I have the following pdf
$f_{XY}(x,y) = \left \{ \begin{matrix} 6\cdot y & \mbox{ if } 0 \leq y \leq x \leq 1 \\ 0 & \mbox{ otherwise } \end{matrix}\right.$
I want to obtain $F_X(x)$
I have done it as
$F_X(x):\int_0^x\int_0^t6sdsdt=\int_0^x3t^2dt=x^3$, $F_{X}(x) = \left \{ \begin{matrix} 0 & \mbox{ if } 0 \geq x \\ x^3 & \mbox{ if } 0 \leq x \leq 1 \\ 1 & \mbox{ otherwise } \end{matrix}\right.$
I was wondering if this result could be obtained by first calculating $F_{XY}(x,y)$ and then taking the limit $\lim_{y \to \infty}$. However, the limits of integration of $F_{XY}(x,y)$ confuse me. Can someone help me?
I have tried to integrate as follows: $\int_0^x\int_0^y6sdsdt=3y^2x$, which would be $3x^3$ if $y \geq x$, so I don't know what to do with that 3.
I think the best way to resolve such confusions is to write density explicitly under the integral.
We are indeed interested in $\int_0^x\, ds \int_0^y\, dt\, f(s, t)$. Assuming $0 \leq x, y \leq 1$, this is equal to $\int_0^x\, ds \int_0^{\min(s, y)}\, dt \, 6t$.
If $y > x$, the later is equal to $\int_0^x\, ds\, 3s^2 = x^3$. This is $F_X$.
If $y < x$, this splits into two parts: $\int_0^y\, ds\int_0^s\, dt\, 6t$ and $\int_y^x\, ds \int_0^y\, dt 6t$. The former is $y^3$, the later is $3y^2(x - y)$. Sanity check: $x = y$ gives the same $x^3$ as above.
To fully specify CDF, we need to include cases $y > 1 \geq x$ (the same as just $1 > y > x$) and $x > 1$ (again splits into two: $y < 1$ or $y \geq 1$), but it's similar to the previous work.