Prove convergence in distribution.
$begingroup$
We have real-valued random variables ${X_n}_{n=1}^infty$, ${Y_n}_{n=1}^infty$, $X$ and $Y$.
$X_n rightarrow X$ in distribution and $Y_n rightarrow Y$ in distribution, respectively. Also, $X$ and $Y$ are independent, as well as $X_n$ and $Y_n$ for any $n ge1$.
Now, we would like to prove $X_n + Y_n rightarrow X+Y$ in distribution. What I have in mind is to first prove $(X_n,Y_n) rightarrow (X,Y)$ in distribution and then invoke the continuous mapping theorem.
To prove $(X_n,Y_n) rightarrow (X,Y)$ in distribution, we only need to prove for every bounded Lipschitz function $f:mathbb{R}^2 rightarrow mathbb{R}$, we have $E[f(X_n,Y_n)] rightarrow E[f(X,Y)]$ as $n rightarrow infty$, by the Portmanteau Theorem. Here's what I ran into problem: $$E[f(X_n,Y_n)] = int_mathbb{R} left( int_{mathbb{R}} f(x,y) mathcal{P}_{X_n}(dx) right) mathcal{P}_{Y_n}(dy),$$ where $mathcal{P}_{X_n}, mathcal{P}_{Y_n}$ are the distributions of $X_n$ and $Y_n$.
How to prove $E[f(X_n,Y_n)] rightarrow E[f(X,Y)]$ from this point?
probability-theory independence weak-convergence
edited Dec 17 '18 at 23:21
Davide Giraudo
127k17154268
asked Dec 17 '18 at 9:52
xixumeixixumei
31818
$endgroup$
add a comment |
$begingroup$
We have real-valued random variables ${X_n}_{n=1}^infty$, ${Y_n}_{n=1}^infty$, $X$ and $Y$.
$X_n rightarrow X$ in distribution and $Y_n rightarrow Y$ in distribution, respectively. Also, $X$ and $Y$ are independent, as well as $X_n$ and $Y_n$ for any $n ge1$.
Now, we would like to prove $X_n + Y_n rightarrow X+Y$ in distribution. What I have in mind is to first prove $(X_n,Y_n) rightarrow (X,Y)$ in distribution and then invoke the continuous mapping theorem.
To prove $(X_n,Y_n) rightarrow (X,Y)$ in distribution, we only need to prove for every bounded Lipschitz function $f:mathbb{R}^2 rightarrow mathbb{R}$, we have $E[f(X_n,Y_n)] rightarrow E[f(X,Y)]$ as $n rightarrow infty$, by the Portmanteau Theorem. Here's what I ran into problem: $$E[f(X_n,Y_n)] = int_mathbb{R} left( int_{mathbb{R}} f(x,y) mathcal{P}_{X_n}(dx) right) mathcal{P}_{Y_n}(dy),$$ where $mathcal{P}_{X_n}, mathcal{P}_{Y_n}$ are the distributions of $X_n$ and $Y_n$.
How to prove $E[f(X_n,Y_n)] rightarrow E[f(X,Y)]$ from this point?
probability-theory independence weak-convergence
edited Dec 17 '18 at 23:21
Davide Giraudo
127k17154268
asked Dec 17 '18 at 9:52
xixumeixixumei
31818
$endgroup$
3
$begingroup$
Use the characteristic functions $$E(e^{it(X_n+Y_n)})$$
$endgroup$
– Did
Dec 17 '18 at 9:55
add a comment |
$begingroup$
We have real-valued random variables ${X_n}_{n=1}^infty$, ${Y_n}_{n=1}^infty$, $X$ and $Y$.
$X_n rightarrow X$ in distribution and $Y_n rightarrow Y$ in distribution, respectively. Also, $X$ and $Y$ are independent, as well as $X_n$ and $Y_n$ for any $n ge1$.
Now, we would like to prove $X_n + Y_n rightarrow X+Y$ in distribution. What I have in mind is to first prove $(X_n,Y_n) rightarrow (X,Y)$ in distribution and then invoke the continuous mapping theorem.
To prove $(X_n,Y_n) rightarrow (X,Y)$ in distribution, we only need to prove for every bounded Lipschitz function $f:mathbb{R}^2 rightarrow mathbb{R}$, we have $E[f(X_n,Y_n)] rightarrow E[f(X,Y)]$ as $n rightarrow infty$, by the Portmanteau Theorem. Here's what I ran into problem: $$E[f(X_n,Y_n)] = int_mathbb{R} left( int_{mathbb{R}} f(x,y) mathcal{P}_{X_n}(dx) right) mathcal{P}_{Y_n}(dy),$$ where $mathcal{P}_{X_n}, mathcal{P}_{Y_n}$ are the distributions of $X_n$ and $Y_n$.
How to prove $E[f(X_n,Y_n)] rightarrow E[f(X,Y)]$ from this point?
probability-theory independence weak-convergence
edited Dec 17 '18 at 23:21
Davide Giraudo
127k17154268
asked Dec 17 '18 at 9:52
xixumeixixumei
31818
$endgroup$
We have real-valued random variables ${X_n}_{n=1}^infty$, ${Y_n}_{n=1}^infty$, $X$ and $Y$.
$X_n rightarrow X$ in distribution and $Y_n rightarrow Y$ in distribution, respectively. Also, $X$ and $Y$ are independent, as well as $X_n$ and $Y_n$ for any $n ge1$.
Now, we would like to prove $X_n + Y_n rightarrow X+Y$ in distribution. What I have in mind is to first prove $(X_n,Y_n) rightarrow (X,Y)$ in distribution and then invoke the continuous mapping theorem.
To prove $(X_n,Y_n) rightarrow (X,Y)$ in distribution, we only need to prove for every bounded Lipschitz function $f:mathbb{R}^2 rightarrow mathbb{R}$, we have $E[f(X_n,Y_n)] rightarrow E[f(X,Y)]$ as $n rightarrow infty$, by the Portmanteau Theorem. Here's what I ran into problem: $$E[f(X_n,Y_n)] = int_mathbb{R} left( int_{mathbb{R}} f(x,y) mathcal{P}_{X_n}(dx) right) mathcal{P}_{Y_n}(dy),$$ where $mathcal{P}_{X_n}, mathcal{P}_{Y_n}$ are the distributions of $X_n$ and $Y_n$.
How to prove $E[f(X_n,Y_n)] rightarrow E[f(X,Y)]$ from this point?
probability-theory independence weak-convergence
probability-theory independence weak-convergence
edited Dec 17 '18 at 23:21
Davide Giraudo
127k17154268
asked Dec 17 '18 at 9:52
xixumeixixumei
31818
edited Dec 17 '18 at 23:21
Davide Giraudo
127k17154268
asked Dec 17 '18 at 9:52
xixumeixixumei
31818
edited Dec 17 '18 at 23:21
Davide Giraudo
127k17154268
edited Dec 17 '18 at 23:21
Davide Giraudo
127k17154268
edited Dec 17 '18 at 23:21
Davide Giraudo
127k17154268
127k17154268
asked Dec 17 '18 at 9:52
xixumeixixumei
31818
asked Dec 17 '18 at 9:52
xixumeixixumei
31818
asked Dec 17 '18 at 9:52
xixumeixixumei
31818
31818
3
$begingroup$
Use the characteristic functions $$E(e^{it(X_n+Y_n)})$$
$endgroup$
– Did
Dec 17 '18 at 9:55
add a comment |
3
$begingroup$
Use the characteristic functions $$E(e^{it(X_n+Y_n)})$$
$endgroup$
– Did
Dec 17 '18 at 9:55
3
3
$begingroup$
Use the characteristic functions $$E(e^{it(X_n+Y_n)})$$
$endgroup$
– Did
Dec 17 '18 at 9:55
$begingroup$
Use the characteristic functions $$E(e^{it(X_n+Y_n)})$$
$endgroup$
– Did
Dec 17 '18 at 9:55
add a comment |
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$begingroup$
Use the characteristic functions $$E(e^{it(X_n+Y_n)})$$
$endgroup$
– Did
Dec 17 '18 at 9:55