Can Convergence in probability in this problem be reinforced to Almost sure convergence
$X_1,X_2,…,X_n$ are independently and identically distributed and $E(X_i)$ exists, $mu_n=E(X_n I(X_n le n)),S_n=sum_{i=1}^n X_i$.
Proof:$$frac{S_n}{n}-mu_noverset{p}{to }0$$
My answer is:
$$frac{S_n}{n}-mu_n=(frac{S_n}{n}-E(X_i))+(E(X_i)-mu_n)$$
According to the law of large numbers, $frac{S_n}{n}-E(X_i)overset{p}{to }0$, and it is easy to proof that $E(X_i)-mu_n to 0$, so the proposition is proved.
Because its form is also very close to the strong law of large numbers, I wonder if $frac{S_n}{n}-mu_noverset{a.s.}{to }0$, too.
Assumptions:
The following proposition has been proven that $$X_noverset{p}{to }X,Y_noverset{p}{to }Y Rightarrow X_n+Y_noverset{p}{to }X+Y$$
It is also established for subtraction, multiplication, and division. However, I don't know if it is established when $X_n,Y_n$ converge almost surely.
probability-theory convergence
asked Dec 8 at 9:22
李子涵
62
add a comment |
$X_1,X_2,…,X_n$ are independently and identically distributed and $E(X_i)$ exists, $mu_n=E(X_n I(X_n le n)),S_n=sum_{i=1}^n X_i$.
Proof:$$frac{S_n}{n}-mu_noverset{p}{to }0$$
My answer is:
$$frac{S_n}{n}-mu_n=(frac{S_n}{n}-E(X_i))+(E(X_i)-mu_n)$$
According to the law of large numbers, $frac{S_n}{n}-E(X_i)overset{p}{to }0$, and it is easy to proof that $E(X_i)-mu_n to 0$, so the proposition is proved.
Because its form is also very close to the strong law of large numbers, I wonder if $frac{S_n}{n}-mu_noverset{a.s.}{to }0$, too.
Assumptions:
The following proposition has been proven that $$X_noverset{p}{to }X,Y_noverset{p}{to }Y Rightarrow X_n+Y_noverset{p}{to }X+Y$$
It is also established for subtraction, multiplication, and division. However, I don't know if it is established when $X_n,Y_n$ converge almost surely.
probability-theory convergence
asked Dec 8 at 9:22
李子涵
62
Yes. it also holds for almost everywhere convergence case. It is a consequence of Continuous mapping theorem.
– Song
Dec 8 at 9:59
If $X_n to X$ almost surely and $Y_n to Y$ almost surely then $X_n+Y_n to X+Y$ almost surely. This hardly requires any proof. It follows from definition of almost sure convergence and the fact that union of two sets of measure $0$ has measure $0$.
– Kavi Rama Murthy
Dec 8 at 12:15
add a comment |
$X_1,X_2,…,X_n$ are independently and identically distributed and $E(X_i)$ exists, $mu_n=E(X_n I(X_n le n)),S_n=sum_{i=1}^n X_i$.
Proof:$$frac{S_n}{n}-mu_noverset{p}{to }0$$
My answer is:
$$frac{S_n}{n}-mu_n=(frac{S_n}{n}-E(X_i))+(E(X_i)-mu_n)$$
According to the law of large numbers, $frac{S_n}{n}-E(X_i)overset{p}{to }0$, and it is easy to proof that $E(X_i)-mu_n to 0$, so the proposition is proved.
Because its form is also very close to the strong law of large numbers, I wonder if $frac{S_n}{n}-mu_noverset{a.s.}{to }0$, too.
Assumptions:
The following proposition has been proven that $$X_noverset{p}{to }X,Y_noverset{p}{to }Y Rightarrow X_n+Y_noverset{p}{to }X+Y$$
It is also established for subtraction, multiplication, and division. However, I don't know if it is established when $X_n,Y_n$ converge almost surely.
probability-theory convergence
asked Dec 8 at 9:22
李子涵
62
$X_1,X_2,…,X_n$ are independently and identically distributed and $E(X_i)$ exists, $mu_n=E(X_n I(X_n le n)),S_n=sum_{i=1}^n X_i$.
Proof:$$frac{S_n}{n}-mu_noverset{p}{to }0$$
My answer is:
$$frac{S_n}{n}-mu_n=(frac{S_n}{n}-E(X_i))+(E(X_i)-mu_n)$$
According to the law of large numbers, $frac{S_n}{n}-E(X_i)overset{p}{to }0$, and it is easy to proof that $E(X_i)-mu_n to 0$, so the proposition is proved.
Because its form is also very close to the strong law of large numbers, I wonder if $frac{S_n}{n}-mu_noverset{a.s.}{to }0$, too.
Assumptions:
The following proposition has been proven that $$X_noverset{p}{to }X,Y_noverset{p}{to }Y Rightarrow X_n+Y_noverset{p}{to }X+Y$$
It is also established for subtraction, multiplication, and division. However, I don't know if it is established when $X_n,Y_n$ converge almost surely.
probability-theory convergence
probability-theory convergence
asked Dec 8 at 9:22
李子涵
62
asked Dec 8 at 9:22
李子涵
62
asked Dec 8 at 9:22
李子涵
62
asked Dec 8 at 9:22
李子涵
62
asked Dec 8 at 9:22
李子涵
62
62
Yes. it also holds for almost everywhere convergence case. It is a consequence of Continuous mapping theorem.
– Song
Dec 8 at 9:59
If $X_n to X$ almost surely and $Y_n to Y$ almost surely then $X_n+Y_n to X+Y$ almost surely. This hardly requires any proof. It follows from definition of almost sure convergence and the fact that union of two sets of measure $0$ has measure $0$.
– Kavi Rama Murthy
Dec 8 at 12:15
add a comment |
Yes. it also holds for almost everywhere convergence case. It is a consequence of Continuous mapping theorem.
– Song
Dec 8 at 9:59
If $X_n to X$ almost surely and $Y_n to Y$ almost surely then $X_n+Y_n to X+Y$ almost surely. This hardly requires any proof. It follows from definition of almost sure convergence and the fact that union of two sets of measure $0$ has measure $0$.
– Kavi Rama Murthy
Dec 8 at 12:15
Yes. it also holds for almost everywhere convergence case. It is a consequence of Continuous mapping theorem.
– Song
Dec 8 at 9:59
Yes. it also holds for almost everywhere convergence case. It is a consequence of Continuous mapping theorem.
– Song
Dec 8 at 9:59
If $X_n to X$ almost surely and $Y_n to Y$ almost surely then $X_n+Y_n to X+Y$ almost surely. This hardly requires any proof. It follows from definition of almost sure convergence and the fact that union of two sets of measure $0$ has measure $0$.
– Kavi Rama Murthy
Dec 8 at 12:15
If $X_n to X$ almost surely and $Y_n to Y$ almost surely then $X_n+Y_n to X+Y$ almost surely. This hardly requires any proof. It follows from definition of almost sure convergence and the fact that union of two sets of measure $0$ has measure $0$.
– Kavi Rama Murthy
Dec 8 at 12:15
add a comment |
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Yes. it also holds for almost everywhere convergence case. It is a consequence of Continuous mapping theorem.
– Song
Dec 8 at 9:59
If $X_n to X$ almost surely and $Y_n to Y$ almost surely then $X_n+Y_n to X+Y$ almost surely. This hardly requires any proof. It follows from definition of almost sure convergence and the fact that union of two sets of measure $0$ has measure $0$.
– Kavi Rama Murthy
Dec 8 at 12:15