Mixing
Sum of sequences of random variables converging in distribution
Even if ${X_n}, {Y_n}$ converges in distribution to $X,Y$, I know that $X_n +Y_n$ need not converge to $X+Y$ . Can it happen that ${X_n+Y_n}$ doesn't converge in distribution to anything at all ?
probability-theory probability-distributions random-variables
asked Nov 20 '18 at 23:25
user521337
9781315
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Even if ${X_n}, {Y_n}$ converges in distribution to $X,Y$, I know that $X_n +Y_n$ need not converge to $X+Y$ . Can it happen that ${X_n+Y_n}$ doesn't converge in distribution to anything at all ?
probability-theory probability-distributions random-variables
asked Nov 20 '18 at 23:25
user521337
9781315
add a comment |
Even if ${X_n}, {Y_n}$ converges in distribution to $X,Y$, I know that $X_n +Y_n$ need not converge to $X+Y$ . Can it happen that ${X_n+Y_n}$ doesn't converge in distribution to anything at all ?
probability-theory probability-distributions random-variables
asked Nov 20 '18 at 23:25
user521337
9781315
Even if ${X_n}, {Y_n}$ converges in distribution to $X,Y$, I know that $X_n +Y_n$ need not converge to $X+Y$ . Can it happen that ${X_n+Y_n}$ doesn't converge in distribution to anything at all ?
probability-theory probability-distributions random-variables
probability-theory probability-distributions random-variables
asked Nov 20 '18 at 23:25
user521337
9781315
asked Nov 20 '18 at 23:25
user521337
9781315
asked Nov 20 '18 at 23:25
user521337
9781315
asked Nov 20 '18 at 23:25
user521337
9781315
asked Nov 20 '18 at 23:25
user521337
9781315
9781315
add a comment |
add a comment |
1 Answer
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Yes, it can happen.
Take $X_1$ Bernoulli ($frac12$) distributed: $mathbb P(X_1=0)=mathbb P(X_1=1)=frac12$, and $X_{2k+1}=X_1$, $X_{2k}=1-X_1$ for $kgeq 1$, $Y_n=X_1$ for all $ngeq 1$. Then both sequences converge in distribution to Bernoulli ($frac12$) distribution, and sum $X_n+Y_n$ does not converges since $X_{2k}+Y_{2k}=1$, $X_{2k+1}+Y_{2k+1}=2X_1$.
answered Nov 21 '18 at 1:07
NCh
6,2682723
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Yes, it can happen.
Take $X_1$ Bernoulli ($frac12$) distributed: $mathbb P(X_1=0)=mathbb P(X_1=1)=frac12$, and $X_{2k+1}=X_1$, $X_{2k}=1-X_1$ for $kgeq 1$, $Y_n=X_1$ for all $ngeq 1$. Then both sequences converge in distribution to Bernoulli ($frac12$) distribution, and sum $X_n+Y_n$ does not converges since $X_{2k}+Y_{2k}=1$, $X_{2k+1}+Y_{2k+1}=2X_1$.
answered Nov 21 '18 at 1:07
NCh
6,2682723
add a comment |
Yes, it can happen.
Take $X_1$ Bernoulli ($frac12$) distributed: $mathbb P(X_1=0)=mathbb P(X_1=1)=frac12$, and $X_{2k+1}=X_1$, $X_{2k}=1-X_1$ for $kgeq 1$, $Y_n=X_1$ for all $ngeq 1$. Then both sequences converge in distribution to Bernoulli ($frac12$) distribution, and sum $X_n+Y_n$ does not converges since $X_{2k}+Y_{2k}=1$, $X_{2k+1}+Y_{2k+1}=2X_1$.
answered Nov 21 '18 at 1:07
NCh
6,2682723
add a comment |
Yes, it can happen.
Take $X_1$ Bernoulli ($frac12$) distributed: $mathbb P(X_1=0)=mathbb P(X_1=1)=frac12$, and $X_{2k+1}=X_1$, $X_{2k}=1-X_1$ for $kgeq 1$, $Y_n=X_1$ for all $ngeq 1$. Then both sequences converge in distribution to Bernoulli ($frac12$) distribution, and sum $X_n+Y_n$ does not converges since $X_{2k}+Y_{2k}=1$, $X_{2k+1}+Y_{2k+1}=2X_1$.
answered Nov 21 '18 at 1:07
NCh
6,2682723
Yes, it can happen.
Take $X_1$ Bernoulli ($frac12$) distributed: $mathbb P(X_1=0)=mathbb P(X_1=1)=frac12$, and $X_{2k+1}=X_1$, $X_{2k}=1-X_1$ for $kgeq 1$, $Y_n=X_1$ for all $ngeq 1$. Then both sequences converge in distribution to Bernoulli ($frac12$) distribution, and sum $X_n+Y_n$ does not converges since $X_{2k}+Y_{2k}=1$, $X_{2k+1}+Y_{2k+1}=2X_1$.
answered Nov 21 '18 at 1:07
NCh
6,2682723
answered Nov 21 '18 at 1:07
NCh
6,2682723
answered Nov 21 '18 at 1:07
NCh
6,2682723
answered Nov 21 '18 at 1:07
NCh
6,2682723
6,2682723
add a comment |
add a comment |
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