How to find the probability density function by the joint distribution?

Question

$f$ is a jointly continuous distribution function $f(x,y)=Cx^2 y^2$ where $0<y<1$,$y^3<x<y^{1/3}$

C is a constant.

Find $f_X (x)$ and $f_{(X|Y)} (x│y)$

This is not a homework but a practice question.

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My work：

$f_X(x)=\int_\infty f(x,y)dy=\int_0^1Cx^2y^2dy=\frac{1}{3}Cx^2$

To find $C$, we assume $\int_0^1\int_{y^3}^{y^{1/3}}f(x,y)dxdy=1$

$C=18$?

Answer 1

You have to write $f$ properly. You can write :

$$ f(x,y) = Cx^2y^2 \mathbb I_{\{0 \leqslant y \leqslant 1 \}} \mathbb I_{\{y^3 \leqslant x \leqslant y^{1/3}\}} $$

Therefore :

$$ f_X(x) = \int_{\mathbb R} f(x,y) dy = \int_{\mathbb R} Cx^2y^2 \mathbb I_{\{0 \leqslant y \leqslant 1 \}} \mathbb I_{\{y^3 \leqslant x \leqslant y^{1/3}\}} dy $$

Which leads to :

$$ f_X(x) = Cx^2 \mathbb I_{\{0 \leqslant x \leqslant 1 \}} \int_0^1 y^2 \mathbb I_{\{y^3 \leqslant x \leqslant y^{1/3}\}} dy $$

Also $ y^3 \leqslant x \leqslant y^{1/3} \Leftrightarrow x^3 \leqslant y \leqslant x^{1/3}$. Therefore :

$$ f_X(x) = Cx^2 \int_{x^3}^{x^{1/3}} y^2 dy = Cx^2 \left[ \frac{y^3}{3}\right]_{x^3}^{x^{1/3}} = C\frac{x^2}{3}(x - x^6) $$

To find $f_{X|Y}$ I suggest you do the same for $f_Y$ and use Bayes formula.