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Free_Kalibri [48]
3 years ago
9

What is 3x-4=18+5x ?

Mathematics
1 answer:
My name is Ann [436]3 years ago
7 0
3x - 4 = 18+5x
-3x             -3x

-4-18 = 18 + 2x
       -18

-22 = 2x

2x/2 = -22/2

x= -22/2

x= -11

Final answer: x = -11
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The joint probability density function of X and Y is given by fX,Y (x, y) = ( 6 7 x 2 + xy 2 if 0 < x < 1, 0 < y < 2
fredd [130]

I'm going to assume the joint density function is

f_{X,Y}(x,y)=\begin{cases}\frac67(x^2+\frac{xy}2\right)&\text{for }0

a. In order for f_{X,Y} to be a proper probability density function, the integral over its support must be 1.

\displaystyle\int_0^2\int_0^1\frac67\left(x^2+\frac{xy}2\right)\,\mathrm dx\,\mathrm dy=\frac67\int_0^2\left(\frac13+\frac y4\right)\,\mathrm dy=1



b. You get the marginal density f_X by integrating the joint density over all possible values of Y:

f_X(x)=\displaystyle\int_0^2f_{X,Y}(x,y)\,\mathrm dy=\boxed{\begin{cases}\frac67(2x^2+x)&\text{for }0

c. We have

P(X>Y)=\displaystyle\int_0^1\int_0^xf_{X,Y}(x,y)\,\mathrm dy\,\mathrm dx=\int_0^1\frac{15}{14}x^3\,\mathrm dx=\boxed{\frac{15}{56}}

d. We have

\displaystyle P\left(X

and by definition of conditional probability,

P\left(Y>\dfrac12\mid X\frac12\text{ and }X

\displaystyle=\dfrac{28}5\int_{1/2}^2\int_0^{1/2}f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=\boxed{\frac{69}{80}}

e. We can find the expectation of X using the marginal distribution found earlier.

E[X]=\displaystyle\int_0^1xf_X(x)\,\mathrm dx=\frac67\int_0^1(2x^2+x)\,\mathrm dx=\boxed{\frac57}

f. This part is cut off, but if you're supposed to find the expectation of Y, there are several ways to do so.

  • Compute the marginal density of Y, then directly compute the expected value.

f_Y(y)=\displaystyle\int_0^1f_{X,Y}(x,y)\,\mathrm dx=\begin{cases}\frac1{14}(4+3y)&\text{for }0

\implies E[Y]=\displaystyle\int_0^2yf_Y(y)\,\mathrm dy=\frac87

  • Compute the conditional density of Y given X=x, then use the law of total expectation.

f_{Y\mid X}(y\mid x)=\dfrac{f_{X,Y}(x,y)}{f_X(x)}=\begin{cases}\frac{2x+y}{4x+2}&\text{for }0

The law of total expectation says

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

We have

E[Y\mid X=x]=\displaystyle\int_0^2yf_{Y\mid X}(y\mid x)\,\mathrm dy=\frac{6x+4}{6x+3}=1+\frac1{6x+3}

\implies E[Y\mid X]=1+\dfrac1{6X+3}

This random variable is undefined only when X=-\frac12 which is outside the support of f_X, so we have

E[Y]=E\left[1+\dfrac1{6X+3}\right]=\displaystyle\int_0^1\left(1+\frac1{6x+3}\right)f_X(x)\,\mathrm dx=\frac87

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Step-by-step explanation:

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Step-by-step explanation:

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