If $Φ$ is the cdf of the standard normal distribution and $φ_i:ℝ→ℝ$ is linear, are the 1st and 2nd...
$begingroup$
Let
$dinmathbb N$ with $d>1$
$lambda^d$ denote the Lebesuge measure on $mathcal Bleft(mathbb R^dright)$
$fin C^2(mathbb R)$ be positive and $$pi(x):=prod_{i=1}^df(x_i);;;text{for }xinmathbb R^d$$
$Phi$ denote the cdf of the standard normal distribution
$ell>0$ and $$sigma^2:=frac{ell^2}{d-1}$$
- $xinmathbb R^d$
$(Omega,mathcal A,operatorname P)$ be a probability space
$Y$ be a $mathbb R^d$-valued random variable on $(Omega,mathcal A,operatorname P)$ with $Ysimmathcal N(0,sigma I)$ (where $I$ denotes the $d$-dimensional identity matrix)
Now, let $$g(z):=frac1{d-1}sum_{i=2}^dleft|frac{f'(z_i)}{f(z_i)}right|^2;;;text{for }zinmathbb R^d,$$ $varepsilon:=g-g(x_1)$ and $$M(z):=Phileft(frac1{{g(x)}^{1/2}}left(frac zell-frac{ell g(x)}2right)right)+e^zPhileft(-frac z{ell{{g(x)}^{1/2}}}-frac{ell{g(x)}^{1/2}}2right);;;text{for }zinmathbb R.$$ Note that there is a real-valued random variable $W$ on $(Omega,mathcal A,operatorname P)$ with $$W(omega)inleft{(1-t)x_1+tY_1(omega):tin[0,1]right}tag1$$ and $$M(varepsilon(Y_1))=M(0)+frac12M(0)g'(x_1)(Y_1-x_1)+frac12left(T(W)right)(Y_1-x_1)^2tag2,$$ where $$T(w):=g''(w)M'(varepsilon(w))+{g'(w)}^2M''(varepsilon(w));;;text{for }winmathbb R.$$
How can we show that $T(W)$ is uniformly bounded with respect to $x$, if the domain of $x$ is restricted to a compact set?
Most probably, the question is highly overcomplicated. I've read that the claim follows by noting that $M'$ and $M''$ are bounded. However, if I'm not terribly wrong, $M'$ and $M''$ are unbounded. And even if they would be bounded, $g'$ and $g''$ (which also occur in the definition of $T$) should be unbounded ... So, I don't see what we need here.
probability-theory normal-distribution
asked Dec 11 '18 at 15:08
0xbadf00d0xbadf00d
1,91341531
$endgroup$
add a comment |
$begingroup$
Let
$dinmathbb N$ with $d>1$
$lambda^d$ denote the Lebesuge measure on $mathcal Bleft(mathbb R^dright)$
$fin C^2(mathbb R)$ be positive and $$pi(x):=prod_{i=1}^df(x_i);;;text{for }xinmathbb R^d$$
$Phi$ denote the cdf of the standard normal distribution
$ell>0$ and $$sigma^2:=frac{ell^2}{d-1}$$
- $xinmathbb R^d$
$(Omega,mathcal A,operatorname P)$ be a probability space
$Y$ be a $mathbb R^d$-valued random variable on $(Omega,mathcal A,operatorname P)$ with $Ysimmathcal N(0,sigma I)$ (where $I$ denotes the $d$-dimensional identity matrix)
Now, let $$g(z):=frac1{d-1}sum_{i=2}^dleft|frac{f'(z_i)}{f(z_i)}right|^2;;;text{for }zinmathbb R^d,$$ $varepsilon:=g-g(x_1)$ and $$M(z):=Phileft(frac1{{g(x)}^{1/2}}left(frac zell-frac{ell g(x)}2right)right)+e^zPhileft(-frac z{ell{{g(x)}^{1/2}}}-frac{ell{g(x)}^{1/2}}2right);;;text{for }zinmathbb R.$$ Note that there is a real-valued random variable $W$ on $(Omega,mathcal A,operatorname P)$ with $$W(omega)inleft{(1-t)x_1+tY_1(omega):tin[0,1]right}tag1$$ and $$M(varepsilon(Y_1))=M(0)+frac12M(0)g'(x_1)(Y_1-x_1)+frac12left(T(W)right)(Y_1-x_1)^2tag2,$$ where $$T(w):=g''(w)M'(varepsilon(w))+{g'(w)}^2M''(varepsilon(w));;;text{for }winmathbb R.$$
How can we show that $T(W)$ is uniformly bounded with respect to $x$, if the domain of $x$ is restricted to a compact set?
Most probably, the question is highly overcomplicated. I've read that the claim follows by noting that $M'$ and $M''$ are bounded. However, if I'm not terribly wrong, $M'$ and $M''$ are unbounded. And even if they would be bounded, $g'$ and $g''$ (which also occur in the definition of $T$) should be unbounded ... So, I don't see what we need here.
probability-theory normal-distribution
asked Dec 11 '18 at 15:08
0xbadf00d0xbadf00d
1,91341531
$endgroup$
add a comment |
$begingroup$
Let
$dinmathbb N$ with $d>1$
$lambda^d$ denote the Lebesuge measure on $mathcal Bleft(mathbb R^dright)$
$fin C^2(mathbb R)$ be positive and $$pi(x):=prod_{i=1}^df(x_i);;;text{for }xinmathbb R^d$$
$Phi$ denote the cdf of the standard normal distribution
$ell>0$ and $$sigma^2:=frac{ell^2}{d-1}$$
- $xinmathbb R^d$
$(Omega,mathcal A,operatorname P)$ be a probability space
$Y$ be a $mathbb R^d$-valued random variable on $(Omega,mathcal A,operatorname P)$ with $Ysimmathcal N(0,sigma I)$ (where $I$ denotes the $d$-dimensional identity matrix)
Now, let $$g(z):=frac1{d-1}sum_{i=2}^dleft|frac{f'(z_i)}{f(z_i)}right|^2;;;text{for }zinmathbb R^d,$$ $varepsilon:=g-g(x_1)$ and $$M(z):=Phileft(frac1{{g(x)}^{1/2}}left(frac zell-frac{ell g(x)}2right)right)+e^zPhileft(-frac z{ell{{g(x)}^{1/2}}}-frac{ell{g(x)}^{1/2}}2right);;;text{for }zinmathbb R.$$ Note that there is a real-valued random variable $W$ on $(Omega,mathcal A,operatorname P)$ with $$W(omega)inleft{(1-t)x_1+tY_1(omega):tin[0,1]right}tag1$$ and $$M(varepsilon(Y_1))=M(0)+frac12M(0)g'(x_1)(Y_1-x_1)+frac12left(T(W)right)(Y_1-x_1)^2tag2,$$ where $$T(w):=g''(w)M'(varepsilon(w))+{g'(w)}^2M''(varepsilon(w));;;text{for }winmathbb R.$$
How can we show that $T(W)$ is uniformly bounded with respect to $x$, if the domain of $x$ is restricted to a compact set?
Most probably, the question is highly overcomplicated. I've read that the claim follows by noting that $M'$ and $M''$ are bounded. However, if I'm not terribly wrong, $M'$ and $M''$ are unbounded. And even if they would be bounded, $g'$ and $g''$ (which also occur in the definition of $T$) should be unbounded ... So, I don't see what we need here.
probability-theory normal-distribution
asked Dec 11 '18 at 15:08
0xbadf00d0xbadf00d
1,91341531
$endgroup$
Let
$dinmathbb N$ with $d>1$
$lambda^d$ denote the Lebesuge measure on $mathcal Bleft(mathbb R^dright)$
$fin C^2(mathbb R)$ be positive and $$pi(x):=prod_{i=1}^df(x_i);;;text{for }xinmathbb R^d$$
$Phi$ denote the cdf of the standard normal distribution
$ell>0$ and $$sigma^2:=frac{ell^2}{d-1}$$
- $xinmathbb R^d$
$(Omega,mathcal A,operatorname P)$ be a probability space
$Y$ be a $mathbb R^d$-valued random variable on $(Omega,mathcal A,operatorname P)$ with $Ysimmathcal N(0,sigma I)$ (where $I$ denotes the $d$-dimensional identity matrix)
Now, let $$g(z):=frac1{d-1}sum_{i=2}^dleft|frac{f'(z_i)}{f(z_i)}right|^2;;;text{for }zinmathbb R^d,$$ $varepsilon:=g-g(x_1)$ and $$M(z):=Phileft(frac1{{g(x)}^{1/2}}left(frac zell-frac{ell g(x)}2right)right)+e^zPhileft(-frac z{ell{{g(x)}^{1/2}}}-frac{ell{g(x)}^{1/2}}2right);;;text{for }zinmathbb R.$$ Note that there is a real-valued random variable $W$ on $(Omega,mathcal A,operatorname P)$ with $$W(omega)inleft{(1-t)x_1+tY_1(omega):tin[0,1]right}tag1$$ and $$M(varepsilon(Y_1))=M(0)+frac12M(0)g'(x_1)(Y_1-x_1)+frac12left(T(W)right)(Y_1-x_1)^2tag2,$$ where $$T(w):=g''(w)M'(varepsilon(w))+{g'(w)}^2M''(varepsilon(w));;;text{for }winmathbb R.$$
How can we show that $T(W)$ is uniformly bounded with respect to $x$, if the domain of $x$ is restricted to a compact set?
Most probably, the question is highly overcomplicated. I've read that the claim follows by noting that $M'$ and $M''$ are bounded. However, if I'm not terribly wrong, $M'$ and $M''$ are unbounded. And even if they would be bounded, $g'$ and $g''$ (which also occur in the definition of $T$) should be unbounded ... So, I don't see what we need here.
probability-theory normal-distribution
probability-theory normal-distribution
asked Dec 11 '18 at 15:08
0xbadf00d0xbadf00d
1,91341531
asked Dec 11 '18 at 15:08
0xbadf00d0xbadf00d
1,91341531
asked Dec 11 '18 at 15:08
0xbadf00d0xbadf00d
1,91341531
asked Dec 11 '18 at 15:08
0xbadf00d0xbadf00d
1,91341531
asked Dec 11 '18 at 15:08
0xbadf00d0xbadf00d
1,91341531
1,91341531
add a comment |
add a comment |
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