Probability: density functions: f(x,y) = |x| if (x,y) is in R; 0 otherwise

[ViGa]

Let the following function of two variables be:
$$f(x,y)=\left\lbrace\begin{array}{l} |x| \text{ si } (x,y)\in R \\ 0 \text{ otherwise} \end{array}\right.$$
where R is the hatched area of Figure 2 (that is, the circular crown formed by the two circles with center (0,0) and radii a and 1, respectively).\\
a)-What must be the value of a for f(x,y) to be a function of the density of a continuous random variable (X,Y)?\\
b)- Calculate P(Y$\geq$ X|X$\geq$0).\\
c)- Find the marginal density functions $f_x(x)$ y $f_y(y)$.\\
d)- Using the results obtained in the previous section, reason if X and Y are independent random variables.

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I want to know if my solutions for parts a) and b) are correct, and how do I do the c) and d)?
a)
$$\int_{R}f(x,y)dxdy=1\Longrightarrow \int_R|x|dxdy=\int_0^{2\pi}\int_a^1r|r \text{cos}\theta|drd\theta=$$=4\int_{0}^{\pi/2}\text{cos}\theta d\theta\int_a^1r^2 dr=4\left[\frac{r^3}{3}\right]_a^1=\frac{4}{3}\left(1-a^3\right)=1\Longrightarrow \boxed{a=\left(\frac{1}{4}\right)^{1/3}}
b) If I do it in polar coordinates, I'm assuming I have to work in that sector:

$$P(Y\geq X|X\geq 0)=\int\int|x|dxdy=\int_{\pi/4}^{\pi/2}\int_a^1r^2 \text{cos}\theta drd\theta=\frac{1}{2}\left[\frac{r^3}{3}\right]^1_a=\frac{1}{6}\left(1-\frac{1}{4}\right)=\frac{3}{24}=\frac{1}{8}$$

 

[blamocur]

You solution for a) looks correct. As for b), I agree with your picture, but your calculation of the integral seems incorrect.

 

[ViGa]

You solution for a) looks correct. As for b), I agree with your picture, but your calculation of the integral seems incorrect.

Yeah, sorry, I misscalculated a few things.
Does this look correct now?

$$P(Y\geq X|X\geq 0)=\frac{P(Y\geq X)}{P(X\geq 0)}=\frac{\int_{\pi/4}^{\pi/2}\int_a^1r^2 \text{cos}\theta drd\theta}{1/2}=2\int_{\pi/4}^{\pi/2}\text{cos}\theta d\theta\int_a^1r^2 dr=2\left[1-\frac{\sqrt{2}}{2}\right]\left[\frac{r^3}{3}\right]^1_a= 2\left(1-\frac{\sqrt{2}}{2}\right)\left(\frac{1}{3}-\frac{1}{12}\right)=0.146\Longrightarrow \boxed{P(Y\geq X|X\geq 0)=0.146}$$
Also, how can I do part c) and d)?

 

[blamocur]

Yeah, sorry, I misscalculated a few things.
Does this look correct now?

$$P(Y\geq X|X\geq 0)=\frac{P(Y\geq X)}{P(X\geq 0)}=\frac{\int_{\pi/4}^{\pi/2}\int_a^1r^2 \text{cos}\theta drd\theta}{1/2}=2\int_{\pi/4}^{\pi/2}\text{cos}\theta d\theta\int_a^1r^2 dr=2\left[1-\frac{\sqrt{2}}{2}\right]\left[\frac{r^3}{3}\right]^1_a= 2\left(1-\frac{\sqrt{2}}{2}\right)\left(\frac{1}{3}-\frac{1}{12}\right)=0.146\Longrightarrow \boxed{P(Y\geq X|X\geq 0)=0.146}$$
Also, how can I do part c) and d)?

Looks good to me.

As for c) -- cannot you simply use the definition, for example from Wikipedia ?