Mixing
Levy's Continuity Theorem on Weak Convergence and Convergence in Distribution
$begingroup$
I am aware of the Levy's Continuity Theorem that states:
Let $(mu_{n})_{n in mathbb N}$ be probability measures on $(mathbb R^{d}, mathcal{B}(mathbb R^{d})$ then:
$mu_{n} xrightarrow{w} muiff hat{mu_{n}}(x)xrightarrow{n to
> infty}hat{mu}(x), forall x in mathbb R^{d}$
where $hat{mu_{n}}$ and $hat{mu}$ are the respective characteristic functions of $mu_{n}$ and $mu$
and in our case $mu_{n} xrightarrow{w} mu$ means that:
$forall f$ continuous and bounded on $mathbb R^{d}$, $int_{mathbb
R^{d}}f(x)dmu_{n}xrightarrow{n to infty}int_{mathbb
R^{d}}f(x)dmu$
I am told that when $d=1$:
$mu_{n} xrightarrow{w} mu iff F_{mu_{n}}(x)xrightarrow{n to infty}F_{mu}(x), forall x in mathbb R$
and this makes sense. But is $d=1$ really necessary, why would this not be the case if $d > 1$? Secondly, does this essentially mean
Weak Convergence $iff$ Convergence in Distribution
If you can help me in further understanding the subject, I would be extremely grateful
probability probability-theory probability-distributions stochastic-processes
asked Jan 30 at 21:12
SABOYSABOY
598311
$endgroup$
add a comment |
$begingroup$
I am aware of the Levy's Continuity Theorem that states:
Let $(mu_{n})_{n in mathbb N}$ be probability measures on $(mathbb R^{d}, mathcal{B}(mathbb R^{d})$ then:
$mu_{n} xrightarrow{w} muiff hat{mu_{n}}(x)xrightarrow{n to
> infty}hat{mu}(x), forall x in mathbb R^{d}$
where $hat{mu_{n}}$ and $hat{mu}$ are the respective characteristic functions of $mu_{n}$ and $mu$
and in our case $mu_{n} xrightarrow{w} mu$ means that:
$forall f$ continuous and bounded on $mathbb R^{d}$, $int_{mathbb
R^{d}}f(x)dmu_{n}xrightarrow{n to infty}int_{mathbb
R^{d}}f(x)dmu$
I am told that when $d=1$:
$mu_{n} xrightarrow{w} mu iff F_{mu_{n}}(x)xrightarrow{n to infty}F_{mu}(x), forall x in mathbb R$
and this makes sense. But is $d=1$ really necessary, why would this not be the case if $d > 1$? Secondly, does this essentially mean
Weak Convergence $iff$ Convergence in Distribution
If you can help me in further understanding the subject, I would be extremely grateful
probability probability-theory probability-distributions stochastic-processes
asked Jan 30 at 21:12
SABOYSABOY
598311
$endgroup$
2
$begingroup$
Your statement for $d=1$ is not correct; weak convergence is equivalent to $F_{mu_n}(x) to F(x)$ for all continuity points $x$ of $F$. Why not $d>1$? Well, in dimension $d>1$ there is simply not intrinsic definition for a distribution function. However, there are general statements which say that it suffices to check the convergence $int f , dmu_n to int f , dmu$ for a smaller class of test functions $f$ (e.g. smooth+bounded). Re your 2nd question: Yes, the two notions are equivalent.
$endgroup$
– saz
Jan 30 at 21:39
add a comment |
$begingroup$
I am aware of the Levy's Continuity Theorem that states:
Let $(mu_{n})_{n in mathbb N}$ be probability measures on $(mathbb R^{d}, mathcal{B}(mathbb R^{d})$ then:
$mu_{n} xrightarrow{w} muiff hat{mu_{n}}(x)xrightarrow{n to
> infty}hat{mu}(x), forall x in mathbb R^{d}$
where $hat{mu_{n}}$ and $hat{mu}$ are the respective characteristic functions of $mu_{n}$ and $mu$
and in our case $mu_{n} xrightarrow{w} mu$ means that:
$forall f$ continuous and bounded on $mathbb R^{d}$, $int_{mathbb
R^{d}}f(x)dmu_{n}xrightarrow{n to infty}int_{mathbb
R^{d}}f(x)dmu$
I am told that when $d=1$:
$mu_{n} xrightarrow{w} mu iff F_{mu_{n}}(x)xrightarrow{n to infty}F_{mu}(x), forall x in mathbb R$
and this makes sense. But is $d=1$ really necessary, why would this not be the case if $d > 1$? Secondly, does this essentially mean
Weak Convergence $iff$ Convergence in Distribution
If you can help me in further understanding the subject, I would be extremely grateful
probability probability-theory probability-distributions stochastic-processes
asked Jan 30 at 21:12
SABOYSABOY
598311
$endgroup$
I am aware of the Levy's Continuity Theorem that states:
Let $(mu_{n})_{n in mathbb N}$ be probability measures on $(mathbb R^{d}, mathcal{B}(mathbb R^{d})$ then:
$mu_{n} xrightarrow{w} muiff hat{mu_{n}}(x)xrightarrow{n to
> infty}hat{mu}(x), forall x in mathbb R^{d}$
where $hat{mu_{n}}$ and $hat{mu}$ are the respective characteristic functions of $mu_{n}$ and $mu$
and in our case $mu_{n} xrightarrow{w} mu$ means that:
$forall f$ continuous and bounded on $mathbb R^{d}$, $int_{mathbb
R^{d}}f(x)dmu_{n}xrightarrow{n to infty}int_{mathbb
R^{d}}f(x)dmu$
I am told that when $d=1$:
$mu_{n} xrightarrow{w} mu iff F_{mu_{n}}(x)xrightarrow{n to infty}F_{mu}(x), forall x in mathbb R$
and this makes sense. But is $d=1$ really necessary, why would this not be the case if $d > 1$? Secondly, does this essentially mean
Weak Convergence $iff$ Convergence in Distribution
If you can help me in further understanding the subject, I would be extremely grateful
probability probability-theory probability-distributions stochastic-processes
probability probability-theory probability-distributions stochastic-processes
asked Jan 30 at 21:12
SABOYSABOY
598311
asked Jan 30 at 21:12
SABOYSABOY
598311
asked Jan 30 at 21:12
SABOYSABOY
598311
asked Jan 30 at 21:12
SABOYSABOY
598311
asked Jan 30 at 21:12
SABOYSABOY
598311
598311
2
$begingroup$
Your statement for $d=1$ is not correct; weak convergence is equivalent to $F_{mu_n}(x) to F(x)$ for all continuity points $x$ of $F$. Why not $d>1$? Well, in dimension $d>1$ there is simply not intrinsic definition for a distribution function. However, there are general statements which say that it suffices to check the convergence $int f , dmu_n to int f , dmu$ for a smaller class of test functions $f$ (e.g. smooth+bounded). Re your 2nd question: Yes, the two notions are equivalent.
$endgroup$
– saz
Jan 30 at 21:39
add a comment |
2
$begingroup$
Your statement for $d=1$ is not correct; weak convergence is equivalent to $F_{mu_n}(x) to F(x)$ for all continuity points $x$ of $F$. Why not $d>1$? Well, in dimension $d>1$ there is simply not intrinsic definition for a distribution function. However, there are general statements which say that it suffices to check the convergence $int f , dmu_n to int f , dmu$ for a smaller class of test functions $f$ (e.g. smooth+bounded). Re your 2nd question: Yes, the two notions are equivalent.
$endgroup$
– saz
Jan 30 at 21:39
2
2
$begingroup$
Your statement for $d=1$ is not correct; weak convergence is equivalent to $F_{mu_n}(x) to F(x)$ for all continuity points $x$ of $F$. Why not $d>1$? Well, in dimension $d>1$ there is simply not intrinsic definition for a distribution function. However, there are general statements which say that it suffices to check the convergence $int f , dmu_n to int f , dmu$ for a smaller class of test functions $f$ (e.g. smooth+bounded). Re your 2nd question: Yes, the two notions are equivalent.
$endgroup$
– saz
Jan 30 at 21:39
$begingroup$
Your statement for $d=1$ is not correct; weak convergence is equivalent to $F_{mu_n}(x) to F(x)$ for all continuity points $x$ of $F$. Why not $d>1$? Well, in dimension $d>1$ there is simply not intrinsic definition for a distribution function. However, there are general statements which say that it suffices to check the convergence $int f , dmu_n to int f , dmu$ for a smaller class of test functions $f$ (e.g. smooth+bounded). Re your 2nd question: Yes, the two notions are equivalent.
$endgroup$
– saz
Jan 30 at 21:39
add a comment |
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$begingroup$
Your statement for $d=1$ is not correct; weak convergence is equivalent to $F_{mu_n}(x) to F(x)$ for all continuity points $x$ of $F$. Why not $d>1$? Well, in dimension $d>1$ there is simply not intrinsic definition for a distribution function. However, there are general statements which say that it suffices to check the convergence $int f , dmu_n to int f , dmu$ for a smaller class of test functions $f$ (e.g. smooth+bounded). Re your 2nd question: Yes, the two notions are equivalent.
$endgroup$
– saz
Jan 30 at 21:39