Find the marginal densities of $X$ and $Y$ knowing $h(x,y)$

I have the distribution with joint density $h$ (with respect to the Lebesgue measure): $$h(x,y)=\frac{3}{2}y 1_{A}(x,y), \ \ A=\{(x,y) \in R^2|0<y, x^2+y^2<1\}$$ And then I have to find the marginal densities for X and Y. While we have that $x^2+y^2<1 \Leftrightarrow y^2<1-x^2 \Leftrightarrow y<\sqrt{1-x^2}$ so $0<y<\sqrt{1-x^2}$ therefore I think that:

$$F_X(x)=\int_{0}^{\sqrt{1-x^2}} \frac{3}{2}ydy=[\frac{3}{4}y^2]_{0}^{\sqrt{1-x^2}}=\frac{3}{4}(\sqrt{1-x^2})^2=\frac{3}{4}(1-x^2)$$

Is that a correct method? Somethings says my out from a simulation I have made that this is wrong. Can anyone correct me And how can I find the marginal distribution of $X$? I must have the upper bound $x<\sqrt{1-y^2}$. But how can I find the lower bound?

Solution 1:

The region is as shown below:

Now, what you have found is the marginal density of $X$ and is correct.

i.e $ ~ \displaystyle f_X(x) = \frac{3}{4} (1-x^2), x \in [- 1, 1]$

To find marginal density of $y$, note that $x$ is bound by the unit circle, both on the left and on the right.

$ \displaystyle f_Y(y) = \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} \frac{3y}{2} dx = 3 y \sqrt{1-y^2}, y \in [0, 1]$

Solution 2:

your joint density is defined over half of the unit disk thus

$$h_X(x)=\int_0^{\sqrt{1-x^2}}\frac{3}{2}y dy=\frac{3}{4}(1-x^2)\times\mathbb{1}_{[-1;1]}(x)$$

and similarly for the other marginal

$$h_Y(y)=\frac{3}{2}y \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}dx=3y\sqrt{1-y^2}\times\mathbb{1}_{[0;1]}(y)$$