MHB Joint probabilty integral boundaries understanding.

[oriel1]

Hi,
I have homework question that I'm trying to solve. But I can't understand the basis.
Here is a picture of the question and what I have done:
View attachment 6986
My question is, How do I set the boundaries for the integral? 1. If I want the whole squart. 2. To sum up areas.

Assume I want to sum the areas, I get difficulties to calculate area "1". By simply calculate integral of (y=x+0.5) from [0 to 0.5] it work, but I need to use double integral, in the picture I did it but the "1/4" seem to be wrong.
Thank you.

 

[Ackbach]

Are you trying to calculate $P(|X+Y|<0.5)$ or $P(|X-Y|<0.5)?$ If the first, then your plot would seem to be incorrect. Here is W|A's plot command: plot |x+y| < 0.5 on [0,2]. You can see that the diagonal goes the other way. This is a tricky integral to set up, because you'll have to break it up into two regions, regardless of the order of integration.

[EDIT] The integral is less tricky than I thought, since you're only interested in the first quadrant. Better command for W|A: plot |x+y| < 0.5, 0<x<2, 0<y<2. It's a triangle. Can you set up that integral?

 

[oriel1]

Are you trying to calculate $P(|X+Y|<0.5)$ or $P(|X-Y|<0.5)?$ If the first, then your plot would seem to be incorrect. Here is W|A's plot command: plot |x+y| < 0.5 on [0,2]. You can see that the diagonal goes the other way. This is a tricky integral to set up, because you'll have to break it up into two regions, regardless of the order of integration.

[EDIT] The integral is less tricky than I thought, since you're only interested in the first quadrant. Better command for W|A: plot |x+y| < 0.5, 0<x<2, 0<y<2. It's a triangle. Can you set up that integral?

Sorry, you right! this |x-y| I confused for a moment.
I'm trying to calculate this one: $P(|X-Y|<0.5)$. Where do I enter that "plot" command? do I need MATLAB or something?

However, I need to find answer. Finding the first quadrant is just part of the process that I think is relevant for solving that problem. Maybe I'm wrong with that way of thinking.

The Solution given is:
$$\int_{0}^{\frac{1}{2}}\int_{0}^{{\frac{1}{2}+x}}\frac{1}{4}dydx$$

I need to get that answer or other answer by myself. But must include nested integrals.
Thank you!

 

[MarkFL]

Sorry, you right! this |x-y| I confused for a moment.
I'm trying to calculate this one: $P(|X-Y|<0.5)$. Where do I enter that "plot" command? do I need MATLAB or something?

W|A refers to Wolfram|Alpha:

plot |x+y| < 0.5 on [0,2]

 

[Ackbach]

Sorry, you right! this |x-y| I confused for a moment.
I'm trying to calculate this one: $P(|X-Y|<0.5)$. Where do I enter that "plot" command? do I need MATLAB or something?

However, I need to find answer. Finding the first quadrant is just part of the process that I think is relevant for solving that problem. Maybe I'm wrong with that way of thinking.

The Solution given is:
$$\int_{0}^{\frac{1}{2}}\int_{0}^{{\frac{1}{2}+x}}\frac{1}{4}\, dy\, dx$$

I need to get that answer or other answer by myself. But must include nested integrals.
Thank you!

I think, then, that you will have to break up the integral into three pieces, because the rule of association changes as you go. The problem is that you don't want to pick up area outside the square $0<x<2, 0<y<2$. If you just integrate from $0$ to $2$, you'll get the wrong value. So here's the necessary setup for the first and second regions. See if you can come up with what you need for the third region.

$$P(|X-Y|<0.5) = \underbrace{\int_{0}^{0.5} \int_{0}^{x+0.5}\frac14 \, dy \, dx}_{\text{first region}} + \underbrace{\int_{0.5}^{1.5} \int_{x-0.5}^{x+0.5}\frac14 \, dy \, dx}_{\text{second region}}
+\text{third region integral}. $$
Does that help? What's the integral for the third region?

 

[oriel1]

I think, then, that you will have to break up the integral into three pieces, because the rule of association changes as you go. The problem is that you don't want to pick up area outside the square $0<x<2, 0<y<2$. If you just integrate from $0$ to $2$, you'll get the wrong value. So here's the necessary setup for the first and second regions. See if you can come up with what you need for the third region.

$$P(|X-Y|<0.5) = \underbrace{\int_{0}^{0.5} \int_{0}^{x+0.5}\frac14 \, dy \, dx}_{\text{first region}} + \underbrace{\int_{0.5}^{1.5} \int_{x-0.5}^{x+0.5}\frac14 \, dy \, dx}_{\text{second region}}
+\text{third region integral}. $$
Does that help? What's the integral for the third region?

That what I did at first, and I got wrong answer. But I found calculation mistake, now it work and give correct answer. The third part need to be:
$$\int_{1.5}^{2}\int_{x-0.5}^{2}f(x,y)dydx$$

Ok, I understand how to set that specific boundaries. But why in the first "1" quadrant area when I calculate it with geometrics methods I get 3/8 but in the integral method it 3/32? even they both correct when sum-up the parts.

Another question, what if asked for P(|X-Y|=k) then k suppose to mobe between zero to four. for that example assuime k=0.5 How then I set the graph?

Thank you very much!

 

[Ackbach]

That what I did at first, and I got wrong answer. But I found calculation mistake, now it work and give correct answer. The third part need to be:
$$\int_{1.5}^{2}\int_{x-0.5}^{2}f(x,y)\, dy\,dx$$

Yep! Hint: I like to use a small spacer: \, in the $\LaTeX$ code to get spaces after $f(x,y)$ and in-between the differentials. I think it looks better.

Ok, I understand how to set that specific boundaries. But why in the first "1" quadrant area when I calculate it with geometrics methods

Not sure what you mean by "geometric methods". You mean non-calculus methods? If so, can you show your calculations?

I get 3/8 but in the integral method it 3/32? even they both correct when sum-up the parts.

Another question, what if asked for P(|X-Y|=k)

I could be wrong, but I think this shape in the $X-Y$ plane is either one or two straight lines, with zero area. Therefore, I would say the probability, with no calculations necessary, is zero.

then k suppose to mobe between zero to four. for that example assuime k=0.5 How then I set the graph?

Thank you very much!