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5

1

$begingroup$

Let

• 
  $fin C^3(mathbb R)$ be positive


• $dinmathbb N$


• $xinmathbb R^d$


• 
  $ell>0$ and $sigma:=ell d^-alpha$ for some $alphain[0,1/2)$
  


• 
  $Z$ be a $mathbb R^d$-valued random variable distributed according to $mathcal N_d(0,I_d)$ and $Y:=x+sigma Z$
  

I want to show that $$liminf_dtoinftyoperatorname Eleft[1wedgeprod_i=1^dfracf(Y_i)f(x_i)right]=0tag1.$$

If necessary, assume that $g:=ln f$ is Lipschitz continuous. We may note that $$prod_i=1^dfracf(y_i)f(x_i)=expsum_i=1^d(g(y_i)-g(x_i));;;textfor all yinmathbb R^d.tag2$$ So, we may be able to show that $$liminf_dtoinftysum_i=1^d(g(Y_i)-g(x_i))=-inftytag3$$ in a suitable mode of convergence.

Is this possible and, if so, how can we conclude $(1)$ from that?

I guess we need to apply Taylor's theorem in a suitable way. If I apply the version of Taylor's theorem that I'm aware of literally, it yields that there is a $mathbb R^d$-valued random variable $W$ with $$W(omega)inlefttsigma Z(omega):tin[0,1]right;;;textfor all omegatag4$$ and $$f(Y)=f(x)+f'(x)sigma Z+fracf''(x)2(sigma Z)^2+frac16f'''(x+W)(sigma Z)^3tag5.$$ But this formulation is somehow awkward.

Remark: I've read a related result claiming that (by the central limit theorem) $ell^2d^-1sum_i=1^dfracf''(x_i)2Z_i^2$ converges in distribution to a Gaussian random variable with $0$ mean and variance $ell^2intfracf$. I would appreciate if someone could comment on how exactly we see this.

---

EDIT: As noted by saz, it seems like the claim is wrong in general. Actually, I was trying to understand the claim on slide 18 here. Therein, $x$ is replaced by a random variable distributed according to $plambda^d$, where $$p(x):=prod_i=1^df(x_i);;;textfor xinmathbb R^d,$$ $lambda^d$ denotes the Lebesgue measure on $mathcal B(mathbb R^d)$ and we assume that $int f(x):rm dxin(0,infty)$.

analysis probability-theory measure-theory asymptotics taylor-expansion

share|cite|improve this question

edited 9 hours ago

0xbadf00d

asked 17 hours ago

0xbadf00d0xbadf00d

1,91441532

$endgroup$

add a comment | 

5

1

$begingroup$

Let

• 
  $fin C^3(mathbb R)$ be positive


• $dinmathbb N$


• $xinmathbb R^d$


• 
  $ell>0$ and $sigma:=ell d^-alpha$ for some $alphain[0,1/2)$
  


• 
  $Z$ be a $mathbb R^d$-valued random variable distributed according to $mathcal N_d(0,I_d)$ and $Y:=x+sigma Z$
  

I want to show that $$liminf_dtoinftyoperatorname Eleft[1wedgeprod_i=1^dfracf(Y_i)f(x_i)right]=0tag1.$$

If necessary, assume that $g:=ln f$ is Lipschitz continuous. We may note that $$prod_i=1^dfracf(y_i)f(x_i)=expsum_i=1^d(g(y_i)-g(x_i));;;textfor all yinmathbb R^d.tag2$$ So, we may be able to show that $$liminf_dtoinftysum_i=1^d(g(Y_i)-g(x_i))=-inftytag3$$ in a suitable mode of convergence.

Is this possible and, if so, how can we conclude $(1)$ from that?

I guess we need to apply Taylor's theorem in a suitable way. If I apply the version of Taylor's theorem that I'm aware of literally, it yields that there is a $mathbb R^d$-valued random variable $W$ with $$W(omega)inlefttsigma Z(omega):tin[0,1]right;;;textfor all omegatag4$$ and $$f(Y)=f(x)+f'(x)sigma Z+fracf''(x)2(sigma Z)^2+frac16f'''(x+W)(sigma Z)^3tag5.$$ But this formulation is somehow awkward.

Remark: I've read a related result claiming that (by the central limit theorem) $ell^2d^-1sum_i=1^dfracf''(x_i)2Z_i^2$ converges in distribution to a Gaussian random variable with $0$ mean and variance $ell^2intfracf$. I would appreciate if someone could comment on how exactly we see this.

---

EDIT: As noted by saz, it seems like the claim is wrong in general. Actually, I was trying to understand the claim on slide 18 here. Therein, $x$ is replaced by a random variable distributed according to $plambda^d$, where $$p(x):=prod_i=1^df(x_i);;;textfor xinmathbb R^d,$$ $lambda^d$ denotes the Lebesgue measure on $mathcal B(mathbb R^d)$ and we assume that $int f(x):rm dxin(0,infty)$.

analysis probability-theory measure-theory asymptotics taylor-expansion

share|cite|improve this question

edited 9 hours ago

0xbadf00d

asked 17 hours ago

0xbadf00d0xbadf00d

1,91441532

$endgroup$

add a comment | 

$begingroup$

Let

• 
  $fin C^3(mathbb R)$ be positive


• $dinmathbb N$


• $xinmathbb R^d$


• 
  $ell>0$ and $sigma:=ell d^-alpha$ for some $alphain[0,1/2)$
  


• 
  $Z$ be a $mathbb R^d$-valued random variable distributed according to $mathcal N_d(0,I_d)$ and $Y:=x+sigma Z$
  

I want to show that $$liminf_dtoinftyoperatorname Eleft[1wedgeprod_i=1^dfracf(Y_i)f(x_i)right]=0tag1.$$

If necessary, assume that $g:=ln f$ is Lipschitz continuous. We may note that $$prod_i=1^dfracf(y_i)f(x_i)=expsum_i=1^d(g(y_i)-g(x_i));;;textfor all yinmathbb R^d.tag2$$ So, we may be able to show that $$liminf_dtoinftysum_i=1^d(g(Y_i)-g(x_i))=-inftytag3$$ in a suitable mode of convergence.

Is this possible and, if so, how can we conclude $(1)$ from that?

I guess we need to apply Taylor's theorem in a suitable way. If I apply the version of Taylor's theorem that I'm aware of literally, it yields that there is a $mathbb R^d$-valued random variable $W$ with $$W(omega)inlefttsigma Z(omega):tin[0,1]right;;;textfor all omegatag4$$ and $$f(Y)=f(x)+f'(x)sigma Z+fracf''(x)2(sigma Z)^2+frac16f'''(x+W)(sigma Z)^3tag5.$$ But this formulation is somehow awkward.

Remark: I've read a related result claiming that (by the central limit theorem) $ell^2d^-1sum_i=1^dfracf''(x_i)2Z_i^2$ converges in distribution to a Gaussian random variable with $0$ mean and variance $ell^2intfracf$. I would appreciate if someone could comment on how exactly we see this.

---

EDIT: As noted by saz, it seems like the claim is wrong in general. Actually, I was trying to understand the claim on slide 18 here. Therein, $x$ is replaced by a random variable distributed according to $plambda^d$, where $$p(x):=prod_i=1^df(x_i);;;textfor xinmathbb R^d,$$ $lambda^d$ denotes the Lebesgue measure on $mathcal B(mathbb R^d)$ and we assume that $int f(x):rm dxin(0,infty)$.

analysis probability-theory measure-theory asymptotics taylor-expansion

share|cite|improve this question

edited 9 hours ago

0xbadf00d

asked 17 hours ago

0xbadf00d0xbadf00d

1,91441532

$endgroup$

share|cite|improve this question

edited 9 hours ago

0xbadf00d

asked 17 hours ago

0xbadf00d0xbadf00d

1,91441532

share|cite|improve this question

edited 9 hours ago

0xbadf00d

asked 17 hours ago

0xbadf00d0xbadf00d

1,91441532

asked 17 hours ago

0xbadf00d0xbadf00d

1,91441532

asked 17 hours ago

0xbadf00d0xbadf00d

1,91441532