Method for finding marginal CDF

Question

I wish to calculate the marginal CDF of a joint probability distribution function. However, I am unsure of the bounds I am supposed to use, and wish to verify it. Suppose I have the expression:

$$f_{XY}(x,y)=x^2y$$ $$-1<x<1$$ $$0<y<\sqrt{3}$$

I wish to calculate the marginal CDF. If I wish to do this, I can apply the relationship:

$$F_X(x) = \lim_{y \to \infty}F_{XY}(x,y)$$

And I calculate the cumulative CDF as follows:

$$F_{XY}(x,y) = \int_0^y \int_{-1}^xu^2v\,du\,dv = \frac{(x^3+1)y^2}{6}$$

I believe these are the correct bounds because $f_{xy}$ is zero with respect to X and Y when x<-1 or y<0. Thus the lower bounds must be -1 and 0. The upper bounds appear proper as well, not the least reason for which is that if I let x=1 and $y=\sqrt{3}$, I end up with F = 1. If I now wish to calculate the marginal CDF with respect to X, I would do:

$$F_X(x) = \lim_{y \to \infty}F_{XY}(x,\infty)$$

But the simply substitution yields an unbound answer, which cannot be the case, since probabilities range from 0 to 1. So my questions are (1) can I calculate the marginal CDF directly from the PDF or marginal PDF and (2) should I use the bound $\sqrt{3}$ when calculating the marginal CDF using the relationship directly above? I believe I should use $\sqrt{3}$, since if I consider the pdf, it is zero when $y>\sqrt{3}$, and since the piecewise nature of the pdf function implies that the probability of Y is 1 when $y>\sqrt{3}$.

Answer 1

First, yes, you can calculate the marginal CDF from the joint PDF:

$$P\left(X \leq x\right) = \int^{x}_{-1} \int^{\sqrt{3}}_0 f_{XY}\left(x,y\right) dy\,dx.$$

Second,

$$\lim_{y\rightarrow\infty} F_{XY}\left(x,y\right) = F\left(x,\sqrt{3}\right)$$

for the reasons that you describe.

Answer 2

Close. You should include the supports and beyond in you CDF. Because the limits towards infinity shall tend up towards $1$, as you noted, so this should be indicated.

$$\begin{align}\mathsf F_{\small X,Y}(x,y)&=\dfrac{\small((x^3+1)\mathbf 1_{-1\leqslant x<1}+2\cdotp\mathbf 1_{1\leqslant x})(y^2\mathbf 1_{0\leqslant y<\surd 3}+3\cdotp\mathbf 1_{\surd 3\leqslant y})}{6}\\[3ex]&=\begin{cases}0&:& x<-1\text{ or }y<0\\(x^3+1)y^2/6&:&-1\leqslant x<1, 0\leqslant y<\surd 3\\y^2/3&:&1\leqslant x, 0\leqslant y<\surd 3\\(x^3+1)/2&:&-1\leqslant x<1,\surd 3\leqslant y\\1&:& 1\leqslant x, \surd 3\leqslant y\end{cases}\end{align}$$

Which, of course, also shows how to obtain the marginal CDF.