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 $y>\sqrt{3}$.
$\endgroup$2 Answers
$\begingroup$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.
$\endgroup$ $\begingroup$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.
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