Recent content by oblixps

  1. O

    MHB Trying to calculate E[(XY)^2] for X, Y ~ N(0, 1) and Cov(X, Y) = p.

    Let X, Y be random variables distributed as N(0, 1) and let Cov(X, Y) = p. Calculate E[X2Y2]. I have that Cov(X, Y) = E[(X - 0)(Y - 0)] = E[XY] = p. I have no idea how to continue on however. I can't think of a way to relate E[X2Y2] with E[XY]. I have considered the Var(XY) but that also...
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    MHB Proving that the sum of 2 measurable functions is measurable

    Amer, could you explain a little more why $s_n - t_n$ must be non decreasing for n > r? I don't really see how that follows from knowing $|s_n(x) - f(x)| < \epsilon$ and $|t_n(x) - f(x)| < \epsilon$. These expressions tell us how to bound $s_n - t_n$ for each n, but doesn't seem to give us a...
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    MHB Proving that the sum of 2 measurable functions is measurable

    I know there are many proofs for this but I am having trouble proving this fact using my book's definition. My book defines first a non negative measurable function f as a function that can be written as the limit of a non decreasing sequence of non-negative simple functions. Then my book...
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    MHB Prove (I + J)/J is isomorphic to I(R/J) as R modules

    ah yes I should have mentioned that my definition of rings include 1. thank you for your answer! i was having trouble showing that the "reverse" of the map i originally suggested was onto but now i understand.
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    MHB Prove (I + J)/J is isomorphic to I(R/J) as R modules

    Let R be a commutative ring and I, J be ideals of R. Show that (I + J)/J is isomorphic to I(R/J) as R modules. I am having trouble coming up with the explicit isomorphism. For I(R/J) I know any element can be expressed as i(r + J) = ir + J by definition of the action of R on R/J. As for (I +...
  6. O

    MHB Show that G is isomorphic to G x G

    ah i see. now forgetting about the group of rationals, if we just let G be the direct product of countably many copies of Z, we would also have G isomorphic to G x G right? it seems like it would be the same argument as in the case for a direct sum.
  7. O

    MHB Show that G is isomorphic to G x G

    i wouldn't have thought of taking the direct sum of infinitely many copies of Z. seems like an interesting example. would the result also be true if we took an infinite direct product instead of direct sum? thanks for your reply!
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    MHB Show that G is isomorphic to G x G

    Let G be the group of positive rational numbers under multiplication. Show G is isomorphic to G x G. i'm not sure how to start this. i am trying to come up with an explicit isomorphim but i can't seem to find one. can someone give me some hints on how to approach this?
  9. O

    MHB Confusion in proof involving lim and liminf

    In a proof showing that the vector space of all functions from a metric space X to the complex numbers C is complete under the supremum norm, there was a line near the end that i was confused about. starting here: \(sup|f - fn| \leq liminf_{m \to +\infty} ||f_n - f_m|| \) the next step then is...
  10. O

    MHB Calculating partial derivatives in different coordinate systems

    let f = x2 + 2y2 and x = rcos(\theta), y = rsin(\theta) . i have \frac{\partial f}{\partial y} (while holding x constant) = 4y . and \frac{\partial f}{\partial y} (while holding r constant) = 2y . i found these partial derivatives by expressing f in terms of only x and y, and then in...
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    MHB Conditions for integral domain to be a PID

    thanks for the counterexample! that helped me see what could happen if that condition wasn't met. i'm still trying to figure out what went wrong in my argument though. By using Zorn's Lemma, i was able to find an element a in I that divides every element in I so it follows that I is contained...
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    MHB Conditions for integral domain to be a PID

    after thinking about it some more, it seems to me that condition #1 follows from condition #2 by taking the ideal I = (a, b) generated by 2 elements a and b. Then from the argument above using condition #2 and Zorn's Lemma, we have I = (a, b) = (c) for some element c that divides all elements of...
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    MHB Conditions for integral domain to be a PID

    this was actually an exercise in Dummit and Foote. in case you have a copy, it is #4 in section 8.2.
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    MHB Conditions for integral domain to be a PID

    Let R be an integral domain. Then R is a PID if the following 2 conditions hold: 1) any 2 elements a, b \in R have a greatest common divisor which can be written as ra + sb for some r, s \in R . 2) If a_1, a_2, ... are nonzero elements of R such that a_{i+1}|a_i for all i, then there is...
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    MHB If X is a subset of G such that xy = yx for all x, y in X, then <X> is Abelian.

    thanks for all your comments! i realize that this problem could of been done easily using the constructive definition, but I wanted to understand how to use the fact that the centralizer was a subgroup. I was initially confused about the jump the author made, but as Opalg pointed out, it was...
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