cases}cxy&\text{for }0\le x,y \le1\\0&\text{otherwise}\end{cases}$

and in particular, f(x, y) = cxy over the unit square [0, 1]², meaning for 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1. (As opposed to the unbounded domain, x ≤ 0 *and* y ≤ 1.)

(a) Find c such that f is a proper density function. This would require

$\displaystyle\int_0^1\int_0^1 cxy\,\mathrm dx\,\mathrm dy=c\left(\int_0^1x\,\mathrm dx\right)\left(\int_0^1y\,\mathrm dy\right)=\frac c{2^2}=1\implies \boxed{c=4}$

(b) Get the marginal density of X by integrating the joint density with respect to y :

$f_X(x)=\displaystyle\int_0^1 4xy\,\mathrm dy=(2xy^2)\bigg|_{y=0}^{y=1}=\begin{cases}2x&\text{for }0\le x\le 1\\0&\text{otherwise}\end{cases}$

(c) Get the marginal density of Y by integrating with respect to x instead:

$f_Y(y)=\displaystyle\int_0^14xy\,\mathrm dx=\begin{cases}2y&\text{for }0\le y\le1\\0&\text{otherwise}\end{cases}$

(d) The conditional distribution of X given Y can obtained by dividing the joint density by the marginal density of Y (which follows directly from the definition of conditional probability):

$f_{X\mid Y}(x\mid y)=\dfrac{f_{X,Y}(x,y)}{f_Y(y)}=\begin{cases}2x&\text{for }0\le x\le 1\\0&\text{otherwise}\end{cases}$

(e) From the definition of expectation:

$E[X]=\displaystyle\int_0^1\int_0^1 x\,f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=4\left(\int_0^1x^2\,\mathrm dx\right)\left(\int_0^1y\,\mathrm dy\right)=\boxed{\frac23}$

$E[Y]=\displaystyle\int_0^1\int_0^1 y\,f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=4\left(\int_0^1x\,\mathrm dx\right)\left(\int_0^1y^2\,\mathrm dy\right)=\boxed{\frac23}$

$E[XY]=\displaystyle\int_0^1\int_0^1xy\,f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=4\left(\int_0^1x^2\,\mathrm dx\right)\left(\int_0^1y^2\,\mathrm dy\right)=\boxed{\frac49}$

(f) Note that the density of X | Y in part (d) identical to the marginal density of X found in (b), so yes, X and Y are indeed independent.

The result in (e) agrees with this conclusion, since E[XY] = E[X] E[Y] (but keep in mind that this is a property of independent random variables; equality alone does not imply independence.)

8 0

3 years ago

Make a list of 5 numbers with a mean of 12 and a mode of 15. Explain if you can

mart [117]

The mean is the average value of the numbers given in the series

The mode is the number that is appears more than any other number given in the series

Step-by-step explanation:

Given challenge is make a list of 5 numbers

The list of 5 number is = 8,10,12,15,15

$Mean value = (8+10+12+15+15)/5 =60/5 = 12$

so required mean is 12 has been computed

Mode = 15 ( the number 12 is appeared in two times 8,10,12,15,15)

Final answer

The required numbers are = 8,10,12,15,15

6 0

3 years ago

The manager of a store records the mode of payment of the 90 customers who visited his store today in the table below.

Rainbow [258]

Answer:

Number 4 is correct.

Step-by-step explanation:

7 0

2 years ago

Help me please peoplee

spayn [35]

it's 9 because 58-49=9

So the answer is 9

<h2>Hope this helps :></h2><h2 /><h2>From Cambridge</h2><h2 /><h2>Please give brainliest</h2>

3 0

2 years ago

What fraction makes the equation true? Enter the answer in the boxes. 5/6x =20/24

Darina [25.2K]

Answer:

x=1

Step-by-step explanation:

To find the value of x, we have to move all the real numbers to the other side.

5/6x = 20/24

We have divide each side by 5/6. Once we do that, we get—

x = 4/4

x = 1

6 0

3 years ago