Proving Convergence in Distribution
up vote
0
down vote
favorite
Say $X_n sim binomial(n,p)$ and $Y_n =frac{X_n}{n}$. How would I go about showing $Z_n = frac{sqrt n (Y_n - p)}{sqrt {Y_n(1-Y_n)}} xrightarrow{D} Z$ if $Z sim N(0,1)$
I was thinking maybe I should try and show convergence in probability first using
some combination of the Central Limit Theorem and Slutsky theorem, but I'm mostly confused on how I would implement them.
Thanks in advance!
probability-theory
edited Dec 2 at 4:53
GNUSupporter 8964民主女神 地下教會
12.6k72445
asked Dec 2 at 4:38
clovis
459
add a comment |
up vote
0
down vote
favorite
Say $X_n sim binomial(n,p)$ and $Y_n =frac{X_n}{n}$. How would I go about showing $Z_n = frac{sqrt n (Y_n - p)}{sqrt {Y_n(1-Y_n)}} xrightarrow{D} Z$ if $Z sim N(0,1)$
I was thinking maybe I should try and show convergence in probability first using
some combination of the Central Limit Theorem and Slutsky theorem, but I'm mostly confused on how I would implement them.
Thanks in advance!
probability-theory
edited Dec 2 at 4:53
GNUSupporter 8964民主女神 地下教會
12.6k72445
asked Dec 2 at 4:38
clovis
459
How is $Z_n$ defined when $Y_n=0$ or $Y_n=1$?
– Michael
Dec 2 at 4:59
2
Assuming $$Z_n = left{ begin{array}{ll} frac{sqrt{n}(Y_n-p)}{sqrt{Y_n(1-Y_n)}} &mbox{ if $Y_n notin {0,1}$} \ 0 & mbox{ otherwise} end{array} right.$$ Then you can note $Y_n$ has the same distribution as $frac{1}{n}sum_{i=1}^n W_i$ with ${W_i}$ iid Bernoulli, and Slutsky directly applies.
– Michael
Dec 2 at 5:07
@Michael I see, so since $W_n$ converges in probability to $W$ then we can say say $Z_n$ does the same, and therefore converges in distribution to $Z$? (I'm not sure how to actually use Slutsky, I only know it as a theorem)
– clovis
Dec 2 at 5:31
1
${W_n}$ does not converge in probability to anything, but it is an i.i.d. process to which you can apply the law of large numbers and the central limit theorem.
– Michael
Dec 2 at 14:41
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Say $X_n sim binomial(n,p)$ and $Y_n =frac{X_n}{n}$. How would I go about showing $Z_n = frac{sqrt n (Y_n - p)}{sqrt {Y_n(1-Y_n)}} xrightarrow{D} Z$ if $Z sim N(0,1)$
I was thinking maybe I should try and show convergence in probability first using
some combination of the Central Limit Theorem and Slutsky theorem, but I'm mostly confused on how I would implement them.
Thanks in advance!
probability-theory
edited Dec 2 at 4:53
GNUSupporter 8964民主女神 地下教會
12.6k72445
asked Dec 2 at 4:38
clovis
459
Say $X_n sim binomial(n,p)$ and $Y_n =frac{X_n}{n}$. How would I go about showing $Z_n = frac{sqrt n (Y_n - p)}{sqrt {Y_n(1-Y_n)}} xrightarrow{D} Z$ if $Z sim N(0,1)$
I was thinking maybe I should try and show convergence in probability first using
some combination of the Central Limit Theorem and Slutsky theorem, but I'm mostly confused on how I would implement them.
Thanks in advance!
probability-theory
probability-theory
edited Dec 2 at 4:53
GNUSupporter 8964民主女神 地下教會
12.6k72445
asked Dec 2 at 4:38
clovis
459
edited Dec 2 at 4:53
GNUSupporter 8964民主女神 地下教會
12.6k72445
asked Dec 2 at 4:38
clovis
459
edited Dec 2 at 4:53
GNUSupporter 8964民主女神 地下教會
12.6k72445
edited Dec 2 at 4:53
GNUSupporter 8964民主女神 地下教會
12.6k72445
edited Dec 2 at 4:53
GNUSupporter 8964民主女神 地下教會
12.6k72445
12.6k72445
asked Dec 2 at 4:38
clovis
459
asked Dec 2 at 4:38
clovis
459
asked Dec 2 at 4:38
clovis
459
459
How is $Z_n$ defined when $Y_n=0$ or $Y_n=1$?
– Michael
Dec 2 at 4:59
2
Assuming $$Z_n = left{ begin{array}{ll} frac{sqrt{n}(Y_n-p)}{sqrt{Y_n(1-Y_n)}} &mbox{ if $Y_n notin {0,1}$} \ 0 & mbox{ otherwise} end{array} right.$$ Then you can note $Y_n$ has the same distribution as $frac{1}{n}sum_{i=1}^n W_i$ with ${W_i}$ iid Bernoulli, and Slutsky directly applies.
– Michael
Dec 2 at 5:07
@Michael I see, so since $W_n$ converges in probability to $W$ then we can say say $Z_n$ does the same, and therefore converges in distribution to $Z$? (I'm not sure how to actually use Slutsky, I only know it as a theorem)
– clovis
Dec 2 at 5:31
1
${W_n}$ does not converge in probability to anything, but it is an i.i.d. process to which you can apply the law of large numbers and the central limit theorem.
– Michael
Dec 2 at 14:41
add a comment |
How is $Z_n$ defined when $Y_n=0$ or $Y_n=1$?
– Michael
Dec 2 at 4:59
2
Assuming $$Z_n = left{ begin{array}{ll} frac{sqrt{n}(Y_n-p)}{sqrt{Y_n(1-Y_n)}} &mbox{ if $Y_n notin {0,1}$} \ 0 & mbox{ otherwise} end{array} right.$$ Then you can note $Y_n$ has the same distribution as $frac{1}{n}sum_{i=1}^n W_i$ with ${W_i}$ iid Bernoulli, and Slutsky directly applies.
– Michael
Dec 2 at 5:07
@Michael I see, so since $W_n$ converges in probability to $W$ then we can say say $Z_n$ does the same, and therefore converges in distribution to $Z$? (I'm not sure how to actually use Slutsky, I only know it as a theorem)
– clovis
Dec 2 at 5:31
1
${W_n}$ does not converge in probability to anything, but it is an i.i.d. process to which you can apply the law of large numbers and the central limit theorem.
– Michael
Dec 2 at 14:41
How is $Z_n$ defined when $Y_n=0$ or $Y_n=1$?
– Michael
Dec 2 at 4:59
How is $Z_n$ defined when $Y_n=0$ or $Y_n=1$?
– Michael
Dec 2 at 4:59
2
2
Assuming $$Z_n = left{ begin{array}{ll} frac{sqrt{n}(Y_n-p)}{sqrt{Y_n(1-Y_n)}} &mbox{ if $Y_n notin {0,1}$} \ 0 & mbox{ otherwise} end{array} right.$$ Then you can note $Y_n$ has the same distribution as $frac{1}{n}sum_{i=1}^n W_i$ with ${W_i}$ iid Bernoulli, and Slutsky directly applies.
– Michael
Dec 2 at 5:07
Assuming $$Z_n = left{ begin{array}{ll} frac{sqrt{n}(Y_n-p)}{sqrt{Y_n(1-Y_n)}} &mbox{ if $Y_n notin {0,1}$} \ 0 & mbox{ otherwise} end{array} right.$$ Then you can note $Y_n$ has the same distribution as $frac{1}{n}sum_{i=1}^n W_i$ with ${W_i}$ iid Bernoulli, and Slutsky directly applies.
– Michael
Dec 2 at 5:07
@Michael I see, so since $W_n$ converges in probability to $W$ then we can say say $Z_n$ does the same, and therefore converges in distribution to $Z$? (I'm not sure how to actually use Slutsky, I only know it as a theorem)
– clovis
Dec 2 at 5:31
@Michael I see, so since $W_n$ converges in probability to $W$ then we can say say $Z_n$ does the same, and therefore converges in distribution to $Z$? (I'm not sure how to actually use Slutsky, I only know it as a theorem)
– clovis
Dec 2 at 5:31
1
1
${W_n}$ does not converge in probability to anything, but it is an i.i.d. process to which you can apply the law of large numbers and the central limit theorem.
– Michael
Dec 2 at 14:41
${W_n}$ does not converge in probability to anything, but it is an i.i.d. process to which you can apply the law of large numbers and the central limit theorem.
– Michael
Dec 2 at 14:41
add a comment |
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3022240%2fproving-convergence-in-distribution%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
How is $Z_n$ defined when $Y_n=0$ or $Y_n=1$?
– Michael
Dec 2 at 4:59
2
Assuming $$Z_n = left{ begin{array}{ll} frac{sqrt{n}(Y_n-p)}{sqrt{Y_n(1-Y_n)}} &mbox{ if $Y_n notin {0,1}$} \ 0 & mbox{ otherwise} end{array} right.$$ Then you can note $Y_n$ has the same distribution as $frac{1}{n}sum_{i=1}^n W_i$ with ${W_i}$ iid Bernoulli, and Slutsky directly applies.
– Michael
Dec 2 at 5:07
@Michael I see, so since $W_n$ converges in probability to $W$ then we can say say $Z_n$ does the same, and therefore converges in distribution to $Z$? (I'm not sure how to actually use Slutsky, I only know it as a theorem)
– clovis
Dec 2 at 5:31
1
${W_n}$ does not converge in probability to anything, but it is an i.i.d. process to which you can apply the law of large numbers and the central limit theorem.
– Michael
Dec 2 at 14:41