# Finding the CDF given marginal PDF's; setting bounds

In this question, I'm having a hard time understanding how specifically to set the bounds for the CDF.

Let $$X$$ and $$Y$$ be independent variables. Find the CDF of $$W=Y/X$$ using the marginal PDFs

$$f_X(x)=2x, f_Y(y)=2y$$ where $$0 \le x,y \le 1$$.

To do this, need to consider two different cases where $$0 \le w \le 1$$ and $$w > 1$$.

In the case that $$0 \le w \le 1$$, the bounds would be $$0$$ and $$1$$:

$$f_W(w)=\int_0^1 x(f_X(x))(f_Y(wx)dx$$

However in the case that $$w > 1$$, the bounds are $$0$$ and $$1/w$$, as shown here:

$$f_W(w)=\int_0^{1/w} x(f_X(x))(f_Y(wx)dx$$

Why is this the case? I'm having a hard time understanding how the bounds are set, as I wouldn't have chosen these.

What if the boundaries for the for the marginal PDFs had changed, like to $$0 \le x \le 2$$, or maybe $$0 \le y \le 2$$; how would this affect the bounds for $$f_W(w)$$?

Thank you!

• please add the self-study tag. – Xi'an Mar 6 at 6:16
• Just added the tag! – Sarina Mar 7 at 2:44

When considering the change of variable from $$(x,y)$$ to $$(x,w)$$, the constraint $$0\le x\le 1\qquad 0\le y\le 1$$becomes$$0\le x\le 1\qquad 0\le xw\le 1$$which can be rewritten as$$0\le x\le 1\qquad 0\le w\le 1/x$$and as$$0\le w\le \infty\qquad 0\le x\le \min\{1,1/\omega\}$$