Mixing
On the limit of a continuous combination of sequences of random variables converging in distribution
Let ${X_n}, {Y_n}$ be sequences of real valued random variables converging in distribution to $X$ and $Y$ respectively. Let $f: mathbb R^2 to mathbb R$ be a continuous function such that ${f(X_n,Y_n)}$ converges in distribution to some random variable $Z$.
Then is it true that $Z$ is identically distributed as $f(X,Y)$ ?
If this is not true in general, is it at least true for the function $f(x,y)=x+y$ ?
probability-theory measure-theory probability-distributions random-variables
asked Nov 20 '18 at 23:10
user521337
9781315
add a comment |
Let ${X_n}, {Y_n}$ be sequences of real valued random variables converging in distribution to $X$ and $Y$ respectively. Let $f: mathbb R^2 to mathbb R$ be a continuous function such that ${f(X_n,Y_n)}$ converges in distribution to some random variable $Z$.
Then is it true that $Z$ is identically distributed as $f(X,Y)$ ?
If this is not true in general, is it at least true for the function $f(x,y)=x+y$ ?
probability-theory measure-theory probability-distributions random-variables
asked Nov 20 '18 at 23:10
user521337
9781315
add a comment |
Let ${X_n}, {Y_n}$ be sequences of real valued random variables converging in distribution to $X$ and $Y$ respectively. Let $f: mathbb R^2 to mathbb R$ be a continuous function such that ${f(X_n,Y_n)}$ converges in distribution to some random variable $Z$.
Then is it true that $Z$ is identically distributed as $f(X,Y)$ ?
If this is not true in general, is it at least true for the function $f(x,y)=x+y$ ?
probability-theory measure-theory probability-distributions random-variables
asked Nov 20 '18 at 23:10
user521337
9781315
Let ${X_n}, {Y_n}$ be sequences of real valued random variables converging in distribution to $X$ and $Y$ respectively. Let $f: mathbb R^2 to mathbb R$ be a continuous function such that ${f(X_n,Y_n)}$ converges in distribution to some random variable $Z$.
Then is it true that $Z$ is identically distributed as $f(X,Y)$ ?
If this is not true in general, is it at least true for the function $f(x,y)=x+y$ ?
probability-theory measure-theory probability-distributions random-variables
probability-theory measure-theory probability-distributions random-variables
asked Nov 20 '18 at 23:10
user521337
9781315
asked Nov 20 '18 at 23:10
user521337
9781315
asked Nov 20 '18 at 23:10
user521337
9781315
asked Nov 20 '18 at 23:10
user521337
9781315
asked Nov 20 '18 at 23:10
user521337
9781315
9781315
add a comment |
add a comment |
2 Answers
2
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Not without independence assumptions (or convergence of joint distributions). Take ${X_n}$ i.i.d. with standard normal distribution, $Y_n=-X_n$ for all $n$ and $f(x,y)=x+y$. Then ${X_n} to X_1$ and ${Y_n} to X_1$ in distribution but $X_n+Y_n to 0$ in distribution.
answered Nov 20 '18 at 23:17
Kavi Rama Murthy
51k31854
add a comment |
For your question to make sense, you need to mention that the vectors $(X,Y)$ and $(X_n, Y_n)$ exist (i.e. $X_n$ and $Y_n$ must live in the same probability space).
If you have the stronger assumption that $(X_n, Y_n)$ converges in distribution to the vector $(X,Y)$, then $f(X_n, Y_n)$ converges in distribution to $f(X,Y)$ by the continuous mapping theorem.
Without this stronger assumption, the claim does not hold, even for $f(x,y) = x + y$, as demonstrated in one of the answers in this question.
answered Nov 20 '18 at 23:17
angryavian
39k23180
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Not without independence assumptions (or convergence of joint distributions). Take ${X_n}$ i.i.d. with standard normal distribution, $Y_n=-X_n$ for all $n$ and $f(x,y)=x+y$. Then ${X_n} to X_1$ and ${Y_n} to X_1$ in distribution but $X_n+Y_n to 0$ in distribution.
answered Nov 20 '18 at 23:17
Kavi Rama Murthy
51k31854
add a comment |
Not without independence assumptions (or convergence of joint distributions). Take ${X_n}$ i.i.d. with standard normal distribution, $Y_n=-X_n$ for all $n$ and $f(x,y)=x+y$. Then ${X_n} to X_1$ and ${Y_n} to X_1$ in distribution but $X_n+Y_n to 0$ in distribution.
answered Nov 20 '18 at 23:17
Kavi Rama Murthy
51k31854
add a comment |
Not without independence assumptions (or convergence of joint distributions). Take ${X_n}$ i.i.d. with standard normal distribution, $Y_n=-X_n$ for all $n$ and $f(x,y)=x+y$. Then ${X_n} to X_1$ and ${Y_n} to X_1$ in distribution but $X_n+Y_n to 0$ in distribution.
answered Nov 20 '18 at 23:17
Kavi Rama Murthy
51k31854
Not without independence assumptions (or convergence of joint distributions). Take ${X_n}$ i.i.d. with standard normal distribution, $Y_n=-X_n$ for all $n$ and $f(x,y)=x+y$. Then ${X_n} to X_1$ and ${Y_n} to X_1$ in distribution but $X_n+Y_n to 0$ in distribution.
answered Nov 20 '18 at 23:17
Kavi Rama Murthy
51k31854
answered Nov 20 '18 at 23:17
Kavi Rama Murthy
51k31854
answered Nov 20 '18 at 23:17
Kavi Rama Murthy
51k31854
answered Nov 20 '18 at 23:17
Kavi Rama Murthy
51k31854
51k31854
add a comment |
add a comment |
For your question to make sense, you need to mention that the vectors $(X,Y)$ and $(X_n, Y_n)$ exist (i.e. $X_n$ and $Y_n$ must live in the same probability space).
If you have the stronger assumption that $(X_n, Y_n)$ converges in distribution to the vector $(X,Y)$, then $f(X_n, Y_n)$ converges in distribution to $f(X,Y)$ by the continuous mapping theorem.
Without this stronger assumption, the claim does not hold, even for $f(x,y) = x + y$, as demonstrated in one of the answers in this question.
answered Nov 20 '18 at 23:17
angryavian
39k23180
add a comment |
For your question to make sense, you need to mention that the vectors $(X,Y)$ and $(X_n, Y_n)$ exist (i.e. $X_n$ and $Y_n$ must live in the same probability space).
If you have the stronger assumption that $(X_n, Y_n)$ converges in distribution to the vector $(X,Y)$, then $f(X_n, Y_n)$ converges in distribution to $f(X,Y)$ by the continuous mapping theorem.
Without this stronger assumption, the claim does not hold, even for $f(x,y) = x + y$, as demonstrated in one of the answers in this question.
answered Nov 20 '18 at 23:17
angryavian
39k23180
add a comment |
For your question to make sense, you need to mention that the vectors $(X,Y)$ and $(X_n, Y_n)$ exist (i.e. $X_n$ and $Y_n$ must live in the same probability space).
If you have the stronger assumption that $(X_n, Y_n)$ converges in distribution to the vector $(X,Y)$, then $f(X_n, Y_n)$ converges in distribution to $f(X,Y)$ by the continuous mapping theorem.
Without this stronger assumption, the claim does not hold, even for $f(x,y) = x + y$, as demonstrated in one of the answers in this question.
answered Nov 20 '18 at 23:17
angryavian
39k23180
For your question to make sense, you need to mention that the vectors $(X,Y)$ and $(X_n, Y_n)$ exist (i.e. $X_n$ and $Y_n$ must live in the same probability space).
If you have the stronger assumption that $(X_n, Y_n)$ converges in distribution to the vector $(X,Y)$, then $f(X_n, Y_n)$ converges in distribution to $f(X,Y)$ by the continuous mapping theorem.
Without this stronger assumption, the claim does not hold, even for $f(x,y) = x + y$, as demonstrated in one of the answers in this question.
answered Nov 20 '18 at 23:17
angryavian
39k23180
answered Nov 20 '18 at 23:17
angryavian
39k23180
answered Nov 20 '18 at 23:17
angryavian
39k23180
answered Nov 20 '18 at 23:17
angryavian
39k23180
39k23180
add a comment |
add a comment |
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