Explain why the empirical distribution function $F_n$ is a reasonable approximation of $F_X$ for large $n$.












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Suppose you have a dataset $x_1, . . . , x_n$ which is a realization of a random sample from a distribution with distribution function $F_X$. Explain why the empirical distribution function $F_n$ is a reasonable approximation of $F_X$ for large $n$.




Can someone please explain or prove this sentence, I really don't know how to explain it, thanks in advance! :)






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  • 1




    $begingroup$
    Look at the Law of Large Numbers.
    $endgroup$
    – user321627
    Jan 26 at 23:32
















0












$begingroup$



Suppose you have a dataset $x_1, . . . , x_n$ which is a realization of a random sample from a distribution with distribution function $F_X$. Explain why the empirical distribution function $F_n$ is a reasonable approximation of $F_X$ for large $n$.




Can someone please explain or prove this sentence, I really don't know how to explain it, thanks in advance! :)






probability statistics







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Look at the Law of Large Numbers.
    $endgroup$
    – user321627
    Jan 26 at 23:32














0












0








0





$begingroup$



Suppose you have a dataset $x_1, . . . , x_n$ which is a realization of a random sample from a distribution with distribution function $F_X$. Explain why the empirical distribution function $F_n$ is a reasonable approximation of $F_X$ for large $n$.




Can someone please explain or prove this sentence, I really don't know how to explain it, thanks in advance! :)






probability statistics







share|cite|improve this question









$endgroup$





Suppose you have a dataset $x_1, . . . , x_n$ which is a realization of a random sample from a distribution with distribution function $F_X$. Explain why the empirical distribution function $F_n$ is a reasonable approximation of $F_X$ for large $n$.




Can someone please explain or prove this sentence, I really don't know how to explain it, thanks in advance! :)






probability statistics




probability statistics






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share|cite|improve this question




share|cite|improve this question










asked Jan 26 at 22:25









Luke MarciLuke Marci

856




856








  • 1




    $begingroup$
    Look at the Law of Large Numbers.
    $endgroup$
    – user321627
    Jan 26 at 23:32














  • 1




    $begingroup$
    Look at the Law of Large Numbers.
    $endgroup$
    – user321627
    Jan 26 at 23:32








1




1




$begingroup$
Look at the Law of Large Numbers.
$endgroup$
– user321627
Jan 26 at 23:32




$begingroup$
Look at the Law of Large Numbers.
$endgroup$
– user321627
Jan 26 at 23:32










1 Answer
1






active

oldest

votes


















1












$begingroup$

First you need to state clearly what is the empirical distribution $F_n$. This is the function $F_ncolon mathbb R longrightarrow [0,1]$ such that
$$F_n(t)=frac1ncdot #{kcolon x_kle t},$$
that is the number of observations smaller or equal than $t$ divided by the total number of observations. Note that since you can't predict the values of each $X_k$ precisely, before taking a sample the value $F_n(t)$ is a random variable (in fact, you get a different random variable for each value of $t$ considered).



The trick here is to define the variables $U_1,U_2,ldots,U_n$ as
$$U_k=left{begin{matrix}1& X_kle t\0 & X_k>t\end{matrix}right.,$$
which are Bernoulli variables of parameter
$$p=P(X_kle t)=F_X(t),$$
for every $k$; in particular, they are identically distributed and have a finite mean and variance.



Since these variables are also clearly independent, the (weak) Law of Large Numbers apply, which means that
$$frac1n sum_{k=1}^n U_kto E(U_k)$$
(as convergence in probability). Check out that the left hand side is actually the same as the empirical distribution at $t$, and by a property of the Bernoulli distribution, the right hand side is the actual distribution of $X$ at $t$. That is,
$$F_n(t)to F_X(t)$$
(in probability). And this happens for any $t$. So in a sense, for bigger and bigger $n$, the empirical distribution values tend to approximate the actual distribution values, for each point $tinmathbb R$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    thank you a lot for the answer and the long explanation, now it's very clear, have a nice day!
    $endgroup$
    – Luke Marci
    Jan 27 at 8:26











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1 Answer
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1












$begingroup$

First you need to state clearly what is the empirical distribution $F_n$. This is the function $F_ncolon mathbb R longrightarrow [0,1]$ such that
$$F_n(t)=frac1ncdot #{kcolon x_kle t},$$
that is the number of observations smaller or equal than $t$ divided by the total number of observations. Note that since you can't predict the values of each $X_k$ precisely, before taking a sample the value $F_n(t)$ is a random variable (in fact, you get a different random variable for each value of $t$ considered).



The trick here is to define the variables $U_1,U_2,ldots,U_n$ as
$$U_k=left{begin{matrix}1& X_kle t\0 & X_k>t\end{matrix}right.,$$
which are Bernoulli variables of parameter
$$p=P(X_kle t)=F_X(t),$$
for every $k$; in particular, they are identically distributed and have a finite mean and variance.



Since these variables are also clearly independent, the (weak) Law of Large Numbers apply, which means that
$$frac1n sum_{k=1}^n U_kto E(U_k)$$
(as convergence in probability). Check out that the left hand side is actually the same as the empirical distribution at $t$, and by a property of the Bernoulli distribution, the right hand side is the actual distribution of $X$ at $t$. That is,
$$F_n(t)to F_X(t)$$
(in probability). And this happens for any $t$. So in a sense, for bigger and bigger $n$, the empirical distribution values tend to approximate the actual distribution values, for each point $tinmathbb R$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    thank you a lot for the answer and the long explanation, now it's very clear, have a nice day!
    $endgroup$
    – Luke Marci
    Jan 27 at 8:26
















1












$begingroup$

First you need to state clearly what is the empirical distribution $F_n$. This is the function $F_ncolon mathbb R longrightarrow [0,1]$ such that
$$F_n(t)=frac1ncdot #{kcolon x_kle t},$$
that is the number of observations smaller or equal than $t$ divided by the total number of observations. Note that since you can't predict the values of each $X_k$ precisely, before taking a sample the value $F_n(t)$ is a random variable (in fact, you get a different random variable for each value of $t$ considered).



The trick here is to define the variables $U_1,U_2,ldots,U_n$ as
$$U_k=left{begin{matrix}1& X_kle t\0 & X_k>t\end{matrix}right.,$$
which are Bernoulli variables of parameter
$$p=P(X_kle t)=F_X(t),$$
for every $k$; in particular, they are identically distributed and have a finite mean and variance.



Since these variables are also clearly independent, the (weak) Law of Large Numbers apply, which means that
$$frac1n sum_{k=1}^n U_kto E(U_k)$$
(as convergence in probability). Check out that the left hand side is actually the same as the empirical distribution at $t$, and by a property of the Bernoulli distribution, the right hand side is the actual distribution of $X$ at $t$. That is,
$$F_n(t)to F_X(t)$$
(in probability). And this happens for any $t$. So in a sense, for bigger and bigger $n$, the empirical distribution values tend to approximate the actual distribution values, for each point $tinmathbb R$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    thank you a lot for the answer and the long explanation, now it's very clear, have a nice day!
    $endgroup$
    – Luke Marci
    Jan 27 at 8:26














1












1








1





$begingroup$

First you need to state clearly what is the empirical distribution $F_n$. This is the function $F_ncolon mathbb R longrightarrow [0,1]$ such that
$$F_n(t)=frac1ncdot #{kcolon x_kle t},$$
that is the number of observations smaller or equal than $t$ divided by the total number of observations. Note that since you can't predict the values of each $X_k$ precisely, before taking a sample the value $F_n(t)$ is a random variable (in fact, you get a different random variable for each value of $t$ considered).



The trick here is to define the variables $U_1,U_2,ldots,U_n$ as
$$U_k=left{begin{matrix}1& X_kle t\0 & X_k>t\end{matrix}right.,$$
which are Bernoulli variables of parameter
$$p=P(X_kle t)=F_X(t),$$
for every $k$; in particular, they are identically distributed and have a finite mean and variance.



Since these variables are also clearly independent, the (weak) Law of Large Numbers apply, which means that
$$frac1n sum_{k=1}^n U_kto E(U_k)$$
(as convergence in probability). Check out that the left hand side is actually the same as the empirical distribution at $t$, and by a property of the Bernoulli distribution, the right hand side is the actual distribution of $X$ at $t$. That is,
$$F_n(t)to F_X(t)$$
(in probability). And this happens for any $t$. So in a sense, for bigger and bigger $n$, the empirical distribution values tend to approximate the actual distribution values, for each point $tinmathbb R$.






share|cite|improve this answer











$endgroup$



First you need to state clearly what is the empirical distribution $F_n$. This is the function $F_ncolon mathbb R longrightarrow [0,1]$ such that
$$F_n(t)=frac1ncdot #{kcolon x_kle t},$$
that is the number of observations smaller or equal than $t$ divided by the total number of observations. Note that since you can't predict the values of each $X_k$ precisely, before taking a sample the value $F_n(t)$ is a random variable (in fact, you get a different random variable for each value of $t$ considered).



The trick here is to define the variables $U_1,U_2,ldots,U_n$ as
$$U_k=left{begin{matrix}1& X_kle t\0 & X_k>t\end{matrix}right.,$$
which are Bernoulli variables of parameter
$$p=P(X_kle t)=F_X(t),$$
for every $k$; in particular, they are identically distributed and have a finite mean and variance.



Since these variables are also clearly independent, the (weak) Law of Large Numbers apply, which means that
$$frac1n sum_{k=1}^n U_kto E(U_k)$$
(as convergence in probability). Check out that the left hand side is actually the same as the empirical distribution at $t$, and by a property of the Bernoulli distribution, the right hand side is the actual distribution of $X$ at $t$. That is,
$$F_n(t)to F_X(t)$$
(in probability). And this happens for any $t$. So in a sense, for bigger and bigger $n$, the empirical distribution values tend to approximate the actual distribution values, for each point $tinmathbb R$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 27 at 9:01

























answered Jan 27 at 5:53









Alejandro Nasif SalumAlejandro Nasif Salum

4,765118




4,765118












  • $begingroup$
    thank you a lot for the answer and the long explanation, now it's very clear, have a nice day!
    $endgroup$
    – Luke Marci
    Jan 27 at 8:26


















  • $begingroup$
    thank you a lot for the answer and the long explanation, now it's very clear, have a nice day!
    $endgroup$
    – Luke Marci
    Jan 27 at 8:26
















$begingroup$
thank you a lot for the answer and the long explanation, now it's very clear, have a nice day!
$endgroup$
– Luke Marci
Jan 27 at 8:26




$begingroup$
thank you a lot for the answer and the long explanation, now it's very clear, have a nice day!
$endgroup$
– Luke Marci
Jan 27 at 8:26


















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