Limit in distribution of increasing variance normal random variable
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Suppose you have a sequence of normal random variables $X_n sim N(0, n)$. Is there any random variable $X$ such that $X_n to X$ in distribution.
Here, I use the following definition: $X_n to X$ in distribution if $P(X_n leq x) to P(X leq x) equiv F_X(x)$ for every $x$ for which $F_x$ is continuous.
functional-analysis stochastic-calculus
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$begingroup$
Suppose you have a sequence of normal random variables $X_n sim N(0, n)$. Is there any random variable $X$ such that $X_n to X$ in distribution.
Here, I use the following definition: $X_n to X$ in distribution if $P(X_n leq x) to P(X leq x) equiv F_X(x)$ for every $x$ for which $F_x$ is continuous.
functional-analysis stochastic-calculus
$endgroup$
add a comment |
$begingroup$
Suppose you have a sequence of normal random variables $X_n sim N(0, n)$. Is there any random variable $X$ such that $X_n to X$ in distribution.
Here, I use the following definition: $X_n to X$ in distribution if $P(X_n leq x) to P(X leq x) equiv F_X(x)$ for every $x$ for which $F_x$ is continuous.
functional-analysis stochastic-calculus
$endgroup$
Suppose you have a sequence of normal random variables $X_n sim N(0, n)$. Is there any random variable $X$ such that $X_n to X$ in distribution.
Here, I use the following definition: $X_n to X$ in distribution if $P(X_n leq x) to P(X leq x) equiv F_X(x)$ for every $x$ for which $F_x$ is continuous.
functional-analysis stochastic-calculus
functional-analysis stochastic-calculus
asked Jan 21 at 18:53
BrisãoBrisão
1799
1799
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$X_n overset{d}{=} sqrt{n} Z$ where $Z sim N(0, 1)$ so
$$P(X_n le x) = P(Z le x / sqrt{n}) to frac{1}{2}.$$
Now show that no CDF can equal $1/2$ wherever it is continuous.
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1 Answer
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1 Answer
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$begingroup$
$X_n overset{d}{=} sqrt{n} Z$ where $Z sim N(0, 1)$ so
$$P(X_n le x) = P(Z le x / sqrt{n}) to frac{1}{2}.$$
Now show that no CDF can equal $1/2$ wherever it is continuous.
$endgroup$
add a comment |
$begingroup$
$X_n overset{d}{=} sqrt{n} Z$ where $Z sim N(0, 1)$ so
$$P(X_n le x) = P(Z le x / sqrt{n}) to frac{1}{2}.$$
Now show that no CDF can equal $1/2$ wherever it is continuous.
$endgroup$
add a comment |
$begingroup$
$X_n overset{d}{=} sqrt{n} Z$ where $Z sim N(0, 1)$ so
$$P(X_n le x) = P(Z le x / sqrt{n}) to frac{1}{2}.$$
Now show that no CDF can equal $1/2$ wherever it is continuous.
$endgroup$
$X_n overset{d}{=} sqrt{n} Z$ where $Z sim N(0, 1)$ so
$$P(X_n le x) = P(Z le x / sqrt{n}) to frac{1}{2}.$$
Now show that no CDF can equal $1/2$ wherever it is continuous.
answered Jan 21 at 19:02
angryavianangryavian
42k23381
42k23381
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