limiting distribution and continuous transformation
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Assume that $sqrt{n}(b-beta)$ converges in distribution to a normal variable with zero mean and some variance S (here $b$ is a random variable and $beta$ is a constant, both are scalar valued). I want to find the limiting distribution of $sqrt{n}(e^{b}-e^beta)$.
I know the conversation property of convergence for continuous transformations, thus if I can rewrite $sqrt{n}(e^{b}-e^beta)$ in terms of $sqrt{n}(b-beta)$ then the limiting distribution is easy to find. But I couldn't write it in terms of $sqrt{n}(b-beta)$. Does anybody have any idea about how to proceed?
edit: I guess the delta method suits well into this case, I will proceed using the delta method.
statistics probability-distributions convergence
asked Nov 19 at 19:16
Osman Bulut
556
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up vote
0
down vote
favorite
Assume that $sqrt{n}(b-beta)$ converges in distribution to a normal variable with zero mean and some variance S (here $b$ is a random variable and $beta$ is a constant, both are scalar valued). I want to find the limiting distribution of $sqrt{n}(e^{b}-e^beta)$.
I know the conversation property of convergence for continuous transformations, thus if I can rewrite $sqrt{n}(e^{b}-e^beta)$ in terms of $sqrt{n}(b-beta)$ then the limiting distribution is easy to find. But I couldn't write it in terms of $sqrt{n}(b-beta)$. Does anybody have any idea about how to proceed?
edit: I guess the delta method suits well into this case, I will proceed using the delta method.
statistics probability-distributions convergence
asked Nov 19 at 19:16
Osman Bulut
556
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Assume that $sqrt{n}(b-beta)$ converges in distribution to a normal variable with zero mean and some variance S (here $b$ is a random variable and $beta$ is a constant, both are scalar valued). I want to find the limiting distribution of $sqrt{n}(e^{b}-e^beta)$.
I know the conversation property of convergence for continuous transformations, thus if I can rewrite $sqrt{n}(e^{b}-e^beta)$ in terms of $sqrt{n}(b-beta)$ then the limiting distribution is easy to find. But I couldn't write it in terms of $sqrt{n}(b-beta)$. Does anybody have any idea about how to proceed?
edit: I guess the delta method suits well into this case, I will proceed using the delta method.
statistics probability-distributions convergence
asked Nov 19 at 19:16
Osman Bulut
556
Assume that $sqrt{n}(b-beta)$ converges in distribution to a normal variable with zero mean and some variance S (here $b$ is a random variable and $beta$ is a constant, both are scalar valued). I want to find the limiting distribution of $sqrt{n}(e^{b}-e^beta)$.
I know the conversation property of convergence for continuous transformations, thus if I can rewrite $sqrt{n}(e^{b}-e^beta)$ in terms of $sqrt{n}(b-beta)$ then the limiting distribution is easy to find. But I couldn't write it in terms of $sqrt{n}(b-beta)$. Does anybody have any idea about how to proceed?
edit: I guess the delta method suits well into this case, I will proceed using the delta method.
statistics probability-distributions convergence
statistics probability-distributions convergence
asked Nov 19 at 19:16
Osman Bulut
556
asked Nov 19 at 19:16
Osman Bulut
556
edited Nov 19 at 19:53
asked Nov 19 at 19:16
Osman Bulut
556
asked Nov 19 at 19:16
Osman Bulut
556
asked Nov 19 at 19:16
Osman Bulut
556
556
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Yes, delta method applies here.
$$sqrt{n}(X_n - theta) xrightarrow{D} N(0, sigma^2)$$
then we have
$$sqrt{n}(g(X_n) - g(theta)) xrightarrow{D} N(0, sigma^2[g'(theta)]^2)$$
if $g'(theta)$ exists and non-zero.
Here, $sigma^2 = S, theta = beta$, $g(x)=exp(x)=g'(x)$.
Hence
$$sqrt{n}(e^{b_n} - e^beta) xrightarrow{D} N(0, Sexp(2beta)).$$
answered Nov 20 at 1:47
Siong Thye Goh
94.9k1462115
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Yes, delta method applies here.
$$sqrt{n}(X_n - theta) xrightarrow{D} N(0, sigma^2)$$
then we have
$$sqrt{n}(g(X_n) - g(theta)) xrightarrow{D} N(0, sigma^2[g'(theta)]^2)$$
if $g'(theta)$ exists and non-zero.
Here, $sigma^2 = S, theta = beta$, $g(x)=exp(x)=g'(x)$.
Hence
$$sqrt{n}(e^{b_n} - e^beta) xrightarrow{D} N(0, Sexp(2beta)).$$
answered Nov 20 at 1:47
Siong Thye Goh
94.9k1462115
add a comment |
up vote
1
down vote
accepted
Yes, delta method applies here.
$$sqrt{n}(X_n - theta) xrightarrow{D} N(0, sigma^2)$$
then we have
$$sqrt{n}(g(X_n) - g(theta)) xrightarrow{D} N(0, sigma^2[g'(theta)]^2)$$
if $g'(theta)$ exists and non-zero.
Here, $sigma^2 = S, theta = beta$, $g(x)=exp(x)=g'(x)$.
Hence
$$sqrt{n}(e^{b_n} - e^beta) xrightarrow{D} N(0, Sexp(2beta)).$$
answered Nov 20 at 1:47
Siong Thye Goh
94.9k1462115
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Yes, delta method applies here.
$$sqrt{n}(X_n - theta) xrightarrow{D} N(0, sigma^2)$$
then we have
$$sqrt{n}(g(X_n) - g(theta)) xrightarrow{D} N(0, sigma^2[g'(theta)]^2)$$
if $g'(theta)$ exists and non-zero.
Here, $sigma^2 = S, theta = beta$, $g(x)=exp(x)=g'(x)$.
Hence
$$sqrt{n}(e^{b_n} - e^beta) xrightarrow{D} N(0, Sexp(2beta)).$$
answered Nov 20 at 1:47
Siong Thye Goh
94.9k1462115
Yes, delta method applies here.
$$sqrt{n}(X_n - theta) xrightarrow{D} N(0, sigma^2)$$
then we have
$$sqrt{n}(g(X_n) - g(theta)) xrightarrow{D} N(0, sigma^2[g'(theta)]^2)$$
if $g'(theta)$ exists and non-zero.
Here, $sigma^2 = S, theta = beta$, $g(x)=exp(x)=g'(x)$.
Hence
$$sqrt{n}(e^{b_n} - e^beta) xrightarrow{D} N(0, Sexp(2beta)).$$
answered Nov 20 at 1:47
Siong Thye Goh
94.9k1462115
answered Nov 20 at 1:47
Siong Thye Goh
94.9k1462115
answered Nov 20 at 1:47
Siong Thye Goh
94.9k1462115
answered Nov 20 at 1:47
Siong Thye Goh
94.9k1462115
94.9k1462115
add a comment |
add a comment |
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