Looking for confirmation on probability questions

I am preparing for an exam and am attempting old problems. However, I have no way of checking my work. I will post it and then my attempts, and I am wondering if anyone could please help to comment on it.

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a) I calculated using double integrals that $k=\frac{1}{7}$

b) I calculated this to be $$f_{x}(X)=\frac{3x+1/2}{7}$$

c) This one I am a bit more confused on,

I know that $$f(Y|X)=\frac{f(x,y)}{f_{x}(X)}$$ for when it is defined.

So now would I just calculate $\int_{0}^{1} y^{2}f(y|x)dy$ ?

In that case, I calculate it to be $\frac{x+1/4}{3x+1/2}=\frac{5}{14}$

(d) I calculated it to be $$\frac{1}{7}(4y^{2}+y)$$

(e) To do this, I would calculate the Covariance, and then calculate there coefficient. ( I am leaving this part for now)

Anyways, any help/suggestions/comments is greatly appreciated. Thank you

In regard to d, when say calculating E(X|Y=y) would it be best to plug in the actual value of y and then evaluate the single integral over the marginal function of X for example?

Solutions Collecting From Web of "Looking for confirmation on probability questions"

a)
$$\int_0^1 \int_0^2 (3x+y) dxdy = 7$$
So your value of $k$ is correct.

b)
$$\int_0^1 \frac{3x+y}{7} dy$$
$$=\left.\frac{6xy+y^2}{14}\right|_0^1$$
$$=\frac{6x+1}{14}$$
And so this is correct too.

c)
$$\int_0^1 y^2\frac{3+y}{7} dy$$
$$\left.\frac{3y^3/3+y^4/4}{7}\right|_0^1$$
$$= \frac{5}{28}$$
Since
$$\int_0^1 \frac{3+y}{7} dy = \frac{1}{2}$$
we adjust the probability distribution and get the answer you already calculated.

d) I’m not sure what you mean with expected values; I assume you read the question incorrectly. It’s asking for the probability that $\frac{x}{y}<2 \implies x<2y$.
$$\int_0^1\int_0^{2y} \frac{3x+y}{7} dx dy = \frac{8}{21}$$

e) In (b), you obtained
$$f_x(x)=\frac{6x+1}{14}$$
To check for independence, calculate $f_y$ using the same methods, and then check if $f(x,y)=f_x(x)f_y(y)$. This is from the definition of statistical independence.