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Richard Hardy
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If r > s ≥ 1 For $r>s\geq1$, convergence in s-$s^{\text{th}}$ mean doesnotdoes not imply convergence in r-$r^{\text{th}}$ mean

I need a counterexample for the problem: Ifif $r > s ≥ 1$$r>s\geq1$, convergence in $s$-th$s^{\text{th}}$ mean does not imply convergence in $r$-th$r^{\text{th}}$ mean.

The definition for convergence in mean is as follows: Let $r≥1$$r\geq1$ be a fixed number. A sequence of random variables $X_1, X_2, X_3,...$ converges in the $r$th$r^{\text{th}}$ mean to a random variable $X$, if

$$\lim_{n\to\infty}E(|X_n−X|^r)=0.$$

If r > s ≥ 1, convergence in s-th mean doesnot imply convergence in r-th mean

I need a counterexample for the problem: If $r > s ≥ 1$, convergence in $s$-th mean does not imply convergence in $r$-th mean.

The definition for convergence in mean is as follows: Let $r≥1$ be a fixed number. A sequence of random variables $X_1, X_2, X_3,...$ converges in the $r$th mean to a random variable $X$, if

$$\lim_{n\to\infty}E(|X_n−X|^r)=0.$$

For $r>s\geq1$, convergence in $s^{\text{th}}$ mean does not imply convergence in $r^{\text{th}}$ mean

I need a counterexample for the problem: if $r>s\geq1$, convergence in $s^{\text{th}}$ mean does not imply convergence in $r^{\text{th}}$ mean.

The definition for convergence in mean is as follows: Let $r\geq1$ be a fixed number. A sequence of random variables $X_1, X_2, X_3,...$ converges in the $r^{\text{th}}$ mean to a random variable $X$, if

$$\lim_{n\to\infty}E(|X_n−X|^r)=0.$$

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Christoph Hanck
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I need a counterexample for the problem: If $r > s ≥ 1$, convergence in $s$-th mean does not imply convergence in $r$-th mean.

The definition for convergence in mean is as follows: Let $r≥1$ be a fixed number. A sequence of random variables $X_1, X_2, X_3,...$ converges in the $r$th mean to a random variable $X$, if

$$\lim_{n\to\infty}E(|X_n−X|^r)=0$$.$$\lim_{n\to\infty}E(|X_n−X|^r)=0.$$

I need a counterexample for the problem: If $r > s ≥ 1$, convergence in $s$-th mean does not imply convergence in $r$-th mean.

The definition for convergence in mean is as follows: Let $r≥1$ be a fixed number. A sequence of random variables $X_1, X_2, X_3,...$ converges in the $r$th mean to a random variable $X$, if

$$\lim_{n\to\infty}E(|X_n−X|^r)=0$$.

I need a counterexample for the problem: If $r > s ≥ 1$, convergence in $s$-th mean does not imply convergence in $r$-th mean.

The definition for convergence in mean is as follows: Let $r≥1$ be a fixed number. A sequence of random variables $X_1, X_2, X_3,...$ converges in the $r$th mean to a random variable $X$, if

$$\lim_{n\to\infty}E(|X_n−X|^r)=0.$$

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Christoph Hanck
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  • 137

I need a counterexample for the problem: If r > s ≥ 1$r > s ≥ 1$, convergence in s$s$-th mean does not imply convergence in r$r$-th mean.

theThe definition for convergence in mean is as follows: Let r≥1$r≥1$ be a fixed number. A sequence of random variables X1, X2, X3, ⋯$X_1, X_2, X_3,...$ converges in the rth$r$th mean to a random variable X$X$, if

limn→∞$E(|Xn−X|^r)=0$$$\lim_{n\to\infty}E(|X_n−X|^r)=0$$.

I need a counterexample for the problem: If r > s ≥ 1, convergence in s-th mean does not imply convergence in r-th mean

the definition for convergence in mean is as follows: Let r≥1 be a fixed number. A sequence of random variables X1, X2, X3, ⋯ converges in the rth mean to a random variable X, if

limn→∞$E(|Xn−X|^r)=0$.

I need a counterexample for the problem: If $r > s ≥ 1$, convergence in $s$-th mean does not imply convergence in $r$-th mean.

The definition for convergence in mean is as follows: Let $r≥1$ be a fixed number. A sequence of random variables $X_1, X_2, X_3,...$ converges in the $r$th mean to a random variable $X$, if

$$\lim_{n\to\infty}E(|X_n−X|^r)=0$$.

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