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#1 Re: Help Me ! » Joint Probability and Independence » 2009-03-24 15:20:20
Anyone?
#2 Help Me ! » Joint Probability and Independence » 2009-03-22 13:37:09
- MarkusD
- Replies: 1
Can someone explain to me why the first is independent and the 2nd is not?
Given the joint probability distribution check if the following is independent:
f(x,y)= 1/4 for:
x=-1, y=-1;
x=-1, y=1;
x=1, y=-1;
x=1, y=1
The example of nonindependence is:
f(x,y)=1/3 for:
x=0, y=0;
x=0, y=1;
x=1, y=1
but Im not sure how one is indenpendent and one isnt when my book just says the def of independence is: f(x1...xn)=f1(x1)*....fn(xn). Any help?
#3 Help Me ! » Joint Probability Question » 2008-01-27 07:25:02
- MarkusD
- Replies: 0
Im having a hard time with joint probability when the probability has something to do with X and Y togther (as in parts b and c). Can anyone help with the following question:
If the joint probability density of X and Y is given by:
f(x,y)= 2 for x>0, y>0, x+y<1
= 0 elsewhere
find:
a) P(X≤1/2, Y≤1/2)
b) P(X+Y>2/3)
c) P(X>2Y)
d) an expression for the values of the joint distribution function of X and Y when x>0, y>0, x+y<1
The answers are 1/2, 5/9, and 1/3 with part d) not being in the back of the book but I think I can figure that out once I know what Im doing for part b and c.
For part a, I just took the double integral from 0 to 1 for x and y and came up with 1/2 once it was all evaluated. Not sure about the others and my book isnt too helpful here.
#4 Help Me ! » Integration Question (Im very rusty!) » 2008-01-21 10:47:33
- MarkusD
- Replies: 2
How do I integrate the following:
∫3x/√(1-x²)dx
I tried integration by parts where u=3x, du=3, v=(1-x²)^-½, dv=-½(1-x²)^(-3/2)-2x(1-x²)^-½
it works out ok, but I get stuck with the ∫vdu part which is ∫3(1-x²)^-½
Any help?
#5 Help Me ! » Question on Quasiconcavity » 2007-11-22 07:26:28
- MarkusD
- Replies: 0
Suppose z=(x1x2)^a
where x1, x2, a>0
How can I prove that this function is quasiconcave without assuming differentiability?
I know that the condition when a function is nondifferentiable is:
f(aX0+(1-a)x1)>= min{f(x0),f(x1)}
But Im not sure how to apply it because of the 'a' exponent. Any tips?
Happy Thanksgiving everyone! 
#6 Help Me ! » Equal Fractions » 2007-11-03 13:50:31
- MarkusD
- Replies: 3
I have a question:
If a/b=c/d (implying ad=bd), then how does a/b=c/d= (a+c)/(b+d)? Can someone verify this?
#7 Help Me ! » Proof with Transitive Closures » 2007-10-07 10:33:25
- MarkusD
- Replies: 1
Im stuck with the following question:
Q: R is a relation on X. Prove if R is symmetric, then Rt is symmetric (Note Rt is the transitive closure of R).
I know that for R to be symmetric, we need xRy and yRx. Also, Rt is the smallest transitive relation containing so Rt is a subset of R.
So in thinking about this, I think I need to show that assuming R is symmetric, that 1) R is a subset of Rt^-1 and that Rt^-1 is transitive. Is this correct? If so, how do I prove 1)?
#8 Re: Help Me ! » Another Question About Sets (Proof) » 2007-09-23 09:45:38
He did a piece wise defintion:
R= {(1,2), (2,1), (1,1), (2,2)}.
This literally means:
1R2
2R1
1R1
2R2We can see that if aRb for any element in there, then it must be that bRa. The only elements in there are 1 and 2, and both 1R2 and 2R1. Likewise, you can verify transitivity. But 3 is not related to 3. Anything that doesn't appear as a tuple in R is not related.
Do you have another example that uses only operators or something non-piece wise. Im just asking because I havent covered piece-wise definitions and am having a hard time with the current tools that I have.
#9 Re: Help Me ! » Another Question About Sets (Proof) » 2007-09-23 09:39:30
Also, just to be sure, the problem with the proof is the bolded phrase?
Let x be an arbitrary element of A. Let y be any element of A such that xRy. Since R is symmetric, it follows that yRx. But then by transitivity, since xRy and yRy, then xRX. Also, since x was random, the proposition xRx for all x in A holds. Thus R is relfexive.
Because I was looking at it further and Im wonder if the phrase "But then by transitivity, since xRy and yRy, then xRX. " is also incorrect. Ive been looking at definitions of transitivity and they always have at least 3 elements (x,y,z) when writing the out the definition. Does transitivity work with two elements as Ive written above?
#10 Re: Help Me ! » Another Question About Sets (Proof) » 2007-09-23 09:18:48
For (B), the relation R defined in HallsofIvys example is symmetric and transitive but not reflexive. For the relation R to be reflexive on A, you must have xRx for all x ∈ A, not just for some x ∈ A. Since A is chosen as the set of natural numbers, in order for R to be reflexive on A, you must have nRn for all natural numbers n, not just for 1 and 2.
Im confused with Ivy's example:
My book says that symmetry is having xRY->yRx.
Transitive is xRy ^ yRz -> xRz
Reflexive is xRx
And so far I have been using relations like >, <, =, U, etc. to verify things. I havent yet had to define a R specifically in any of the work that Ive done.
Ivy defined R as some ordered pairs but Im not seeing how 1R2->2R1 unless R is the equality operator (which is transitive adn reflexive).
#11 Re: Help Me ! » Another Question About Sets (Proof) » 2007-09-23 06:24:43
OK here is the full question:
Suppose R is a relation on A, and deinfe a relation S on P(A) as follows:
S={(X,Y)in P(A)xP(A)| Ex in X, Ey in Y(xRy)}
Note: P is a power set, and E=exist
If R is transitive, then R then so is S.
Part A) Is the theorem correct? Prove the theorem above or provide a counter example.
Part B) Now suppose R is still a relation on A. If R is symmetric and transitive then R is reflexive. Is this theorem correct? Verify by proof.
Im concerned with part B.
I wrote a proof but clearly its wrong (can it be fixed?). Apparently the theorem is wrong too right?
Ivy, is the example you provided not relfexive also because of the first two ordered pairs?
#12 Re: Help Me ! » Another Question About Sets (Proof) » 2007-09-22 11:43:08
First, make sure you have the question right. Are you trying to prove R or S is reflexive? I don't see any reason to define S if you are just trying to prove that R is reflexive. In either case, the statement is false. So you find a counter example, like Halls has done.
Its R in this case because the question has multiple parts that relate to S. Im just stuck on this part regarding the proof.
#13 Re: Help Me ! » Another Question About Sets (Proof) » 2007-09-22 09:36:55
The typo I was referring to is here:
But then by transitivity, since xRy and yRy
Now back to the problem with your proof.
Ahh ok, youre right, that was a typo.
Let y be any element of A such that xRy.
What if such a y doesn't exist? Perhaps there is a reason that it must, but you certainly haven't given it.
So do I have to try to prove that theres a y in A such that xRy or should I be trying an avenue when trying to prove that R is reflexive?
#14 Re: Help Me ! » Another Question About Sets (Proof) » 2007-09-21 12:20:27
You've got a bit of a typo, but I'm assuming you meant yRx there.
No thats the way its given. Part of the problem is to see if the theorem is correct.
Next, write a proof.
So it's time to reflect on your proof. What did your proof use? Remember, one of the most common pitfalls is to assume the existence of something that might not exist.
Not sure what you mean by what my proof used. I took the part that said If R is symmetric and transitive and assumed to try and come to a conclusion. I drag at writing proofs 
#15 Re: Help Me ! » Another Question About Sets (Proof) » 2007-09-20 15:39:18
Also, I wrote the following proof to say it is but can someone check my work to see if I wrote a correct proof?
Let x be an arbitrary element of A. Let y be any element of A such that xRy. Since R is symmetric, it follows that yRx. But then by transitivity, since xRy and yRy, then xRX. Also, since x was random, the proposition xRx for all x in A holds. Thus R is relfexive.
#16 Help Me ! » Another Question About Sets (Proof) » 2007-09-20 15:32:51
- MarkusD
- Replies: 17
Im dealing with the following Theorem:
Suppose R is a relation on A, and deinfe a relation S on P(A) as follows:
S={(X,Y)in P(A)xP(A)| Ex in X, Ey in Y(xRy)}
If R is symmetric and transitive, then R is reflexive.
Note: P is a power set, and E=exist
Is this theorem correct? If not why?
#17 Re: Help Me ! » Composite of Two Relations » 2007-09-19 16:28:18
5 goes to 4 in S^-1, and 4 goes to both 5 and 6 in S. So 5 goes to 5 and 5 goes to 6. The same thing happens with 6.
How does that happen. I thought that you take the first elements of S and pair them with the second elements in S^-1. Since there are 4 pairs, you should get 4 pairs at the end right?
#18 Help Me ! » Composite of Two Relations » 2007-09-19 14:38:55
- MarkusD
- Replies: 2
Hello everyone, I have the following:
S={(4,5),(4,6),(5,4),(6,6)}
What is SoS^-1?
Now I wrote that S^-1={(5,4),(6,4),(4,5),(6,6)}
So I should have:
{(4,4), (4,4), (5,5), (6,6)}
but the back of my book says that I should (5,6) and (6,5) as part of the solution. Can anyone explain?
#19 Re: Help Me ! » Need a Hint » 2007-09-16 08:12:06
That's correct, but there's an easier way of doing it. Try to alter the two given fractions so that their denominators are both k! (n-k+1)!.
OK I did the following:
Using the fact that k!=k(k-1)! and that (n-k+1)! = (n-k+1)(n-k)! I was able to get a common denominator and then have n!(n+1) on the numerator which equals the other side.
#20 Re: Help Me ! » Need a Hint » 2007-09-16 07:31:06
Hmm, what exactly do you not find rigorous about it? Here are the statements:
1. Number of combinations of choosing k out of n objects is C(n, k)
2. Number of such choices which contain a specific element x: C(n-1, k-1)
3. Number of such choices which don't contain a specific element x: C(n-1, k)x can not be both in and out of a set, so (2) and (3) are mutually exclusive. As x must be either in or out of a set, the sum of these will result in the total of all the choices.
Nevermind, I had to google combinatorial proof and double counting. I thought you were giving me instructions on how to follow through algebraically. But now I have another question, whats the difference between bijective proof and double counting? They seem to be very related.
#21 Re: Help Me ! » Need a Hint » 2007-09-16 06:49:52
Ricky, what you wrote doesnt seem all too clear to me at least in writing it mathematically.
Clear as in understandable, or clear as in rigorous?
Implementing it rigorously.
#22 Re: Help Me ! » Need a Hint » 2007-09-16 05:17:25
You're correct so far. From here, you just need to add up your two fractions.
You'll need to fiddle with them a bit to give them the same denominator, but once you've done that everything should fall into place.
Heres what I got when I tried that, I got stuck because perhaps theres a property that Im missing that allows me to cancel things. But in adding the two givens and setting a common denominator yields:
n!{[(k-1)!(n-k+1)!+k!(n-k)! / [k!(n-k)!(k-1)!(n-k+1)!]}
Ricky, what you wrote doesnt seem all too clear to me at least in writing it mathematically.
#23 Help Me ! » Need a Hint » 2007-09-15 17:36:12
- MarkusD
- Replies: 10
Ok last question of the night.
I am trying to prove that for n>=k>=1 the following holds:
C(n+1,k)=C(n,k)+C(n, k-1)
Now Ive set this problem by organizing my givens by using the definition of a combination and set a goal:
Givens:
n>=k>=1
C(n,k)= n!/[k!(n-k)!]
C(n,k-1)=n!/[(k-1)!(n-k+1)!]
Goal:
(n+1)!/[k!(n-k+1)!] (which equals C(n+1,k))
Now from here, Im stuck. I working it out algebraically but that didnt really work. Are my givens correct? And if so, is there anything Im over looking as far as a given? Any hints on how to proceed?
#24 Re: Help Me ! » Adding and Subtracting from a Set » 2007-09-15 16:54:33
Ahh ok. Very clear now
#25 Help Me ! » Adding and Subtracting from a Set » 2007-09-15 14:11:50
- MarkusD
- Replies: 2
Why does the following not hold:
(AUB)\B=A
It would seem that adding something to a set and then taking it away would leave the original set, but my book tells me this doesnt hold.
I was able to prove (AUB)\BCA but am curious why it does not equal A.