Convergence in distribution of $frac{1}{n}max(X_n,Y_n)$, where $X_ntext{ }U(-n-5,4n-5)$, $Y_ntext{...
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Convergence in distribution of $frac{1}{n}max(X_n,Y_n)$, where $X_ntext{ is }U(-n-5,4n-5)$, $Y_ntext{ is }Poiss(10n)$. My idea is to look how $X_n, Y_n$ behave as n tends to infinity, so my first observation is that probability of $X_n$ being infinity is positive as n tends to infinity, however poisson distribution "converges to zero" as n tends to infinity. So I think the only important factor in maximum is the uniform variable, but I might be wrong. Can anyone relate?
The PDF of $frac{1}{n}max{(X_n,Y_n)}$ is $$frac{tn+n+5}{5n}e^{-10n}sum_{k=0}^{floor{tn}}frac{10n^{k}}{k!}$$, so we are up to calculating the limit of it when n tends to infinity... $$text{After a while I think this limit is zero.}$$
probability-theory probability-distributions convergence weak-convergence
asked Dec 4 '18 at 10:01
ryszard egginkryszard eggink
308110
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show 1 more comment
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Convergence in distribution of $frac{1}{n}max(X_n,Y_n)$, where $X_ntext{ is }U(-n-5,4n-5)$, $Y_ntext{ is }Poiss(10n)$. My idea is to look how $X_n, Y_n$ behave as n tends to infinity, so my first observation is that probability of $X_n$ being infinity is positive as n tends to infinity, however poisson distribution "converges to zero" as n tends to infinity. So I think the only important factor in maximum is the uniform variable, but I might be wrong. Can anyone relate?
The PDF of $frac{1}{n}max{(X_n,Y_n)}$ is $$frac{tn+n+5}{5n}e^{-10n}sum_{k=0}^{floor{tn}}frac{10n^{k}}{k!}$$, so we are up to calculating the limit of it when n tends to infinity... $$text{After a while I think this limit is zero.}$$
probability-theory probability-distributions convergence weak-convergence
asked Dec 4 '18 at 10:01
ryszard egginkryszard eggink
308110
$endgroup$
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Excellent. Have you tried computing the cdf of $1/n max(X_n, Y_n)$ ?
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– Stockfish
Dec 4 '18 at 10:35
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I would do that, had I known the CDF of poisson distribution.
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– ryszard eggink
Dec 4 '18 at 10:37
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Well you could just look it up en.wikipedia.org/wiki/Poisson_distribution
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– Stockfish
Dec 4 '18 at 10:55
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Is the calculation correct? I do not think I can do the cancelation of this sum with $e^{-10n}$
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– ryszard eggink
Dec 4 '18 at 11:04
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You need an indicator function of the set $tn in [-n-5, 4n-5]$. In the sum it should be $(10n)^k$.
$endgroup$
– Stockfish
Dec 5 '18 at 9:31
|
show 1 more comment
$begingroup$
Convergence in distribution of $frac{1}{n}max(X_n,Y_n)$, where $X_ntext{ is }U(-n-5,4n-5)$, $Y_ntext{ is }Poiss(10n)$. My idea is to look how $X_n, Y_n$ behave as n tends to infinity, so my first observation is that probability of $X_n$ being infinity is positive as n tends to infinity, however poisson distribution "converges to zero" as n tends to infinity. So I think the only important factor in maximum is the uniform variable, but I might be wrong. Can anyone relate?
The PDF of $frac{1}{n}max{(X_n,Y_n)}$ is $$frac{tn+n+5}{5n}e^{-10n}sum_{k=0}^{floor{tn}}frac{10n^{k}}{k!}$$, so we are up to calculating the limit of it when n tends to infinity... $$text{After a while I think this limit is zero.}$$
probability-theory probability-distributions convergence weak-convergence
asked Dec 4 '18 at 10:01
ryszard egginkryszard eggink
308110
$endgroup$
Convergence in distribution of $frac{1}{n}max(X_n,Y_n)$, where $X_ntext{ is }U(-n-5,4n-5)$, $Y_ntext{ is }Poiss(10n)$. My idea is to look how $X_n, Y_n$ behave as n tends to infinity, so my first observation is that probability of $X_n$ being infinity is positive as n tends to infinity, however poisson distribution "converges to zero" as n tends to infinity. So I think the only important factor in maximum is the uniform variable, but I might be wrong. Can anyone relate?
The PDF of $frac{1}{n}max{(X_n,Y_n)}$ is $$frac{tn+n+5}{5n}e^{-10n}sum_{k=0}^{floor{tn}}frac{10n^{k}}{k!}$$, so we are up to calculating the limit of it when n tends to infinity... $$text{After a while I think this limit is zero.}$$
probability-theory probability-distributions convergence weak-convergence
probability-theory probability-distributions convergence weak-convergence
asked Dec 4 '18 at 10:01
ryszard egginkryszard eggink
308110
asked Dec 4 '18 at 10:01
ryszard egginkryszard eggink
308110
edited Dec 4 '18 at 16:38
ryszard eggink
asked Dec 4 '18 at 10:01
ryszard egginkryszard eggink
308110
asked Dec 4 '18 at 10:01
ryszard egginkryszard eggink
308110
asked Dec 4 '18 at 10:01
ryszard egginkryszard eggink
308110
308110
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Excellent. Have you tried computing the cdf of $1/n max(X_n, Y_n)$ ?
$endgroup$
– Stockfish
Dec 4 '18 at 10:35
$begingroup$
I would do that, had I known the CDF of poisson distribution.
$endgroup$
– ryszard eggink
Dec 4 '18 at 10:37
$begingroup$
Well you could just look it up en.wikipedia.org/wiki/Poisson_distribution
$endgroup$
– Stockfish
Dec 4 '18 at 10:55
$begingroup$
Is the calculation correct? I do not think I can do the cancelation of this sum with $e^{-10n}$
$endgroup$
– ryszard eggink
Dec 4 '18 at 11:04
$begingroup$
You need an indicator function of the set $tn in [-n-5, 4n-5]$. In the sum it should be $(10n)^k$.
$endgroup$
– Stockfish
Dec 5 '18 at 9:31
|
show 1 more comment
$begingroup$
Excellent. Have you tried computing the cdf of $1/n max(X_n, Y_n)$ ?
$endgroup$
– Stockfish
Dec 4 '18 at 10:35
$begingroup$
I would do that, had I known the CDF of poisson distribution.
$endgroup$
– ryszard eggink
Dec 4 '18 at 10:37
$begingroup$
Well you could just look it up en.wikipedia.org/wiki/Poisson_distribution
$endgroup$
– Stockfish
Dec 4 '18 at 10:55
$begingroup$
Is the calculation correct? I do not think I can do the cancelation of this sum with $e^{-10n}$
$endgroup$
– ryszard eggink
Dec 4 '18 at 11:04
$begingroup$
You need an indicator function of the set $tn in [-n-5, 4n-5]$. In the sum it should be $(10n)^k$.
$endgroup$
– Stockfish
Dec 5 '18 at 9:31
$begingroup$
Excellent. Have you tried computing the cdf of $1/n max(X_n, Y_n)$ ?
$endgroup$
– Stockfish
Dec 4 '18 at 10:35
$begingroup$
Excellent. Have you tried computing the cdf of $1/n max(X_n, Y_n)$ ?
$endgroup$
– Stockfish
Dec 4 '18 at 10:35
$begingroup$
I would do that, had I known the CDF of poisson distribution.
$endgroup$
– ryszard eggink
Dec 4 '18 at 10:37
$begingroup$
I would do that, had I known the CDF of poisson distribution.
$endgroup$
– ryszard eggink
Dec 4 '18 at 10:37
$begingroup$
Well you could just look it up en.wikipedia.org/wiki/Poisson_distribution
$endgroup$
– Stockfish
Dec 4 '18 at 10:55
$begingroup$
Well you could just look it up en.wikipedia.org/wiki/Poisson_distribution
$endgroup$
– Stockfish
Dec 4 '18 at 10:55
$begingroup$
Is the calculation correct? I do not think I can do the cancelation of this sum with $e^{-10n}$
$endgroup$
– ryszard eggink
Dec 4 '18 at 11:04
$begingroup$
Is the calculation correct? I do not think I can do the cancelation of this sum with $e^{-10n}$
$endgroup$
– ryszard eggink
Dec 4 '18 at 11:04
$begingroup$
You need an indicator function of the set $tn in [-n-5, 4n-5]$. In the sum it should be $(10n)^k$.
$endgroup$
– Stockfish
Dec 5 '18 at 9:31
$begingroup$
You need an indicator function of the set $tn in [-n-5, 4n-5]$. In the sum it should be $(10n)^k$.
$endgroup$
– Stockfish
Dec 5 '18 at 9:31
|
show 1 more comment
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$begingroup$
Excellent. Have you tried computing the cdf of $1/n max(X_n, Y_n)$ ?
$endgroup$
– Stockfish
Dec 4 '18 at 10:35
$begingroup$
I would do that, had I known the CDF of poisson distribution.
$endgroup$
– ryszard eggink
Dec 4 '18 at 10:37
$begingroup$
Well you could just look it up en.wikipedia.org/wiki/Poisson_distribution
$endgroup$
– Stockfish
Dec 4 '18 at 10:55
$begingroup$
Is the calculation correct? I do not think I can do the cancelation of this sum with $e^{-10n}$
$endgroup$
– ryszard eggink
Dec 4 '18 at 11:04
$begingroup$
You need an indicator function of the set $tn in [-n-5, 4n-5]$. In the sum it should be $(10n)^k$.
$endgroup$
– Stockfish
Dec 5 '18 at 9:31