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How does the inequality $|hat V_1-V|le max_1le ile nBig|frac11+tildelambda g_iBig|^2cdot|frac1nsum_i=1^ng_i^2- V|$ hold?

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1

$begingroup$

Suppose we have
$$frac1nsum_i=1^nfracg_i1+lambda g_i=0.$$
Since $frac11+ax=1-fracax(1+axi)^2$, $xi$ is between $0$ and $x$, we obtain
$$frac1nsum_i=1^nfracg_i1+lambda g_i=frac1nsum_i=1^ng_icdotBig[1-fraclambda g_i(1+tildelambda g_i)^2Big]=frac1nsum_i=1^ng_i-lambdacdotfrac1nsum_i=1^nfracg_i^2(1+tildelambda g_i)^2=0,$$
where $tildelambda$ is between $0$ and $lambda$.
Denote $hat V_1=frac1nsum_i=1^nfracg_i^2(1+tildelambda g_i)^2$, we have
$$lambda=hat V_1^-1cdotfrac1nsum_i=1^ng_i$$

Suppose $frac1nsum_i=1^ng_i^2oversetpto V$ and $max_1le ile n|lambda g_i|oversetpto 0$.
I saw in one paper that
$$|hat V_1-V|le max_1le ile nBig|frac11+tildelambda g_iBig|^2cdot|frac1nsum_i=1^ng_i^2- V|oversetpto0$$

My question is that how does this inequality hold? Does it have any trick?

---

If we use the approximation $frac11+axapprox 1-ax$, we chave
$$frac1nsum_i=1^nfracg_i1+lambda g_i=frac1nsum_i=1^ng_icdot(1-lambda g_i)=frac1nsum_i=1^ng_i-lambdacdotfrac1nsum_i=1^ng_i^2=0,$$
Denote $hat V_2=frac1nsum_i=1^ng_i^2$, under the assumption, we can directly obtain that
$$lambda=hat V_2^-1cdotfrac1nsum_i=1^ng_i ~~mboxand~~ hat V_2oversetpto V.$$

Is this observation equavalent to the inequality above?

probability probability-theory inequality convergence random-variables

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asked Mar 29 at 22:30

J.MikeJ.Mike

343110

$endgroup$

add a comment | 

1

$begingroup$

Suppose we have
$$frac1nsum_i=1^nfracg_i1+lambda g_i=0.$$
Since $frac11+ax=1-fracax(1+axi)^2$, $xi$ is between $0$ and $x$, we obtain
$$frac1nsum_i=1^nfracg_i1+lambda g_i=frac1nsum_i=1^ng_icdotBig[1-fraclambda g_i(1+tildelambda g_i)^2Big]=frac1nsum_i=1^ng_i-lambdacdotfrac1nsum_i=1^nfracg_i^2(1+tildelambda g_i)^2=0,$$
where $tildelambda$ is between $0$ and $lambda$.
Denote $hat V_1=frac1nsum_i=1^nfracg_i^2(1+tildelambda g_i)^2$, we have
$$lambda=hat V_1^-1cdotfrac1nsum_i=1^ng_i$$

Suppose $frac1nsum_i=1^ng_i^2oversetpto V$ and $max_1le ile n|lambda g_i|oversetpto 0$.
I saw in one paper that
$$|hat V_1-V|le max_1le ile nBig|frac11+tildelambda g_iBig|^2cdot|frac1nsum_i=1^ng_i^2- V|oversetpto0$$

My question is that how does this inequality hold? Does it have any trick?

---

If we use the approximation $frac11+axapprox 1-ax$, we chave
$$frac1nsum_i=1^nfracg_i1+lambda g_i=frac1nsum_i=1^ng_icdot(1-lambda g_i)=frac1nsum_i=1^ng_i-lambdacdotfrac1nsum_i=1^ng_i^2=0,$$
Denote $hat V_2=frac1nsum_i=1^ng_i^2$, under the assumption, we can directly obtain that
$$lambda=hat V_2^-1cdotfrac1nsum_i=1^ng_i ~~mboxand~~ hat V_2oversetpto V.$$

Is this observation equavalent to the inequality above?

probability probability-theory inequality convergence random-variables

share|cite|improve this question

asked Mar 29 at 22:30

J.MikeJ.Mike

343110

$endgroup$

add a comment | 

$begingroup$

Suppose we have
$$frac1nsum_i=1^nfracg_i1+lambda g_i=0.$$
Since $frac11+ax=1-fracax(1+axi)^2$, $xi$ is between $0$ and $x$, we obtain
$$frac1nsum_i=1^nfracg_i1+lambda g_i=frac1nsum_i=1^ng_icdotBig[1-fraclambda g_i(1+tildelambda g_i)^2Big]=frac1nsum_i=1^ng_i-lambdacdotfrac1nsum_i=1^nfracg_i^2(1+tildelambda g_i)^2=0,$$
where $tildelambda$ is between $0$ and $lambda$.
Denote $hat V_1=frac1nsum_i=1^nfracg_i^2(1+tildelambda g_i)^2$, we have
$$lambda=hat V_1^-1cdotfrac1nsum_i=1^ng_i$$

Suppose $frac1nsum_i=1^ng_i^2oversetpto V$ and $max_1le ile n|lambda g_i|oversetpto 0$.
I saw in one paper that
$$|hat V_1-V|le max_1le ile nBig|frac11+tildelambda g_iBig|^2cdot|frac1nsum_i=1^ng_i^2- V|oversetpto0$$

My question is that how does this inequality hold? Does it have any trick?

---

If we use the approximation $frac11+axapprox 1-ax$, we chave
$$frac1nsum_i=1^nfracg_i1+lambda g_i=frac1nsum_i=1^ng_icdot(1-lambda g_i)=frac1nsum_i=1^ng_i-lambdacdotfrac1nsum_i=1^ng_i^2=0,$$
Denote $hat V_2=frac1nsum_i=1^ng_i^2$, under the assumption, we can directly obtain that
$$lambda=hat V_2^-1cdotfrac1nsum_i=1^ng_i ~~mboxand~~ hat V_2oversetpto V.$$

Is this observation equavalent to the inequality above?

probability probability-theory inequality convergence random-variables

share|cite|improve this question

asked Mar 29 at 22:30

J.MikeJ.Mike

343110

$endgroup$

share|cite|improve this question

asked Mar 29 at 22:30

J.MikeJ.Mike

343110

share|cite|improve this question

asked Mar 29 at 22:30

J.MikeJ.Mike

343110

asked Mar 29 at 22:30

J.MikeJ.Mike

343110

asked Mar 29 at 22:30

J.MikeJ.Mike

343110