[strictlyplaid]

How can I be the first post in a thread I didn't start?

[realmeatychunks]

Why isn't my Post there? Strange.

Okay, well here is what I tried to post:


5.12 from Wackerly et al.

Suppose that the random variables X and Y have joint probability density function f(x,y) given by

f(x,y) = 6(x^2)y for 0≤x≤y, x+y≤2
0 elsewhere.

a) Verify that this is a valid joint density function.

This means that f(x,y) is greater than 0 for all x,y which is trivial. The second part, is that the double integral of f(x,y) over all space is equal to 1. I'm having trouble setting the limits for the integration, at first I tried setting the limits for the x integration from 0 to y and for y from 0 to 2, which yielded the wrong result (64/5). I've tried several others and cannot get the integral equal 1. Does anyone have any insight as to what the limits of integration should be and why?

Thanks in advance,
Greg

(edited to fix the function formating [thanks hyperb0le] and to make my post a little clearer)

[11011001]

edit: neat

[hyperb0le]

Is the function supposed to be f(x,y) = 6x^(2y) or f(x,y) = 6x^2 * y?

EDIT: Judging by your result (64/5), I assume that it is f(x,y) = 6x^2 * y. My initial instinct is to set the limits on the y integration from 0 to (2-x) and set the limits for x from 0 to 2 (x and y axis are bounds for the region, as is y=2-x (or x+y=2), and so is the vertical line x=2 where y=2-x intersects the x-axis) and evaluate the y integral first. However, that results in 16/5... I may be missing something, as I am very tired, but those seem like the correct limits. Is there any possiblity that the function is not a valid joint density function?

[realmeatychunks]

Ah yes, sorry about the 'format' of the function, you are correct, it is 6(x^2)y. I had written it as this, but did not correct it when i reposted it. I suppose it is possible that it is not a valid distribution function, but the next part of the question asks for the probability that X + Y < 1.

This makes me think it is more probable (haha) that it is indeed a valid distribution function and there's just something I'm not getting.

[strictlyplaid]

Originally Posted by realmeatychunks

Why isn't my Post there? Strange.

Okay, well here is what I tried to post:


5.12 from Wackerly et al.

Suppose that the random variables X and Y have joint probability density function f(x,y) given by

f(x,y) = 6(x^2)y for 0≤x≤y, x+y≤2
0 elsewhere.

a) Verify that this is a valid joint density function.

This means that f(x,y) is greater than 0 for all x,y which is trivial. The second part, is that the double integral of f(x,y) over all space is equal to 1. I'm having trouble setting the limits for the integration, at first I tried setting the limits for the x integration from 0 to y and for y from 0 to 2, which yielded the wrong result (64/5). I've tried several others and cannot get the integral equal 1. Does anyone have any insight as to what the limits of integration should be and why?

Thanks in advance,
Greg

(edited to fix the function formating [thanks hyperb0le] and to make my post a little clearer)

My double integral skills are a little weak, but if I'm picturing this correctly, you're integrating over a triangle-shaped area in Real(2) space. You set the bounds of y-integration from x to 2-x, and the bounds of x integration from 0 to 1 (the point at which these two lines intersect). Essentially, you're adding up a series of slices of a triangle over the desired area; plotting this out makes it clear what you're doing. Maple tells me that this integral = 1, the desired answer.

Hope that helps!

[ghporter]

Originally Posted by strictlyplaid

How can I be the first post in a thread I didn't start?

Stupid database tricks, of course! I'm sorry to say that while the version upgrade has been sorted out, the server is still an issue.

I was going to make this a probability joke (quantum states and all that) but it just seemed too off topic.

[realmeatychunks]

Thank you for the explanation strictlyplaid. I think you are correct.

Your help is much appreciated.

Greg

[tie]

Keep the thread alive! What if we have a distribution on AxB such that p(a,b) = Theta(p(a)p(b)). Do there exist distributions q_s and q_s' (parametrized by s) and t(s), such that p(a,b) = integral t(s) q_s(a) q_s'(b) such that q_s(a) = Theta(p(a)), q_s'(b) = Theta(p(b)) and of course integral t(s) = 1?

[strictlyplaid]

Originally Posted by realmeatychunks

Thank you for the explanation strictlyplaid. I think you are correct.

Your help is much appreciated.

Greg

No problem!

[11011001]

Originally Posted by realmeatychunks

Why isn't my Post there? Strange.
5.12 from Wackerly et al.

I'm going to guess that you might be from Calgary?

[realmeatychunks]

Nope, I'm not from Calgary, what gave you that impression?

I go to university in Montréal, but am originally from New Hampshire.

[11011001]

Originally Posted by realmeatychunks

Nope, I'm not from Calgary, what gave you that impression?

I go to university in Montréal, but am originally from New Hampshire.

I have the same textbook, though I realize that many different universities use it, there was that slim chance. At least you are in Canada. Booya!

[chabig]

Originally Posted by strictlyplaid

My double integral skills are a little weak, but if I'm picturing this correctly, you're integrating over a triangle-shaped area in Real(2) space. You set the bounds of y-integration from x to 2-x, and the bounds of x integration from 0 to 1 (the point at which these two lines intersect). Essentially, you're adding up a series of slices of a triangle over the desired area; plotting this out makes it clear what you're doing. Maple tells me that this integral = 1, the desired answer.

Hope that helps!

Aren't we talking about an equilateral triangle with vertices at (0,0), (1,1), and (2,0)? If so, then I would calculate 2 integrals; one covering 0≤x≤1 (integrate y from 0 to x), and the other covering 1≤x≤2 (integrate y from 0 to 2-x).

Chris

[strictlyplaid]

Originally Posted by chabig

Aren't we talking about an equilateral triangle with vertices at (0,0), (1,1), and (2,0)? If so, then I would calculate 2 integrals; one covering 0≤x≤1 (integrate y from 0 to x), and the other covering 1≤x≤2 (integrate y from 0 to 2-x).

Chris

Actually, the triangle has vertices (0,0), (0,2), and (1,1). Your integral is also improper (it integrates to more than one.)

Incidentally, a more elegant solution to the integration problem that you are defining would be to integrate x from y to 2-y, then integrate y from 0 to 1.

[iREZ]

what a bunch of nerds!!! (fixes jock strap)

[chabig]

Originally Posted by strictlyplaid

Actually, the triangle has vertices (0,0), (0,2), and (1,1). Your integral is also improper (it integrates to more than one.)

Incidentally, a more elegant solution to the integration problem that you are defining would be to integrate x from y to 2-y, then integrate y from 0 to 1.

I think I disagree. Your triangle would satisfy the inequalities x≥0, x≥y, and x+y≤2. But the problem given specified x≥0, x≤y, and x+y≤2. And it could be solved with by integrating x from y=x to y=2-y, while y ranges from 0 to 1.

Still, my integrals evaluate to 2.2, which either means I made a mistake with the integration, or the given function is not a valid joint density function. Since I've done the integration twice using different methods and I get the same result, I tend to think the latter is true.

Chris

[strictlyplaid]

Originally Posted by chabig

I think I disagree. Your triangle would satisfy the inequalities x≥0, x≥y, and x+y≤2. But the problem given specified x≥0, x≤y, and x+y≤2. And it could be solved with by integrating x from y=x to y=2-y, while y ranges from 0 to 1.

Still, my integrals evaluate to 2.2, which either means I made a mistake with the integration, or the given function is not a valid joint density function. Since I've done the integration twice using different methods and I get the same result, I tend to think the latter is true.

Chris

This part of your reply is wrong: "Your triangle would satisfy the inequalities x≥0, x≥y, and x+y≤2. But the problem given specified x≥0, x≤y, and x+y≤2." Plot the lines y=x and y=2-x out on the Cartesian plane. shade everything above y=x (i.e., y>x) and everything below y=2-x (i.e., x+y<2 or otherwise written y<2-x). The overlap is the triangle described by vertices (0,0), (0,2), (1,1). This is the region that satisfies the conditions of the problem. The fact that my area integrates to 1 is further evidence; note that the problem asks you to prove THAT the density is proper, not determine WHETHER it is.

[chabig]

Originally Posted by strictlyplaid

This part of your reply is wrong: "Your triangle would satisfy the inequalities x≥0, x≥y, and x+y≤2. But the problem given specified x≥0, x≤y, and x+y≤2." Plot the lines y=x and y=2-x out on the Cartesian plane. shade everything above y=x (i.e., y>x) and everything below y=2-x (i.e., x+y<2 or otherwise written y<2-x). The overlap is the triangle described by vertices (0,0), (0,2), (1,1). This is the region that satisfies the conditions of the problem. The fact that my area integrates to 1 is further evidence; note that the problem asks you to prove THAT the density is proper, not determine WHETHER it is.

I agree with your method of plotting y=x and y=2-x. I was doing the same thing. But I misread the original problem and shaded the area below y=x, not the area above it. And I do agree that the integral now equals 1.

Chris

[strictlyplaid]

Originally Posted by chabig

I agree with your method of plotting y=x and y=2-x. I was doing the same thing. But I misread the original problem and shaded the area below y=x, not the area above it. And I do agree that the integral now equals 1.

Chris

No prob, getting the inequalities right was the only really tricky part of the problem and it took me a try or two as well. :-)

[chabig]

So the only question remaining is, did Greg get it?



Chris

[realmeatychunks]

Originally Posted by chabig

So the only question remaining is, did Greg get it?



Chris

Nice image, LaTeX?

I do understand it, after reading the explanation and drawing out the integration region. Thanks a lot guys!

One further question though, if I were to take the marginal distribution function of y, defined as

fy = int(f(x,y),x,-inf,inf) i.e. integrate x over all space and hold y constant,

then would the integral int(f(x,y),x,0,y) be correct? (I don't think it is...) Oi, it's been awhile since i've done something like this. Maybe I need to perform a transformation of variables or something.

I know that the marginal distribution function of x = fx = int(f(x,y),y,x,2-x) which makes sense given our previous work (and it is confirmed by being a beta distribution as the book asks you to prove).

Regards,
Greg

[chabig]

Originally Posted by realmeatychunks

Nice image, LaTeX?

No. Grapher (ships with OS X Tiger). I just wrote the formula and then took a screen capture using cmd-shift-4.

One further question though, if I were to take the marginal distribution function of y, defined as

fy = int(f(x,y),x,-inf,inf) i.e. integrate x over all space and hold y constant,

then would the integral int(f(x,y),x,0,y) be correct? (I don't think it is...) Oi, it's been awhile since i've done something like this. Maybe I need to perform a transformation of variables or something.

I know that the marginal distribution function of x = fx = int(f(x,y),y,x,2-x) which makes sense given our previous work (and it is confirmed by being a beta distribution as the book asks you to prove).

I think you're right. If the general equation is:



Then the specific equation would be:



I think you have to split the integral into two regions, though, because you have to define the limits of integration differently depending upon whether y is less than or greater than 1.

Chris

[bstone]

[jcadam]

I took Calc I, II, & III, as well as linear algebra, discrete math/set theory, and adv. probability in College. I have not used them in the 3 years since I graduated. And I have consumed enough alcohol since then that the neurons that were storing that knowledge are most likely dead.

[realmeatychunks]

Originally Posted by jcadam

I took Calc I, II, & III, as well as linear algebra, discrete math/set theory, and adv. probability in College. I have not used them in the 3 years since I graduated. And I have consumed enough alcohol since then that the neurons that were storing that knowledge are most likely dead.

What field are you in?

[jcadam]

Oh, there you go getting me started <RANT>

I majored in Computer Science (minor in Math).

Used an Army scholarship to fund my education, so now I'm Uncle Sam's indentured slave until May 2006 (almost there w00t). I'm a signal corps officer (communications), which means I could be doing something IT/tech related right now, but currently the Army is making maximum use of my BS in Comp Sci. by having me do Human Resources management (I'm an S-1) type work.

I'm looking forward to getting out so I can be a software engineer, if my unused 4 year old degree isn't totally worthless by now.

</RANT>