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 $f_Y$.

  • $\begingroup$ 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. $\endgroup$ – Joe Mar 20 at 18:23
  • $\begingroup$ @Joe The domain is the real line. The support (where the density is non zero) is an interval. $\endgroup$ – yurnero Mar 20 at 19:24

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