Weak LLN problem












0














Suppose $(X_n)$ is a sequence of r.v's satisfying $P(X_n=pmln (n))=frac{1}{2}$ for each $n=1,2dots$. I am trying to show that $(X_n)$ satisfies the weak LLN.



The idea is to show that $P(overline{X_n}>varepsilon)$ tends to 0, but i am unsure how to do so.






probability probability-theory stochastic-processes







share|cite|improve this question




















  • 1




    Are they independent?
    – John_Wick
    Dec 2 '18 at 6:59
















0














Suppose $(X_n)$ is a sequence of r.v's satisfying $P(X_n=pmln (n))=frac{1}{2}$ for each $n=1,2dots$. I am trying to show that $(X_n)$ satisfies the weak LLN.



The idea is to show that $P(overline{X_n}>varepsilon)$ tends to 0, but i am unsure how to do so.






probability probability-theory stochastic-processes







share|cite|improve this question




















  • 1




    Are they independent?
    – John_Wick
    Dec 2 '18 at 6:59














0












0








0







Suppose $(X_n)$ is a sequence of r.v's satisfying $P(X_n=pmln (n))=frac{1}{2}$ for each $n=1,2dots$. I am trying to show that $(X_n)$ satisfies the weak LLN.



The idea is to show that $P(overline{X_n}>varepsilon)$ tends to 0, but i am unsure how to do so.






probability probability-theory stochastic-processes







share|cite|improve this question















Suppose $(X_n)$ is a sequence of r.v's satisfying $P(X_n=pmln (n))=frac{1}{2}$ for each $n=1,2dots$. I am trying to show that $(X_n)$ satisfies the weak LLN.



The idea is to show that $P(overline{X_n}>varepsilon)$ tends to 0, but i am unsure how to do so.






probability probability-theory stochastic-processes




probability probability-theory stochastic-processes






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 2 '18 at 6:54









GNUSupporter 8964民主女神 地下教會

12.8k72445




12.8k72445










asked Dec 2 '18 at 6:53









user593295

627




627








  • 1




    Are they independent?
    – John_Wick
    Dec 2 '18 at 6:59














  • 1




    Are they independent?
    – John_Wick
    Dec 2 '18 at 6:59








1




1




Are they independent?
– John_Wick
Dec 2 '18 at 6:59




Are they independent?
– John_Wick
Dec 2 '18 at 6:59










1 Answer
1






active

oldest

votes


















3














If $X_n$ are independent, then $Var(X_n)=E(X_n^2)=(log n)^2.$ Then $P(|bar X_n|>epsilon)leq Var(bar X_n)/epsilon^2=frac{1}{n^2epsilon^2}sum_{i=1}^{n}Var(X_i)=frac{1}{n^2epsilon^2}sum_{i=1}^{n}(log i)^2leq frac{(log n)^2}{nepsilon^2}rightarrow 0.$






share|cite|improve this answer





















  • How did you compute the expectation (variance) like that? can i see the computation? Thanks again for the response
    – user593295
    Dec 2 '18 at 15:25










  • $E(X_n)=0$ and $Var(X_n)=E(X_n^2)-E^2(X_n)=E(X_n^2)=(log n)^2$
    – John_Wick
    Dec 2 '18 at 15:49










  • $Var(bar X_n)=Var(1/nsum_{i=1}^n X_i)=1/n^2sum_i Var(X_i)$ if $X_i$'s are independent.
    – John_Wick
    Dec 2 '18 at 15:50











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1 Answer
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1 Answer
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3














If $X_n$ are independent, then $Var(X_n)=E(X_n^2)=(log n)^2.$ Then $P(|bar X_n|>epsilon)leq Var(bar X_n)/epsilon^2=frac{1}{n^2epsilon^2}sum_{i=1}^{n}Var(X_i)=frac{1}{n^2epsilon^2}sum_{i=1}^{n}(log i)^2leq frac{(log n)^2}{nepsilon^2}rightarrow 0.$






share|cite|improve this answer





















  • How did you compute the expectation (variance) like that? can i see the computation? Thanks again for the response
    – user593295
    Dec 2 '18 at 15:25










  • $E(X_n)=0$ and $Var(X_n)=E(X_n^2)-E^2(X_n)=E(X_n^2)=(log n)^2$
    – John_Wick
    Dec 2 '18 at 15:49










  • $Var(bar X_n)=Var(1/nsum_{i=1}^n X_i)=1/n^2sum_i Var(X_i)$ if $X_i$'s are independent.
    – John_Wick
    Dec 2 '18 at 15:50
















3














If $X_n$ are independent, then $Var(X_n)=E(X_n^2)=(log n)^2.$ Then $P(|bar X_n|>epsilon)leq Var(bar X_n)/epsilon^2=frac{1}{n^2epsilon^2}sum_{i=1}^{n}Var(X_i)=frac{1}{n^2epsilon^2}sum_{i=1}^{n}(log i)^2leq frac{(log n)^2}{nepsilon^2}rightarrow 0.$






share|cite|improve this answer





















  • How did you compute the expectation (variance) like that? can i see the computation? Thanks again for the response
    – user593295
    Dec 2 '18 at 15:25










  • $E(X_n)=0$ and $Var(X_n)=E(X_n^2)-E^2(X_n)=E(X_n^2)=(log n)^2$
    – John_Wick
    Dec 2 '18 at 15:49










  • $Var(bar X_n)=Var(1/nsum_{i=1}^n X_i)=1/n^2sum_i Var(X_i)$ if $X_i$'s are independent.
    – John_Wick
    Dec 2 '18 at 15:50














3












3








3






If $X_n$ are independent, then $Var(X_n)=E(X_n^2)=(log n)^2.$ Then $P(|bar X_n|>epsilon)leq Var(bar X_n)/epsilon^2=frac{1}{n^2epsilon^2}sum_{i=1}^{n}Var(X_i)=frac{1}{n^2epsilon^2}sum_{i=1}^{n}(log i)^2leq frac{(log n)^2}{nepsilon^2}rightarrow 0.$






share|cite|improve this answer












If $X_n$ are independent, then $Var(X_n)=E(X_n^2)=(log n)^2.$ Then $P(|bar X_n|>epsilon)leq Var(bar X_n)/epsilon^2=frac{1}{n^2epsilon^2}sum_{i=1}^{n}Var(X_i)=frac{1}{n^2epsilon^2}sum_{i=1}^{n}(log i)^2leq frac{(log n)^2}{nepsilon^2}rightarrow 0.$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 2 '18 at 7:07









John_Wick

1,366111




1,366111












  • How did you compute the expectation (variance) like that? can i see the computation? Thanks again for the response
    – user593295
    Dec 2 '18 at 15:25










  • $E(X_n)=0$ and $Var(X_n)=E(X_n^2)-E^2(X_n)=E(X_n^2)=(log n)^2$
    – John_Wick
    Dec 2 '18 at 15:49










  • $Var(bar X_n)=Var(1/nsum_{i=1}^n X_i)=1/n^2sum_i Var(X_i)$ if $X_i$'s are independent.
    – John_Wick
    Dec 2 '18 at 15:50


















  • How did you compute the expectation (variance) like that? can i see the computation? Thanks again for the response
    – user593295
    Dec 2 '18 at 15:25










  • $E(X_n)=0$ and $Var(X_n)=E(X_n^2)-E^2(X_n)=E(X_n^2)=(log n)^2$
    – John_Wick
    Dec 2 '18 at 15:49










  • $Var(bar X_n)=Var(1/nsum_{i=1}^n X_i)=1/n^2sum_i Var(X_i)$ if $X_i$'s are independent.
    – John_Wick
    Dec 2 '18 at 15:50
















How did you compute the expectation (variance) like that? can i see the computation? Thanks again for the response
– user593295
Dec 2 '18 at 15:25




How did you compute the expectation (variance) like that? can i see the computation? Thanks again for the response
– user593295
Dec 2 '18 at 15:25












$E(X_n)=0$ and $Var(X_n)=E(X_n^2)-E^2(X_n)=E(X_n^2)=(log n)^2$
– John_Wick
Dec 2 '18 at 15:49




$E(X_n)=0$ and $Var(X_n)=E(X_n^2)-E^2(X_n)=E(X_n^2)=(log n)^2$
– John_Wick
Dec 2 '18 at 15:49












$Var(bar X_n)=Var(1/nsum_{i=1}^n X_i)=1/n^2sum_i Var(X_i)$ if $X_i$'s are independent.
– John_Wick
Dec 2 '18 at 15:50




$Var(bar X_n)=Var(1/nsum_{i=1}^n X_i)=1/n^2sum_i Var(X_i)$ if $X_i$'s are independent.
– John_Wick
Dec 2 '18 at 15:50


















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