# Calculate $f_X(x)$ and $f_Y(y)$ given a pair $(X,Y)$ of continuous random variables with a joint PDF of…

Calculate $$f_X(x)$$ and $$f_Y(y)$$ given a pair (X,Y) of continuous random variables with a joint PDF of:

$$f(x,y)=$$ $$\begin{cases} 3 & 0\leq x \leq 1 & 0\leq y \leq x^2 \\ 0 & \text{otherwise} \end{cases}$$

This problem was given to me as a review for an upcoming exam.

My current workings:

$$f_X(x) = \int_{-\infty}^{\infty} f(x,y) dy$$

$$f_Y(y) = \int_{-\infty}^{\infty} f(x,y) dx$$

I'm not exactly sure how to use the f(x,y) in the integral. For $$f_X(x)$$ do I plug in $$x^2$$ into the integral and 1 for $$f_Y(y)$$? If someone can point in the correct direction on what to integrate I should be able to continue from there.

Updated attempt:

$$f_X(x) = \int_{0}^{x^2} 3 dy = 3x^2$$

$$f_Y(y) = \int_{\sqrt{y}}^{1} 3 dx = 3-3\sqrt{y}$$

$$f_X(x) =$$ $$\begin{cases} 3x^2 & 0\leq x \leq 1 \\ 0 & \text{otherwise} \end{cases}$$

$$f_Y(y) =$$ $$\begin{cases} 3-3\sqrt{y} & 0\leq y \leq 1 \\ 0 & \text{otherwise} \end{cases}$$

according the the definition of $$f_{XY}(x,y)$$ we obtain $$f_X(x)=\int_{y=0}^{y=x^2}f_{XY}(x,y)dy$$and$$f_Y(y)=\int_{x=\sqrt y}^{x=1}f_{XY}(x,y)dy$$
For $$x<0$$ or $$x>1$$, it's clear that $$f_X=0$$ because for such $$x$$, $$f(x,y)=0$$ for all $$y$$. For $$0\leq x\leq 1$$, we have: $$f_X(x)=\int_{-\infty}^\infty f(x,y)dy=\int_{0}^{x^2} 3dy=3x^2.$$ You can do something similar for $$f_Y$$.
• Once I solve the integrals, is the domain for each of the functions basically given? So $0 \leq x\leq 1$ and $0 \leq y\leq 1$ (Because x can only be from 0 to 1, so the maximum value for y would be $1^2=1$) I've updated my attempt to demonstrate my comment. – Joe Mar 20 at 18:23