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I have been hunting around on the internet for days to find an example I can easily relate to but without successes. It would really appreciate some help with the following;

Joint pdf;

$f(x,y)=2(x+y)$ for $0<x<1$ and $0<y<x$, $0$ otherwise.

I found the conditional pdf to be;

$f(y|x)=2(x+y)/3x^2$ for $0<x<1$ and $0<y<x$, $0$ otherwise.

  • Question;

I now need to find;

$Pr( X + Y >= 1)$

I believe I need to sub;

$f(y|x)=2(x+y)/3x^2$, into;

$f_Z(z) =\int_{0}^z f_{Y|X}(z-x|x)f_{X}(x)\,\mathrm dx$

Which I get $f_Z(z) = 2z^2$

However I am not sure what the support for $f_Z(z)$ should be? What is the general procedure to figure the support range out for transformations?

In a previous post it was suggested to think of the support range as $0<y<x<1$ but I haven't been able to see past this.

Note this is just a practice exam question.

Tejay Lovelock
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  • Please add the `[self-study]` tag & read its [wiki](http://stats.stackexchange.com/tags/self-study/info). – gung - Reinstate Monica Jun 17 '16 at 12:21
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    Make a drawing showing the event $X+Y\ge 1$ as a region $R$ in the $x,y$-plane, then solve the double integral $\iint_R f(x,y) dy dx$. – Jarle Tufto Jun 17 '16 at 12:53
  • Well thinking about that another way, would this be correct? $1 - PR(X+Y<1) = \int_{0}^{1}\int_{0}^{1-x} 2*(x+y) dydx = \frac{1}{3}$ – Tejay Lovelock Jun 17 '16 at 13:20
  • By deleting your previous version of this question, you have made the comment thread inaccessible to most readers. When you make changes, please just edit your question rather than creating new versions. In the meantime, you will find problems *exactly* like this one solved, with extensive explanations and illustrations, at http://stats.stackexchange.com/questions/180715 and by searching for terms like [pictures of integration](http://stats.stackexchange.com/search?q=integration+picture+votes%3A1+is%3Aanswer). – whuber Jun 17 '16 at 13:52
  • Hi whuber, even with the link you have provided, I am struggling to five a comparable example. The examples I am finding are when both x and y have independent ranges. I've gotten quite frustrated with this problem as you can probably gather. Just really looking forward to a clear cut answer. Thanks... – Tejay Lovelock Jun 17 '16 at 14:28
  • I think the above won't work because I am not taking into account y < x. So for X + Y > 1 and y – Tejay Lovelock Jun 17 '16 at 15:56

1 Answers1

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You have the joint distribution so you only need to integrate over the region desired for the given probability. That region looks like this:

enter image description here

The shaded area fulfills the conditions you are looking for, specifically $$Pr(X+Y\geq1)=\int_{.5}^{1}\int_{1-x}^{x} 2(x+y) dydx \approx .6667$$

Edit:

I am using the image to conceptually understand the boundaries of $x$ & $y$ for the desired probability. This means that the red region contains all of the possible values of $x$ & $y$ as if $0<x<1$, $0<y<1$. Although we want $y<x$ so $y$ can only come arbitrarily close to $x$ (so $x$ is our upper bound for $y$). Since we always want $x+y>1$, then we need $y>1-x$ (our lower bound for $y$). We know that $y<x$, so the only possible values of $x+y>1$ are when $x>.5$ (our lower bound for $x$). Then $x$ can be up to $1$ (our upper bound).

Travis
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  • (1) Could you explain the relevance of shading the entire unit square? (2) I suspect the OP might appreciate an explanation of how you found the integration region, because that seems to be the conceptual block behind the question. – whuber Jun 23 '16 at 14:11
  • Hi Travis thanks for taking the time to answer. I've got a firmer grip on how to go about these questions now. Thank you for your time. – Tejay Lovelock Jun 24 '16 at 06:28