Is the limit of a sequence of characteristic functions of probability measures a characteristic function of a...
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Given a sequence of probability measure $P_n$ on $mathbb R$ with the Boreal sigma algebra, define $f_n = int e^{ixt}dp_n$, and suppose that $f_n to f$ for some $f$ bounded measurable, is it true that there exists some $p$ such that $f = int e^{ixt}dp$?
My intuition tells me that this is false, but I cannot come up with an example. The example that I tried: Let $p_n$ be the Gaussian with mean $0$, variance $frac{1}{n^2}$, then the characteristic function converges to the constant $1$. However, $1$ is the characteristic function of some dirac delta function.
real-analysis probability
asked Dec 7 '18 at 0:28
pennypenny
635
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add a comment |
$begingroup$
Given a sequence of probability measure $P_n$ on $mathbb R$ with the Boreal sigma algebra, define $f_n = int e^{ixt}dp_n$, and suppose that $f_n to f$ for some $f$ bounded measurable, is it true that there exists some $p$ such that $f = int e^{ixt}dp$?
My intuition tells me that this is false, but I cannot come up with an example. The example that I tried: Let $p_n$ be the Gaussian with mean $0$, variance $frac{1}{n^2}$, then the characteristic function converges to the constant $1$. However, $1$ is the characteristic function of some dirac delta function.
real-analysis probability
asked Dec 7 '18 at 0:28
pennypenny
635
$endgroup$
add a comment |
$begingroup$
Given a sequence of probability measure $P_n$ on $mathbb R$ with the Boreal sigma algebra, define $f_n = int e^{ixt}dp_n$, and suppose that $f_n to f$ for some $f$ bounded measurable, is it true that there exists some $p$ such that $f = int e^{ixt}dp$?
My intuition tells me that this is false, but I cannot come up with an example. The example that I tried: Let $p_n$ be the Gaussian with mean $0$, variance $frac{1}{n^2}$, then the characteristic function converges to the constant $1$. However, $1$ is the characteristic function of some dirac delta function.
real-analysis probability
asked Dec 7 '18 at 0:28
pennypenny
635
$endgroup$
Given a sequence of probability measure $P_n$ on $mathbb R$ with the Boreal sigma algebra, define $f_n = int e^{ixt}dp_n$, and suppose that $f_n to f$ for some $f$ bounded measurable, is it true that there exists some $p$ such that $f = int e^{ixt}dp$?
My intuition tells me that this is false, but I cannot come up with an example. The example that I tried: Let $p_n$ be the Gaussian with mean $0$, variance $frac{1}{n^2}$, then the characteristic function converges to the constant $1$. However, $1$ is the characteristic function of some dirac delta function.
real-analysis probability
real-analysis probability
asked Dec 7 '18 at 0:28
pennypenny
635
asked Dec 7 '18 at 0:28
pennypenny
635
asked Dec 7 '18 at 0:28
pennypenny
635
asked Dec 7 '18 at 0:28
pennypenny
635
asked Dec 7 '18 at 0:28
pennypenny
635
635
add a comment |
add a comment |
1 Answer
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Instead of having Gaussian with decreasing variance, let $p_n$ be Gaussian with mean 0 and variance $n$. This gives the counterexample you are looking for. In general the continuity theorem states that if the limiting function $f$ is continuous, then there exists such a $p$, but not otherwise.
answered Dec 7 '18 at 1:29
zoidbergzoidberg
1,070113
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$begingroup$
Instead of having Gaussian with decreasing variance, let $p_n$ be Gaussian with mean 0 and variance $n$. This gives the counterexample you are looking for. In general the continuity theorem states that if the limiting function $f$ is continuous, then there exists such a $p$, but not otherwise.
answered Dec 7 '18 at 1:29
zoidbergzoidberg
1,070113
$endgroup$
add a comment |
$begingroup$
Instead of having Gaussian with decreasing variance, let $p_n$ be Gaussian with mean 0 and variance $n$. This gives the counterexample you are looking for. In general the continuity theorem states that if the limiting function $f$ is continuous, then there exists such a $p$, but not otherwise.
answered Dec 7 '18 at 1:29
zoidbergzoidberg
1,070113
$endgroup$
add a comment |
$begingroup$
Instead of having Gaussian with decreasing variance, let $p_n$ be Gaussian with mean 0 and variance $n$. This gives the counterexample you are looking for. In general the continuity theorem states that if the limiting function $f$ is continuous, then there exists such a $p$, but not otherwise.
answered Dec 7 '18 at 1:29
zoidbergzoidberg
1,070113
$endgroup$
Instead of having Gaussian with decreasing variance, let $p_n$ be Gaussian with mean 0 and variance $n$. This gives the counterexample you are looking for. In general the continuity theorem states that if the limiting function $f$ is continuous, then there exists such a $p$, but not otherwise.
answered Dec 7 '18 at 1:29
zoidbergzoidberg
1,070113
answered Dec 7 '18 at 1:29
zoidbergzoidberg
1,070113
answered Dec 7 '18 at 1:29
zoidbergzoidberg
1,070113
answered Dec 7 '18 at 1:29
zoidbergzoidberg
1,070113
1,070113
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