Show $Z_{n} xrightarrow{d} mathcal{N}(0,1)$
$begingroup$
Let $(X_{k})_{k in mathbb N}$ a sequence of independent random variables and $F_{k}=F_{X_{k}}$ the respective cdf functions of $(X_{k})_{k in mathbb N}$ that are both continuous and strictly increasing.
Show that for $Z_{n}:=frac{1}{sqrt{n}}sum_{k=1}^{n}(1+log{(1-F_{k}(X_{k})))}$
$Z_{n} xrightarrow{d} mathcal{N}(0,1)$
My ideas:
I've realized there are two ways to show convergence in distribution, namely
through the
$i)$ Central Limit Theorem
$ii)$ or the very definition of the Cumulative distribution function
I suspect in this case, given the definition of $Z_{n}$, the CLT will be useful
It is clear since $(F_{k})_{k in mathbb N}$ are continuous that $(F_{k}(X_{k}))_{k in mathbb N}$ are independent random variables
The only problem is I see no way to get $(1+log{(1-F_{k}(X_{k})))}$ into the form $frac{F_{k}(X_{k})-mathbb E [F_{k}(X_{k})]}{sigma}$ that is necessary for the CLT.
Any ideas?
probability probability-theory probability-distributions independence central-limit-theorem
asked Jan 14 at 23:15
SABOYSABOY
600311
$endgroup$
add a comment |
$begingroup$
Let $(X_{k})_{k in mathbb N}$ a sequence of independent random variables and $F_{k}=F_{X_{k}}$ the respective cdf functions of $(X_{k})_{k in mathbb N}$ that are both continuous and strictly increasing.
Show that for $Z_{n}:=frac{1}{sqrt{n}}sum_{k=1}^{n}(1+log{(1-F_{k}(X_{k})))}$
$Z_{n} xrightarrow{d} mathcal{N}(0,1)$
My ideas:
I've realized there are two ways to show convergence in distribution, namely
through the
$i)$ Central Limit Theorem
$ii)$ or the very definition of the Cumulative distribution function
I suspect in this case, given the definition of $Z_{n}$, the CLT will be useful
It is clear since $(F_{k})_{k in mathbb N}$ are continuous that $(F_{k}(X_{k}))_{k in mathbb N}$ are independent random variables
The only problem is I see no way to get $(1+log{(1-F_{k}(X_{k})))}$ into the form $frac{F_{k}(X_{k})-mathbb E [F_{k}(X_{k})]}{sigma}$ that is necessary for the CLT.
Any ideas?
probability probability-theory probability-distributions independence central-limit-theorem
asked Jan 14 at 23:15
SABOYSABOY
600311
$endgroup$
$begingroup$
Hint: first show that $F_k(X_k)$ has the uniform distribution on the interval $[0,1]$. Then use this to calculate the mean and variance of $1+log(1-F_k(X_k))$.
$endgroup$
– Mike Earnest
Jan 14 at 23:34
add a comment |
$begingroup$
Let $(X_{k})_{k in mathbb N}$ a sequence of independent random variables and $F_{k}=F_{X_{k}}$ the respective cdf functions of $(X_{k})_{k in mathbb N}$ that are both continuous and strictly increasing.
Show that for $Z_{n}:=frac{1}{sqrt{n}}sum_{k=1}^{n}(1+log{(1-F_{k}(X_{k})))}$
$Z_{n} xrightarrow{d} mathcal{N}(0,1)$
My ideas:
I've realized there are two ways to show convergence in distribution, namely
through the
$i)$ Central Limit Theorem
$ii)$ or the very definition of the Cumulative distribution function
I suspect in this case, given the definition of $Z_{n}$, the CLT will be useful
It is clear since $(F_{k})_{k in mathbb N}$ are continuous that $(F_{k}(X_{k}))_{k in mathbb N}$ are independent random variables
The only problem is I see no way to get $(1+log{(1-F_{k}(X_{k})))}$ into the form $frac{F_{k}(X_{k})-mathbb E [F_{k}(X_{k})]}{sigma}$ that is necessary for the CLT.
Any ideas?
probability probability-theory probability-distributions independence central-limit-theorem
asked Jan 14 at 23:15
SABOYSABOY
600311
$endgroup$
Let $(X_{k})_{k in mathbb N}$ a sequence of independent random variables and $F_{k}=F_{X_{k}}$ the respective cdf functions of $(X_{k})_{k in mathbb N}$ that are both continuous and strictly increasing.
Show that for $Z_{n}:=frac{1}{sqrt{n}}sum_{k=1}^{n}(1+log{(1-F_{k}(X_{k})))}$
$Z_{n} xrightarrow{d} mathcal{N}(0,1)$
My ideas:
I've realized there are two ways to show convergence in distribution, namely
through the
$i)$ Central Limit Theorem
$ii)$ or the very definition of the Cumulative distribution function
I suspect in this case, given the definition of $Z_{n}$, the CLT will be useful
It is clear since $(F_{k})_{k in mathbb N}$ are continuous that $(F_{k}(X_{k}))_{k in mathbb N}$ are independent random variables
The only problem is I see no way to get $(1+log{(1-F_{k}(X_{k})))}$ into the form $frac{F_{k}(X_{k})-mathbb E [F_{k}(X_{k})]}{sigma}$ that is necessary for the CLT.
Any ideas?
probability probability-theory probability-distributions independence central-limit-theorem
probability probability-theory probability-distributions independence central-limit-theorem
asked Jan 14 at 23:15
SABOYSABOY
600311
asked Jan 14 at 23:15
SABOYSABOY
600311
asked Jan 14 at 23:15
SABOYSABOY
600311
asked Jan 14 at 23:15
SABOYSABOY
600311
asked Jan 14 at 23:15
SABOYSABOY
600311
600311
$begingroup$
Hint: first show that $F_k(X_k)$ has the uniform distribution on the interval $[0,1]$. Then use this to calculate the mean and variance of $1+log(1-F_k(X_k))$.
$endgroup$
– Mike Earnest
Jan 14 at 23:34
add a comment |
$begingroup$
Hint: first show that $F_k(X_k)$ has the uniform distribution on the interval $[0,1]$. Then use this to calculate the mean and variance of $1+log(1-F_k(X_k))$.
$endgroup$
– Mike Earnest
Jan 14 at 23:34
$begingroup$
Hint: first show that $F_k(X_k)$ has the uniform distribution on the interval $[0,1]$. Then use this to calculate the mean and variance of $1+log(1-F_k(X_k))$.
$endgroup$
– Mike Earnest
Jan 14 at 23:34
$begingroup$
Hint: first show that $F_k(X_k)$ has the uniform distribution on the interval $[0,1]$. Then use this to calculate the mean and variance of $1+log(1-F_k(X_k))$.
$endgroup$
– Mike Earnest
Jan 14 at 23:34
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hints: Let $U_k=F_k(X_k)$. Verify that $U_k$ 's are i.i.d with uniform distribution on $(0,1)$. Apply standard form of CLT.
answered Jan 14 at 23:42
Kavi Rama MurthyKavi Rama Murthy
76.6k53471
$endgroup$
$begingroup$
Surely $U_{k}=F_{k}(X_{k})=P(X_{k}leq X_{k})=1$ So, in order to find the distribution we look at the cdf of $U_{k}$, namely $F_{F_{k}(X_{k})}(c)=P(F_{k}(X_{k})leq c)=P(1 < c)$, how can I glean the distribution of $U_{k}$ from this?
$endgroup$
– SABOY
Jan 15 at 0:28
2
$begingroup$
$P{F_k(X_k) leq t}=P{ X_k leq F_k^{-1}(t)}= F_k(F_k^{-1}(t))=t$ for all $t in (0,1)$
$endgroup$
– Kavi Rama Murthy
Jan 15 at 0:36
$begingroup$
Why is it necessary that $F_{k}$ is strictly increasing and continuous?
$endgroup$
– SABOY
Jan 18 at 21:05
$begingroup$
Is it so that $F_{k}^{-1}(t)$ exists?
$endgroup$
– SABOY
Jan 18 at 21:11
$begingroup$
@SABOY Yes, that is the reason.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 23:07
add a comment |
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
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%2f3073881%2fshow-z-n-xrightarrowd-mathcaln0-1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hints: Let $U_k=F_k(X_k)$. Verify that $U_k$ 's are i.i.d with uniform distribution on $(0,1)$. Apply standard form of CLT.
answered Jan 14 at 23:42
Kavi Rama MurthyKavi Rama Murthy
76.6k53471
$endgroup$
$begingroup$
Surely $U_{k}=F_{k}(X_{k})=P(X_{k}leq X_{k})=1$ So, in order to find the distribution we look at the cdf of $U_{k}$, namely $F_{F_{k}(X_{k})}(c)=P(F_{k}(X_{k})leq c)=P(1 < c)$, how can I glean the distribution of $U_{k}$ from this?
$endgroup$
– SABOY
Jan 15 at 0:28
2
$begingroup$
$P{F_k(X_k) leq t}=P{ X_k leq F_k^{-1}(t)}= F_k(F_k^{-1}(t))=t$ for all $t in (0,1)$
$endgroup$
– Kavi Rama Murthy
Jan 15 at 0:36
$begingroup$
Why is it necessary that $F_{k}$ is strictly increasing and continuous?
$endgroup$
– SABOY
Jan 18 at 21:05
$begingroup$
Is it so that $F_{k}^{-1}(t)$ exists?
$endgroup$
– SABOY
Jan 18 at 21:11
$begingroup$
@SABOY Yes, that is the reason.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 23:07
add a comment |
$begingroup$
Hints: Let $U_k=F_k(X_k)$. Verify that $U_k$ 's are i.i.d with uniform distribution on $(0,1)$. Apply standard form of CLT.
answered Jan 14 at 23:42
Kavi Rama MurthyKavi Rama Murthy
76.6k53471
$endgroup$
$begingroup$
Surely $U_{k}=F_{k}(X_{k})=P(X_{k}leq X_{k})=1$ So, in order to find the distribution we look at the cdf of $U_{k}$, namely $F_{F_{k}(X_{k})}(c)=P(F_{k}(X_{k})leq c)=P(1 < c)$, how can I glean the distribution of $U_{k}$ from this?
$endgroup$
– SABOY
Jan 15 at 0:28
2
$begingroup$
$P{F_k(X_k) leq t}=P{ X_k leq F_k^{-1}(t)}= F_k(F_k^{-1}(t))=t$ for all $t in (0,1)$
$endgroup$
– Kavi Rama Murthy
Jan 15 at 0:36
$begingroup$
Why is it necessary that $F_{k}$ is strictly increasing and continuous?
$endgroup$
– SABOY
Jan 18 at 21:05
$begingroup$
Is it so that $F_{k}^{-1}(t)$ exists?
$endgroup$
– SABOY
Jan 18 at 21:11
$begingroup$
@SABOY Yes, that is the reason.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 23:07
add a comment |
$begingroup$
Hints: Let $U_k=F_k(X_k)$. Verify that $U_k$ 's are i.i.d with uniform distribution on $(0,1)$. Apply standard form of CLT.
answered Jan 14 at 23:42
Kavi Rama MurthyKavi Rama Murthy
76.6k53471
$endgroup$
Hints: Let $U_k=F_k(X_k)$. Verify that $U_k$ 's are i.i.d with uniform distribution on $(0,1)$. Apply standard form of CLT.
answered Jan 14 at 23:42
Kavi Rama MurthyKavi Rama Murthy
76.6k53471
answered Jan 14 at 23:42
Kavi Rama MurthyKavi Rama Murthy
76.6k53471
answered Jan 14 at 23:42
Kavi Rama MurthyKavi Rama Murthy
76.6k53471
answered Jan 14 at 23:42
Kavi Rama MurthyKavi Rama Murthy
76.6k53471
76.6k53471
$begingroup$
Surely $U_{k}=F_{k}(X_{k})=P(X_{k}leq X_{k})=1$ So, in order to find the distribution we look at the cdf of $U_{k}$, namely $F_{F_{k}(X_{k})}(c)=P(F_{k}(X_{k})leq c)=P(1 < c)$, how can I glean the distribution of $U_{k}$ from this?
$endgroup$
– SABOY
Jan 15 at 0:28
2
$begingroup$
$P{F_k(X_k) leq t}=P{ X_k leq F_k^{-1}(t)}= F_k(F_k^{-1}(t))=t$ for all $t in (0,1)$
$endgroup$
– Kavi Rama Murthy
Jan 15 at 0:36
$begingroup$
Why is it necessary that $F_{k}$ is strictly increasing and continuous?
$endgroup$
– SABOY
Jan 18 at 21:05
$begingroup$
Is it so that $F_{k}^{-1}(t)$ exists?
$endgroup$
– SABOY
Jan 18 at 21:11
$begingroup$
@SABOY Yes, that is the reason.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 23:07
add a comment |
$begingroup$
Surely $U_{k}=F_{k}(X_{k})=P(X_{k}leq X_{k})=1$ So, in order to find the distribution we look at the cdf of $U_{k}$, namely $F_{F_{k}(X_{k})}(c)=P(F_{k}(X_{k})leq c)=P(1 < c)$, how can I glean the distribution of $U_{k}$ from this?
$endgroup$
– SABOY
Jan 15 at 0:28
2
$begingroup$
$P{F_k(X_k) leq t}=P{ X_k leq F_k^{-1}(t)}= F_k(F_k^{-1}(t))=t$ for all $t in (0,1)$
$endgroup$
– Kavi Rama Murthy
Jan 15 at 0:36
$begingroup$
Why is it necessary that $F_{k}$ is strictly increasing and continuous?
$endgroup$
– SABOY
Jan 18 at 21:05
$begingroup$
Is it so that $F_{k}^{-1}(t)$ exists?
$endgroup$
– SABOY
Jan 18 at 21:11
$begingroup$
@SABOY Yes, that is the reason.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 23:07
$begingroup$
Surely $U_{k}=F_{k}(X_{k})=P(X_{k}leq X_{k})=1$ So, in order to find the distribution we look at the cdf of $U_{k}$, namely $F_{F_{k}(X_{k})}(c)=P(F_{k}(X_{k})leq c)=P(1 < c)$, how can I glean the distribution of $U_{k}$ from this?
$endgroup$
– SABOY
Jan 15 at 0:28
$begingroup$
Surely $U_{k}=F_{k}(X_{k})=P(X_{k}leq X_{k})=1$ So, in order to find the distribution we look at the cdf of $U_{k}$, namely $F_{F_{k}(X_{k})}(c)=P(F_{k}(X_{k})leq c)=P(1 < c)$, how can I glean the distribution of $U_{k}$ from this?
$endgroup$
– SABOY
Jan 15 at 0:28
2
2
$begingroup$
$P{F_k(X_k) leq t}=P{ X_k leq F_k^{-1}(t)}= F_k(F_k^{-1}(t))=t$ for all $t in (0,1)$
$endgroup$
– Kavi Rama Murthy
Jan 15 at 0:36
$begingroup$
$P{F_k(X_k) leq t}=P{ X_k leq F_k^{-1}(t)}= F_k(F_k^{-1}(t))=t$ for all $t in (0,1)$
$endgroup$
– Kavi Rama Murthy
Jan 15 at 0:36
$begingroup$
Why is it necessary that $F_{k}$ is strictly increasing and continuous?
$endgroup$
– SABOY
Jan 18 at 21:05
$begingroup$
Why is it necessary that $F_{k}$ is strictly increasing and continuous?
$endgroup$
– SABOY
Jan 18 at 21:05
$begingroup$
Is it so that $F_{k}^{-1}(t)$ exists?
$endgroup$
– SABOY
Jan 18 at 21:11
$begingroup$
Is it so that $F_{k}^{-1}(t)$ exists?
$endgroup$
– SABOY
Jan 18 at 21:11
$begingroup$
@SABOY Yes, that is the reason.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 23:07
$begingroup$
@SABOY Yes, that is the reason.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 23:07
add a comment |
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.
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%2f3073881%2fshow-z-n-xrightarrowd-mathcaln0-1%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
$begingroup$
Hint: first show that $F_k(X_k)$ has the uniform distribution on the interval $[0,1]$. Then use this to calculate the mean and variance of $1+log(1-F_k(X_k))$.
$endgroup$
– Mike Earnest
Jan 14 at 23:34