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Look, I know that [math]\frac{1}{n}\sum_{i=1}^nX_i^2 = E[X^2][/math] is the second sample moment, and [math]E[X^2]= \mu^2+\sigma^2[/math] so for rth-mean convergence, I have [math] \lim_{n \rightarrow \infty}E[|Z_n-(\mu^2+\sigma^2)|]=0[/math], and if converges in rth-mean implies that converges in probability.

The other idea is

[math]\lim_{n \rightarrow \infty}P(|Z_n-(\mu^2+\sigma^2)|\geq\epsilon)=\lim_{n \rightarrow \infty}P(|E[X^2]-(\mu^2+\sigma^2)|\geq\epsilon)[/math][math]=\lim_{n \rightarrow \infty}P(\mu^2+\sigma^2-(\mu^2+\sigma^2)\geq\epsilon)=\lim_{n \rightarrow \infty}P(0\geq\epsilon)\rightarrow0[/math]

some of the ideas are right?

The other idea is

[math]\lim_{n \rightarrow \infty}P(|Z_n-(\mu^2+\sigma^2)|\geq\epsilon)=\lim_{n \rightarrow \infty}P(|E[X^2]-(\mu^2+\sigma^2)|\geq\epsilon)[/math][math]=\lim_{n \rightarrow \infty}P(\mu^2+\sigma^2-(\mu^2+\sigma^2)\geq\epsilon)=\lim_{n \rightarrow \infty}P(0\geq\epsilon)\rightarrow0[/math]

some of the ideas are right?

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i.e. Are you sure you can directly use the result

[math] \lim_{n \to +\infty}E[|Z_n - (\mu + \sigma^2)|] = 0 [/math]

without proving it?

For the second idea, the step

[math] \lim_{n\to+\infty} \Pr\{|Z_n - (\mu + \sigma^2)| \geq \epsilon\}

= \lim_{n\to+\infty} \Pr\{|E[Z_n] - (\mu + \sigma^2)| \geq \epsilon\} [/math]

is wrong. As a quick check: the RHS is independent of the limit [math] n [/math], and as you see there is no random variable inside the probability and it is actually equal to zero.

[math] \lim_{n \rightarrow \infty}P(|Z_n-(\mu^2+\sigma^2)|\geq\epsilon)\leq\frac{1}{\epsilon}E[Z_n-(\mu^2+\sigma^2|]=\frac{1}{\epsilon}(E[Z_n]-E[\sigma^2+\mu^2])[/math]

where [math]E[Z_n]=E[\frac{1}{n}\sum_{i=1}^nX_i^2]=\frac{1}{n}E[\sum_{i=1}^nX_i^2]=\frac{1}{n}*nE[X_1^2]=E[X_1^2]=\mu^2+\sigma^2[/math]

so

[math] \lim_{n \rightarrow \infty}P(|Z_n-(\mu^2+\sigma^2)|\geq\epsilon)\leq\frac{1}{\epsilon}(E[Z_n]-E[\sigma^2+\mu^2])=\frac{1}{\epsilon}((\mu^2+\sigma^2)-(\mu^2+\sigma^2))[/math][math]=\frac{1}{\epsilon}(0)\rightarrow0[/math]

This is right?

Anyway it is a particular case of chebyshev's inequality?

[math]P(|X-a|\geq\epsilon)\leq\frac{1}{\epsilon^p}E[|X-a|^p][/math]

where [math]E[Z_n]=E[\frac{1}{n}\sum_{i=1}^nX_i^2]=\frac{1}{n}E[\sum_{i=1}^nX_i^2]=\frac{1}{n}*nE[X_1^2]=E[X_1^2]=\mu^2+\sigma^2[/math]

so

[math] \lim_{n \rightarrow \infty}P(|Z_n-(\mu^2+\sigma^2)|\geq\epsilon)\leq\frac{1}{\epsilon}(E[Z_n]-E[\sigma^2+\mu^2])=\frac{1}{\epsilon}((\mu^2+\sigma^2)-(\mu^2+\sigma^2))[/math][math]=\frac{1}{\epsilon}(0)\rightarrow0[/math]

This is right?

Anyway it is a particular case of chebyshev's inequality?

[math]P(|X-a|\geq\epsilon)\leq\frac{1}{\epsilon^p}E[|X-a|^p][/math]

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My last try

If [math]S_n=\sum_{i=1}^nX_i^2[/math] and [math]E[\frac{1}{n}\sum_{i=1}^nX_i^2]=\frac{1}{n}*nE[X_1^2]=\mu^2+\sigma^2[/math], so

[math]P(|Z_n-(\mu^2+\sigma^2|\geq\epsilon)=P(|\frac{S_n-E[S_n]|}{n}\geq\epsilon)=P(S_n-E[S_n])\geq n\epsilon)\leq\frac{Var(S_n)}{n^2\epsilon^2}\rightarrow0[/math] when [math]n\rightarrow\infty[/math]

So [math]Z_n \rightarrow^p \mu^2+\sigma^2[/math]

Is that right? If not please solve for me, because I do not know what to do

If [math]S_n=\sum_{i=1}^nX_i^2[/math] and [math]E[\frac{1}{n}\sum_{i=1}^nX_i^2]=\frac{1}{n}*nE[X_1^2]=\mu^2+\sigma^2[/math], so

[math]P(|Z_n-(\mu^2+\sigma^2|\geq\epsilon)=P(|\frac{S_n-E[S_n]|}{n}\geq\epsilon)=P(S_n-E[S_n])\geq n\epsilon)\leq\frac{Var(S_n)}{n^2\epsilon^2}\rightarrow0[/math] when [math]n\rightarrow\infty[/math]

So [math]Z_n \rightarrow^p \mu^2+\sigma^2[/math]

Is that right? If not please solve for me, because I do not know what to do

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