Probability integral not converging

[ArcanaNoir]

Probability integral not converging :(

Homework Statement

for the joint probability density function:
$$f(x,y)= \begin{cases} y & 0 \leq x, \, y\leq1 \\ \frac{1}{4} (2-y) & 0\leq x\leq 1, \, 0\leq y\leq 2 \\ 0 & \text{elsewhere} \end{cases}$$
find the following:
a) $f_1 (x)$
b) $f_2 (y)$
c) $F(x,y)$

there are others but maybe this will be enough to help me see what the solution to my actual problem is, problem being that my integrals are not converging and I don't know what to do with that.

Homework Equations

a) $$f_1 (x)= \int_{-\infty }^\infty \! f(x,y) \, \mathrm{d}y$$
Where the integral is integrated over the range space of y.

b) $$f_2 (y)= \int_{-\infty }^\infty \! f(x,y) \, \mathrm{d}x$$
Where the integral is integrated over the range space of x.

c) $$F(x,y)= \int_{-\infty }^y \int_{-\infty }^x \! f(s,t) \, \mathrm{d} s \mathrm{d} t$$
Again, the lower bound is replaced by the lower bound of the range space for the appropriate variables.

The Attempt at a Solution

a) $$f_1 (x) = \int_{-\infty }^1 y \, \mathrm{d}y = \frac{1}{2} y^2 |_{-\infty }^1 = \frac{1}{2} -\frac{1}{2} \infty$$ ?

Lets start there. My time is up on the library computer.

 

[gb7nash]

Homework Statement

for the joint probability density function:
$$f(x,y)= \begin{cases} y & 0 \leq x, \, y\leq1 \\ \frac{1}{4} (2-y) & 0\leq x\leq 1, \, 0\leq y\leq 2 \end{cases}$$

I'm not sure if this will help much, but you have contradicting information in your problem statement. You're defining:

f(x,y) = y for 0 ≤ x, y ≤ 1
f(x,y) = .25(2-y) for 0 ≤ x ≤ 1, 0 ≤ y ≤ 2.

These two regions intersect.

 

[ArcanaNoir]

(have I mentioned before how much I hate my erroneous book? Especially pdfs that don't sum to 1. Those are among my favorites...)

I'm going to email my professor about the bounds. I will post back when he responds.

 

[I like Serena]

Hi Arcana!

I propose you modify the problem to:
$$f(x,y)= \begin{cases} y & 0 \leq x, \, y\leq1 \\ 2-y & 0\leq x\leq 1, \, 1\leq y\leq 2 \\ 0 & \text{elsewhere} \end{cases}$$

I expect that this was intended, and at least that gives you the opportunity to exercise your problem solving skills without delay.

 

[ArcanaNoir]

Did you mean to drop the $\frac{1}{4}$ ? Also, even if I accept these bounds, what should I do about the integral that's not converging? I don't think the new bounds change it.

 

[I like Serena]

Yes, I intended to drop the 1/4.
Perhaps (somewhat later) you can tell me why?Perhaps you should change the bounds on your integral?

Since f(x,y)=0 for negative values of y, it does not seem right that you integrate from -infty.

 

[ArcanaNoir]

Yes, I intended to drop the 1/4.
Perhaps (somewhat later) you can tell me why?

Uh, without doing extra math since I have way to much to do already (the lenghtiest problem set for this class, ever), I'm guessing you dropped the 1/4 because when you do then it sums to 1 like a real pdf should. Haha. We don't have many pdfs that actually obey the properties of a pdf in this book... *grumble grumble* I guess the ability to check your work by using such properties is a luxury we've learned to do without in this class. What a detriment. *sigh*

Perhaps you should change the bounds on your integral?

Since f(x,y)=0 for negative values of y, it does not seem right that you integrate from -infty.

Oh yeah. no negative probabilities. That's just silly! :P

[STRIKE]Tell you what though, I'm going to change the function to the bounds $0 \leq x, \, y\leq 1$ and $0\leq x\leq 1, \, 1\leq y\leq 2$ but leave the 1/4, for the reason that the bounds are stupid if the overlap, and I have many more things to work out about this distribution that may be really stupid if the bounds are like that, but I'll levae the 1/4 because we work on pdfs that are big fat lies all the time anyway, and the grader will probably have worked it out with the 1/4.[/STRIKE]

Edit: no, forget it. I'll change it all the way you said. I'll just email the prof again and tell him I did.

 

[I like Serena]

Uh, without doing extra math since I have way to much to do already (the lenghtiest problem set for this class, ever), I'm guessing you dropped the 1/4 because when you do then it sums to 1 like a real pdf should. Haha. We don't have many pdfs that actually obey the properties of a pdf in this book... *grumble grumble* I guess the ability to check your work by using such properties is a luxury we've learned to do without in this class. What a detriment. *sigh*

Right! It gives you the unique opportunity to really dive into the problems and rectify them.
When you are able to do that, you will truly understand the material!

Next time I will just ask you if you can find the mistake in the problem, and perhaps give a hint for that!

Oh yeah. no negative probabilities. That's just silly! :P

Edit: no, forget it. I'll change it all the way you said. I'll just email the prof again and tell him I did.

Yep. And good!

 

[ArcanaNoir]

Okay, problem. First, I can't tell my prof I'm dropping the 1/4 without justification, and I can't make the math work out to justify.
To prove that the pdf sums to 1 when I don't include 1/4, I tried to do:

$$\int_0^1 \int_0^{\infty } y \; \mathrm{d} x \, \mathrm{d} y + \int_1^2 \int_0^1 2-y \; \mathrm{d} x \, \mathrm{d} y$$
But the first integral goes like this:
$$\int_0^1 \int_0^{\infty } y \; \mathrm{d} x \, \mathrm{d} y = \int_0^1 (xy |_0^{\infty } ) \; \mathrm{d} y$$ and I'll save myself the tedious typing, you can see the non-convergence problem I'm having. Reversing the order of integration didn't help.


Also, I solved part (a) by adjusting the bounds, but I'm still stuck on part b), which is essential the problem I just said about the total sum anyway. Basically copy and pasted this from above:
b) $$\int_0^{\infty } y \; \mathrm{d} x = xy |_0^{\infty }$$

 

[I like Serena]

What would f(x,y) be for large values of x? Say x > 1?

 

[ArcanaNoir]

great big values of x, when y is still less than 1, are still only =y. How does that affect the integral?

 

[I like Serena]

Nope. They're not.
Check your definition of f(x,y).
x and y are simultaneously bound.

 

[ArcanaNoir]

when $x \geq 0 \: , \text{and} \: 0 \leq y \leq 1 \: , f(x,y)=y$
Thus, for example, $f(1274689,\frac{1}{2} )=\frac{1}{2}$

Isn't it?

 

[I like Serena]

That's not what was intended.
Read it like this:
0 ≤ (x,y) ≤ 1.

That is, 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1.
It's a common shorthand.

 

[ArcanaNoir]

*expletive*

Okay, I think I can solve the rest of this now that we re-defined and further clarified the *expletive* problem.

Thank you, you're a life saver, really. You've prevented my suicide at least four times now. j/k :P

 

[ArcanaNoir]

ps,
can you take a look at my other question? people are using big words at me again...

 

[ArcanaNoir]

alas, further problem with this problem. I'm doing part c.
first, I do two integrals:
$$\int_0^y \int_0^x t \; \mathrm{d} s \, \mathrm{d} t = \int_0^y xt \; \mathrm{d} t = \frac{1}{2} xy^2$$
$$\int_1^y \int_0^x 2-y \; \mathrm{d} s \mathrm{d} t = \int_1^y x(2-t) \; \mathrm{d} t = \int_1^y 2x-tx \; \mathrm{d} t = (2xt-\frac{1}{2} xt^2)|_1^y =$$
$$(2xy-\frac{1}{2}xy^2)-(2x-\frac{1}{2} x)=2xy-\frac{1}{2} xy^2-2x+\frac{1}{2} x=2xy-\frac{1}{2} xy^2-\frac{3}{2}x$$
Now, $$F(x,y)= \begin{cases} 0 & x \: \text{or} \: y \leq 0 \\ \frac{1}{2} xy^2 & 0 \leq x,y \leq 1 \\ 2xy-\frac{1}{2} xy^2 -\frac{3}{2} x & 0 \leq x \leq 1 \: \text{and} \: 1\leq y\leq 2 \end{cases}$$

Now, when one or both variable is above it's bound, you just plug in the maximum value of the variable, right? so the rest of F(x,y) looks like this:
$$\begin{cases} \frac{1}{2} y^2 & x>1 \: \text{and} \: 0\leq y \leq 1 \\ 2y-\frac{1}{2} y^2-\frac{3}{2} & x>1 \: \text{and} \: 1\leq y \leq 2 \\ \frac{1}{2} x & y>2 \: \text{and} \: 0\leq x \leq 1 \\ \frac{1}{2} & y>2 \: \text{and} \: x>2 \end{cases}$$

But, $F(\infty ,\infty )$ should equal 1, not 1/2

What gives?

 

[I like Serena]

Haven't checked your entire calculation yet, but I know you'll have to divide the problem up into 4x3 cases, like this:

$$\begin{array}{| c | c | c | c |} \hline \\ & x < 0 & 0 \le x < 1 & x \ge 1 \\ \hline \\ y < 0 & & & \\ 0 \le y < 1 & & & \\ 1 \le y < 2 & & & \\ y \ge 2 & & & \\ \hline \end{array}$$

Can you calculate F(x,y) for each of the cells?
Some of them will be the same.

 

[ArcanaNoir]

I think I have that table covered. I have seven cases, because x or y is less than zero is one case, so count 5 cases on your table as one case, and 12-5 is 7. Anyway, the problem lies with the case x>1 and y>2. I get 1/2 but it should be 1.

 

[ArcanaNoir]

should I be adding the integral of the first piece, y, and the integral of the second piece, 2-y, when both are maximum? that would be 1.

 

[I like Serena]

You've taken the integral for y from 1 to y.
But you left out the part where y is between 0 and 1.

In other words, if y > 1 then:
$$\def \d{\textrm{ d}} \int_{-\infty}^{y} f(x,y) \d y = \int_0^1 f(x,y) \d y + \int_1^y f(x,y) \d y$$

Btw, note the nifty use of \def\d{\textrm{ d}}.
The beauty of it is, that you have to do it only once in one post.

 

[ArcanaNoir]

Btw, note the nifty use of \def\d{\textrm{ d}}.
The beauty of it is, that you have to do it only once in one post.

Hah, was my latex showing? :P Thanks for the tip. This stuff takes WAY to long for me to input, I do need to learn some shorter ways of doing it.

 

[I like Serena]

Most people don't make the effort to make the "d" upright.

And did you like the table I made for you?
I just wish there was an easier way to make it.

 

[ArcanaNoir]

I liked your pretty table, but I already had all the cases covered. Just covered kinda wrong. Should I add a 1/2 x to my $x>1 \: \text{and} \: 1\leq y \leq 2$ case?

As for making my d's upright, I was actually thinking today that I put too much effort into my latex, since it took me almost 30 minutes to type each of my longer posts.
I sure hope practice makes speedy. I'm in the "tedious" phase.

 

[I like Serena]

I liked your pretty table, but I already had all the cases covered. Just covered kinda wrong. Should I add a 1/2 x to my $x>1 \: \text{and} \: 1\leq y \leq 2$ case?

Yes, you should add 1/2 x.

As for making my d's upright, I was actually thinking today that I put too much effort into my latex, since it took me almost 30 minutes to type each of my longer posts.
I sure hope practice makes speedy. I'm in the "tedious" phase.

Latex is designed to type a formula as "natural" as possible.
Still, it takes me quite a bit of time too, but usually thinking about a problem takes much more time.
And I do like neat formulas!

 

[ArcanaNoir]

Just got my prof's email. He says something about how swell my effort is... and that whatever I put for the answer will be fine, it's only enough to show effort. *grumble grumble* I hate the *effort* answers. This whole problem is a joke! It completely falls apart at the conditional stage. Anyway, I put some answers down, which are as close to resembling a good answer as one can get with such a squirrely problem. Moving on now. Thanks for the help and support ILS!

Edit: Oh god, I just realized I'll have to see this problem again. As if the 10 part initial question wasn't stupid enough, there is a subsequent 7 part question, and then even another three part question. Lame lame lame.