Probability integral not converging

[ArcanaNoir]

Probability integral not converging :(

Homework Statement

for the joint probability density function:
f(x,y)= \begin{cases}<br /> y &amp; 0 \leq x, \, y\leq1 \\<br /> \frac{1}{4} (2-y) &amp; 0\leq x\leq 1, \, 0\leq y\leq 2 \\<br /> 0 &amp; \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}<br /> y &amp; 0 \leq x, \, y\leq1 \\<br /> \frac{1}{4} (2-y) &amp; 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} <br /> y &amp; 0 \leq x, \, y\leq1 \\ <br /> 2-y &amp; 0\leq x\leq 1, \, 1\leq y\leq 2 \\ <br /> 0 &amp; \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 &amp; x \: \text{or} \: y \leq 0 \\ \frac{1}{2} xy^2 &amp; 0 \leq x,y \leq 1 \\ 2xy-\frac{1}{2} xy^2 -\frac{3}{2} x &amp; 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 &amp; x&gt;1 \: \text{and} \: 0\leq y \leq 1 \\ 2y-\frac{1}{2} y^2-\frac{3}{2} &amp; x&gt;1 \: \text{and} \: 1\leq y \leq 2 \\ \frac{1}{2} x &amp; y&gt;2 \: \text{and} \: 0\leq x \leq 1 \\ \frac{1}{2} &amp; y&gt;2 \: \text{and} \: x&gt;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 |}<br /> \hline \\<br /> &amp; x &lt; 0 &amp; 0 \le x &lt; 1 &amp; x \ge 1 \\<br /> \hline \\<br /> y &lt; 0 &amp; &amp; &amp; \\<br /> 0 \le y &lt; 1 &amp; &amp; &amp; \\<br /> 1 \le y &lt; 2 &amp; &amp; &amp; \\<br /> y \ge 2 &amp; &amp; &amp; \\ \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}}<br /> \int_{-\infty}^{y} f(x,y) \d y = \int_0^1 f(x,y) \d y + \int_1^y f(x,y) \d y<br />

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&gt;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&gt;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.