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Hello people I have a small confusion regarding a given problem.

Lets see it a bit.

We have given that:

$f(x,y) =\begin{cases} x y \text{ if } 0 \leq x \leq 2 \text{ and } 0\leq y\leq 1 \text{ , }\\ 0 \text{ else. }\end{cases}$

So, I have to find this probability $P(X/2 < Y < X)$

To begin with, I found $f$, $y$.

$f_x(x)=x/2$ and $f_y(y)= 2y$
So as you can see $f_x$ and $f_y$ are independent as $f_x \cdot f_y = f(x,y)$.

Now, my main problem is that I am not sure that integral limits should I take!

I have done quite a paper work already, and my "best bet" is this:

  • If $x$ belongs to $(0,1)$ then I take integral from $\int_{x/2}^x 2y dy= \frac 3 4 x^2.$ So finally, to calculate this interval, we take integral $\int_0^1 \frac 3 4 x^2\cdot \frac x 2 dx=3/32. \tag{Eq 1}$

  • If $x$ belongs to $ (1,2)$ then, $ \int_0^1 2y dy= 1.$ Basically I should take also another integral from $1$ to $y$, but that one equals to $0$. So finally, to calculate this interval, we take $\int_1^2 x/2 dx = 3/4 \tag{Eq 2}$

To sum up just add Eq1 + Eq2.

As you can see, I cant understand if my second bullet is correct, and why? (Maybe 1st bullet is wrong as well, but at least it makes more sense.

Thanks for your time Gents

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    Many if not most people find their own questions very interesting. I don't think placing this very subjective assessement into the title adds valuable information about the question to it. Also, note that the tags are visible whereever the title is visible, so having "probability" as the only substantial word in the title makes it entirely redundant. Please consider replacing the title by one that more specifically summarizes the question.2012-11-13

2 Answers 2

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In the second bullet, if $x\in (1,2)$, then the integral limit for $y$ should be $(\frac{x}{2},1)$.

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    +1. This is the exact point to draw the OP's attention to to allow them to complete successfully their attempt.2012-11-13
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We can find the answer by integrating the joint density function over a suitable region. You have correctly identified the integration that needs to be done. Presumably you have drawn a picture. (I probably could not solve the problem without drawing a picture.) The region of integration is the triangle with corners $(0,0)$, $(2,1)$, and $(1,1)$.

For the integration the way you did it, integrating first with respect to $x$, yes, things break up naturally into two parts, $x=0$ to $1$ and $x=1$ to $2$. The first part was done correctly.

For the second part, we want $\int_{x=1}^2 \left(\int_{y=x/2}^1 xy\,dy\right)\,dx.$

Another way: We do not need to break up the region! Integrate first with respect to $x$, which goes from $y$ to $2y$. Then integrate with respect to $y$. We get $\int_{y=0}^1 \left(\int_{x=y}^{2y} xy\,dx\right)\,dy.$ The calculations are straightforward.

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    Andre Nicolas, I just saw your answer. Thank you to begin with. I found it as 9/32, and it seems correct. But I have a major question regarding this second part. We put the integration limits from x/2 to 1 , alright I can understand this. But shouldn't we actually put limits from 0 to x/2 , and then from x/2 to 1 , and then from 1 to 2 , although the limits from 1 to 2 would produce zero in this case , since X<2 ? Thank you!2012-11-13