Show $Z_{n} xrightarrow{d} mathcal{N}(0,1)$












0












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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
















0












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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














0












0








0





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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












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1 Answer
1






active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














1












1








1





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















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