Recent content by OhMyMarkov
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MHB Joint cumulative distribution of dependent variables
:confused:- OhMyMarkov
- Post #3
- Forum: Set Theory, Logic, Probability, Statistics
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MHB Joint cumulative distribution of dependent variables
Hello everyone! The problem: $X,Y,Z$ are random variables that are dependent and uniformly-distributed in $[0,1]$, and let $\alpha$ be a given number in $[0,1]$. I am asked to compute the following: $\text{Pr}(X+Y+Z>\alpha \;\;\; \& \;\;\; X+Y\leq \alpha)$ What I have so far...- OhMyMarkov
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- Dependent variables Distribution Joint Variables
- Replies: 5
- Forum: Set Theory, Logic, Probability, Statistics
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MHB Discrete-continuous random variable
Hello everyone! I'm looking at the following random variables: $Z_1$ is normally distributed with zero mean and variance $\sigma _1 ^2$ $Z_2$ is normally distributed with zero mean and variance $\sigma _2 ^2$ $B$ is Bernoulli with probability of success $p$. $X$ is a random variable that...- OhMyMarkov
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- Random Random variable Variable
- Replies: 5
- Forum: Set Theory, Logic, Probability, Statistics
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MHB Least upper bound - greatest lower bound duality
Hello everyone! There's a point I didn't get in Rudin's theorem 1.11 that says: Suppose S is an ordered set with the LUB property, and B $\subset$ S, B is not empty and B is bounded below. Let L be the set of lower bounds of B. Then a = sup L exists in S, and a - inf B. In particular inf B...- OhMyMarkov
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- Bound Duality Upper bound
- Replies: 1
- Forum: Topology and Analysis
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MHB Showing that nth root of c_n is equal to nth root of c_n+1 in the limit
Ah excuse me LaTeX typo: I meant I'm trying to show that: $\lim \sup \sqrt[n]{c_{n+1}} = \lim \sup \sqrt[n]{c_{n}}$ I fixed it in the thread- OhMyMarkov
- Post #3
- Forum: Topology and Analysis
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MHB Showing that nth root of c_n is equal to nth root of c_n+1 in the limit
Hello everyone! I'm trying to show that $\lim \sup \sqrt[n]{c_{n+1}}=\lim \sup \sqrt[n]{c_n}$ This is my attempt: $\lim \sup \sqrt[n]{c_{n+1}} = \lim \sup \sqrt[m-1]{c_m}=\lim \sup c_m \; ^{\frac{1}{m}}c_m \; ^{\frac{1}{m(m-1)}}$ I'm stuck here, I think I must use some exponential property...- OhMyMarkov
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- Limit Root
- Replies: 2
- Forum: Topology and Analysis
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MHB On proving sup A is less than sup B when A is in B
Ok... (1) Suppose $\beta < \alpha$, since $\alpha = \sup A$, any number $t < \alpha$ is not an upper bound of $A$ by the definition of the least upper bound (2) But $\beta$ is an upper bound of $B$, so $\forall x \in B$, $x \leq \beta$, in particular, every $x\in A, x \leq \beta$ so that...- OhMyMarkov
- Post #3
- Forum: Topology and Analysis
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MHB On proving sup A is less than sup B when A is in B
Hello everybody! I want to prove that if $A\subset B$, then $\sup A \leq \sup B$. I'm taking the exhausting approach of considering cases in proving this: First let $\alpha = \sup A, \beta = \sup B$ (1) If $\alpha \in A, \alpha \in B,$ so $\alpha \leq \beta$ (2) If $\alpha \notin A, \alpha...- OhMyMarkov
- Thread
- Replies: 2
- Forum: Topology and Analysis
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MHB Proving lim sqrt(n) alpha^n is 0
Hello everyone! I want to prove that $\lim \sqrt(n) \alpha ^n \rightarrow 0$ whenever $0 <\alpha < 1$. I got the following proof: (1) Write $\alpha$ as $\alpha = 1/x$ where $x > 1$. (2) $\sqrt{n} \alpha ^n = \displaystyle...- OhMyMarkov
- Thread
- Replies: 1
- Forum: Topology and Analysis
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MHB Showing that lim_sup a_n b_n = lim a_n lim_sup b_n$
Hello everyone! :) I'm trying to prove the following: $a_n$ and $b_n$ are real sequences. If $a_n \rightarrow M \in R$, and $\lim \sup a_n b_n = L$ then $\lim \sup a_n b_n = \lim a_n \cdot \lim \sup b_n$. This is what I got so far: (1) Let $c_n = a_n b_n$, and let $C_n$ = $\sup \{c_k \; | \...- OhMyMarkov
- Thread
- Replies: 1
- Forum: Topology and Analysis
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MHB Supremum of f(x) on [a,b] vs. supremum of f(x+c) on [a+c, b+c]
Re: Supremum of $f(x)$ on $[a,b]$ vs. supremum of $f(x+c)$ on $[a+c,b+c]$ Hello caffeinemachine, thanks for replying (twice) I may have sounded rude repeating "any suggestions?" twice, but that was due to an internet problem. I did exactly what you suggested the 1st time :)- OhMyMarkov
- Post #7
- Forum: Topology and Analysis
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MHB Supremum of f(x) on [a,b] vs. supremum of f(x+c) on [a+c, b+c]
Re: Supremum of $f(x)$ on $[a,b]$ vs. supremum of $f(x+c)$ on $[a+c,b+c]$ Yes that was a typo, any suggestions?- OhMyMarkov
- Post #5
- Forum: Topology and Analysis
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MHB Supremum of f(x) on [a,b] vs. supremum of f(x+c) on [a+c, b+c]
Re: Supremum of $f(x)$ on $[a,b]$ vs. supremum of $f(x+c)$ on $[a+c,b+c]$ Yes that was a typo... Any answers?- OhMyMarkov
- Post #3
- Forum: Topology and Analysis
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MHB Supremum of f(x) on [a,b] vs. supremum of f(x+c) on [a+c, b+c]
Hello everyone! I'm really stuck on this one, it looks so obvious, but I can't prove it: Let $\alpha = \sup _{x\in [a,b]} f(x)$, how can I show that $\alpha = \sup _{x\in [a+c,b+c]} f(x+c)$? Thanks!- OhMyMarkov
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- Supremum
- Replies: 7
- Forum: Topology and Analysis
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MHB Differentiating an integral wrt a function
Hello ILikeSerena, thanks for replying! Okay, now I have the book, please let me give out the exact statement: $h(f )$ is a concave function over a convex set. We form the functional: \begin{equation} \displaystyle J(f )= -\int f\ln f + \lambda _0 \int f + \sum _k \lambda _k \int f r_k...- OhMyMarkov
- Post #3
- Forum: Topology and Analysis