Find PDF of Y from (constant) joint PDF

Question

Given the following joint PDF:

$$ f_{X,Y}(t,s) = \begin{cases} \frac{2}{3} & 0\leq t\leq 1, -1\leq s\leq t \\ 0 & otherwise \end{cases} $$

I need to find $f_y(s)$

So according to defintion:

$f_Y(s)=\int_{-\infty}^{\infty}f_{X,Y}(t,s) dt = \frac{2}{3}\int_{0}^{1} dt = \frac{2}{3}$

Which is TOTALLY wrong since it's not even a valid density (doesn't integrate to 1). I believe I have some component missing here, but I can't see what. I guess it has something with the fact that $s$ is bounded by $t$ in some of the region, but I can't understand what to do with that.

Could anyone please enlighten me? I really really wish to understand what's wrong with my way of thinking here.

Thanks!

Answer 1

The integration is not from $t= 0$ to $t= 1$. For $s< 0$, the integration is from $t= 0$ to $t= 1$ but for $s\ge 0$ the integration is from $t= s$ to $t= 1$.

Answer 2

$$f_Y(s)=\int_{-\infty}^{\infty}f_{X,Y}(t,s) dt = \int_{s}^{1} \frac{2}{3} dt = \frac{2}{3}(1 - s),$$ which is indeed a proper density function for $s > 0$.

Note that for $s \le 0$ we have $f_Y(s)$ as you've calculated, hence

$$f_Y(s) = \begin{cases}\frac{2}{3}(1 - s), s \ge 0 \\ \frac{2}{3}, s < 0 \end{cases}$$