I totally agree with the comments, to complete the question but fortunately, I remembered I've encountered a similar problem in the past lessons of our class.

To complete the question:

1. The y- intercept of the linear function
2
2. The y-intercept of the exponential function
1
3. The number of points of intersection.
0
4. The quadrant of the solution (s)
1/2

I do hope these are the missing given in the question.
Listed below are the answers.

1. 0 3*0 =0

2. 1 2^0=1

3. 2 (they are: x = 0.4578223732320551 & x = 3.313178380475635)

4. 1/2

6 0

4 years ago

Let X and Y have the joint density f(x, y) = e −y , for 0 ≤ x ≤ y. (a) Find Cov(X, Y ) and the correlation of X and Y . (b) Find

adoni [48]

a. I assume the following definitions for covariance and correlation:

$\mathrm{Cov}[X,Y]=E[(X-E[X])(Y-E[Y])]=E[XY]-E[X]E[Y]$

$\mathrm{Corr}[X,Y]=\dfrac{\mathrm{Cov}[X,Y]}{\sqrt{\mathrm{Var}[X]\mathrm{Var}[Y]}}$

Recall that

$E[g(X,Y)]=\displaystyle\iint_{\Bbb R^2}g(x,y)f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy$

where $f_{X,Y}$ is the joint density, which allows us to easily compute the necessary expectations (a.k.a. first moments):

$E[XY]=\displaystyle\int_0^\infty\int_0^yxye^{-y}\,\mathrm dx\,\mathrm dy=3$

$E[X]=\displaystyle\int_0^\infty\int_0^yxe^{-y}\,\mathrm dx\,\mathrm dy=1$

$E[Y]=\displaystyle\int_0^\infty\int_0^yye^{-y}\,\mathrm dx=2$

Also, recall that the variance of a random variable $X$ is defined by

$\mathrm{Var}[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2$

We use the previous fact to find the second moments:

$E[X^2]=\displaystyle\int_0^\infty\int_0^yx^2e^{-y}\,\mathrm dx\,\mathrm dy=2$

$E[Y^2]=\displaystyle\int_0^\infty\int_0^yy^2e^{-y}\,\mathrm dx\,\mathrm dy=6$

Then the variances are

$\mathrm{Var}[X]=2-1^2=1$

$\mathrm{Var}[Y]=6-2^2=2$

Putting everything together, we find the covariance to be

$\mathrm{Cov}[X,Y]=3-1\cdot2\implies\boxed{\mathrm{Cov}[X,Y]=1}$

and the correlation to be

$\mathrm{Corr}[X,Y]=\dfrac1{\sqrt{1\cdot2}}\implies\boxed{\mathrm{Corr}[X,Y]=\dfrac1{\sqrt2}}$

b. To find the conditional expectations, first find the conditional densities. Recall that

$f_{X,Y}=f_{X\mid Y}(x\mid y)f_Y(y)=f_{Y\mid X}(y\mid x)f_X(x)$

where $f_{X\mid Y}$ is the conditional density of $X$ given $Y$ , and $f_X$ is the marginal density of $X$ .

The law of total probability gives us a way to obtain the marginal densities:

$f_X(x)=\displaystyle\int_x^\infty e^{-y}\,\mathrm dy=\begin{cases}e^{-x}&\text{for }x\ge0\\0&\text{otherwise}\end{cases}$

$f_Y(y)=\displaystyle\int_0^ye^{-y}\,\mathrm dx=\begin{cases}ye^{-y}&\text{for }y\ge0\\0&\text{otherwise}\end{cases}$

Then it follows that the conditional densities are

$f_{X\mid Y}(x\mid y)=\begin{cases}\frac1y&\text{for }0\le x$

$f_{Y\mid X}(y\mid x)=\begin{cases}e^{x-y}&\text{for }0\le x$

Then the conditional expectations are

$E[X\mid Y=y]=\displaystyle\int_0^y\frac xy\,\mathrm dy\implies\boxed{E[X\mid Y=y]=\frac y2}$

$E[Y\mid X=x]=\displaystyle\int_x^\infty ye^{x-y}\,\mathrm dy\implies\boxed{E[Y\mid X=x]=x+1}$

c. I don't know which theorems are mentioned here, but it's probably safe to assume they are the laws of total expectation (LTE) and variance (LTV), which say

$E[X]=E[E[X\mid Y]]$

$\mathrm{Var}[X]=E[\mathrm{Var}[X\mid Y]]+\mathrm{Var}[E[X\mid Y]]$

We've found that $E[X\mid Y]=\frac Y2$ and $E[Y\mid X]=X+1$ , so that by the LTE,

$E[X]=E[E[X\mid Y]]=E\left[\dfrac Y2\right]\implies E[Y]=2E[X]$

$E[Y]=E[E[Y\mid X]]=E[X+1]\implies E[Y]=E[X]+1$

$\implies2E[X]=E[X]+1\implies\boxed{E[X]=1}$

Next, we have

$\mathrm{Var}[X\mid Y]=E[X^2\mid Y]-E[X\mid Y]^2=\dfrac{Y^2}3-\left(\dfrac Y2\right)^2\implies\mathrm{Var}[X\mid Y]=\dfrac{Y^2}{12}$

where the second moment is computed via

$E[X^2\mid Y=y]=\displaystyle\int_0^y\frac{x^2}y\,\mathrm dx=\frac{y^2}3$

In turn, this gives

$E\left[\dfrac{Y^2}{12}\right]=\displaystyle\int_0^\infty\int_0^y\frac{y^2e^{-y}}{12}\,\mathrm dx\,\mathrm dy\implies E[\mathrm{Var}[X\mid Y]]=\frac12$

$\mathrm{Var}[E[X\mid Y]]=\mathrm{Var}\left[\dfrac Y2\right]=\dfrac{\mathrm{Var}[Y]}4\implies\mathrm{Var}[E[X\mid Y]]=\dfrac12$

$\implies\mathrm{Var}[X]=\dfrac12+\dfrac12\implies\boxed{\mathrm{Var}[X]=1}$

5 0

3 years ago

Gene is playing a game with a bag of marbles. 3 of the marbles are blue, 4 are green, and 7 are yellow. See below for awarded pr

jek_recluse [69]

Answer:

Step-by-step explanation:

Gene is playing a game with a bag of marbles. If 3 of the marbles are blue, 4 are green, and 7 are yellow and awarded prices for the marbles are $2 green $0.5 yellow $4 blue, the expected payout for Gens game is expressed as shown;

If a blue marble costs $4, 3 blue marbles will cost 3*$4 = $12

If a green marble costs $2, 4 green marbles will cost 4*$2 = $8.0

If a yellow marble costs $0.5, 7 yellow marbles will cost 7*$0.5 = $3.5

Total payout for Gene's game will be the equivalent to $12+ $8 + $3.5 = $23.5.

<em>Hence Gene expected cost will be $21.5</em>

5 0

3 years ago