# Tag Info

## New answers tagged inequality

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### Probability inequality in Markov chain

$(EZ_n)^{2}=(EZ_n1_{Z_n >0})^{2}\leq EZ_n^{2}E1_{Z_n >0}^{2}=P(Z_n >0) EZ_n^{2}$. (I have used Cauchy-Schwarz-Bunyakovski inequality).
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### let $I_k=\int_0^\infty\frac{x^k}{p(x)}dx$ for which $k$ is $I_k$ smaller?

Perhaps, it is possible to reduce the comparison problem only to the comparison of $\int x^k$ by applying the product rule to get rid of the influence of $p(x)$ on $x^k$ under the integral sign. Then ...
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### How to show that $\ln(2) > \frac{2}{3}$

$\ln(2) > \frac{2}{3}$ is equivalent to $e^2 < 8$, so we need an upper bound for $e$. A standard trick is to replace the “tail” of the exponential series by a geometric series, which can be ...
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### How to show that $\ln(2) > \frac{2}{3}$

You can define $\ln 2$ by $\displaystyle \ln 2 = \int_1^2 \frac 1t \, dt$. Any Riemann sum for this integral with right endpoints will be less than $\ln 2$ because $\dfrac 1t$ is decreasing. In ...
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### Converse to Jensen's inequality for $1/x$ on a positive bounded interval?

We want to determine an upper bound for the function $$h(x_1, \ldots, x_n) = \frac{1}{n^2} \sum_{k=1}^n x_k\sum_{k=1}^n \frac{1}{x_k} = \sum_{k, l=1}^n \frac{x_k}{x_l}$$ on the hypercube $[a, b]^n$....
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### generalization of Markov's inequality

Hint:- For positive numbers, is $a>b$ equivalent to $a^{p}>b^{p}$ ? If yes then what can you say about the event $\{|X|>c\}$ and $\{|X|^{p}>c^{p}\}$ ? . What can you then say about the ...
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### Probability Question, sanity check

My understanding of your problem is, there are effectively two random variables representing squares $X$ and $X_n$. Suppose each random variable is the 'mass' of a square. $X + \epsilon$ is always a ...
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### Permutations with inequalities constraint - analytical solution

Apologies for prematurely closing your question as a duplicate. Although counting the things you want – linear extensions – for general posets is indeed hard, the special case of chained inequalities ...
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### If $\frac{1}{4} \leq x^2 + y^2 \leq 1$, then is it true that $0 \leq \left|-(x^2+y^2)-1\right|\leq \frac{3}{4}$?

Inequality (1) is correct; you have reflected your given inequality across the origin. Inequality (2) is correct; you have shifted (1) in the negative direction. When working with inequalities, it is ...
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### Showing that $\lfloor \sqrt{n^2+b}\rfloor=n \iff 0 \le b < 2n+1$

\begin{align} 0\leq b<2n+1 &\iff n^2 \leq n^2+b <n^2+2n+1\\ &\iff n^2\leq n^2+b <(n+1)^2 \\ &\iff n\leq \sqrt{n^2+b} <n+1 \\ &\iff \lfloor\sqrt{n^2+b} \rfloor = n \\ \end{...
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### Adding constants to the numerator and denominator of a fraction

No, $\frac{1}{2} \leq \frac{3}{4}$ and $\frac{4}{2} = \frac{2}{1}$, but $\frac{1+4}{2+2} = \frac{5}{4} > \frac{3+2}{4+1}=\frac{5}{5}$
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### Adding constants to the numerator and denominator of a fraction

This is not correct. For example: $$\frac{1}{2}=0.5 < 0.66\approx \frac{1}{1.5}.$$ If you add $\frac{5}{5}=\frac{1}{1}$ you get $$\frac{6}{7} \approx 0.85 > 0.8=\frac{2}{2.5}$$
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### Adding constants to the numerator and denominator of a fraction

No: $\frac{2}{1}\leq\frac{1}{\frac 12}$ (in fact, this is an equality) but $\frac{2+2}{1+1}\not\leq\frac{1+1}{\frac 12+\frac 12}$.
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Let us discuss the calculation over a parsing tree example from the book. The formula $\phi$ is $(p_{1}\wedge (\neg(p_{0}\rightarrow p_{2})))$, depicted as: So, the number of nodes, $n$, is $6$. The ...