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Given two independent variables $X$ and $Y$, with marginal pdfs $f_X(x)=2x, 0 \le x \le 1$ and $f_Y(y)=1, 0 \le y \le 1$, calculate $P(\frac{Y}{X} > 2)$. So this can be written as $P(Y>2X)$,

and can be solved by solving the following integral: $\int_0^1 \int_0^{y/2}f_X(x)f_Y(y)dxdy$

However when switching the $dx$ and $dy$ it seems to me that this could also be solved this way: $\int_0^1 \int_{2x}^{1}f_X(x)f_Y(y)dydx$

But solving the integral, this doesn't give the right answer. Why not? Aren't they both defining the same area?

Thank you so much!

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  • $\begingroup$ The duplicate is the same question with slightly different limits involved. See the first comment to it as well. $\endgroup$
    – whuber ♦
    Commented Mar 22, 2019 at 21:11
  • $\begingroup$ Sorry about that! It seems like I'm having the same kind of questions! $\endgroup$
    – Sarina
    Commented Mar 22, 2019 at 21:36
  • $\begingroup$ It looks like the advice I provided to another poster at stats.stackexchange.com/questions/398993/… might help you out too: when you find you're asking the same questions about the same fundamental concept or technique, you're probably best off consulting a good textbook, because you need to cope with issues that go deeper than a mere difficulty solving one or two problems. $\endgroup$
    – whuber ♦
    Commented Mar 23, 2019 at 13:51
  • $\begingroup$ Thank you! I've just read my textbook so many times over and still having trouble with a certain aspect of it obviously! I'll try looking at some other resources. Thanks! $\endgroup$
    – Sarina
    Commented Mar 24, 2019 at 21:54

1 Answer 1

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If you draw the region of integration, you’ll see that it’s bounded by the lines $y=2x, y=1, x=0$. These intersect at $(0,0),(0,1),(1/2,1)$. So, $x$ is not bigger than $1/2$. Intuitively, if $x>1/2$, how can $y$ be bigger than $2x$, while it is also smaller than $1$? Thus, your first integral has limits $0\rightarrow 1/2$.

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  • $\begingroup$ Wow! Thank you so much! Do you think the best way to figure this out is to draw it out? It seems like that's the only way, right? $\endgroup$
    – Sarina
    Commented Mar 22, 2019 at 20:32
  • $\begingroup$ It’s not the only way, but I think it is the best way :) $\endgroup$
    – gunes
    Commented Mar 22, 2019 at 20:34

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