Doesn't *identically distributed* imply *independent*?
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What the title says. If
- I draw a random value $x_1 sim mathcal{N}(mu, sigma)$
- a minute later, I draw another $x_2 sim mathcal{N}(mu, sigma)$
they come from identical distributions. Is there any scenario in which that does not necessarily imply that they are independent?
random-variables independence
asked Dec 14 '18 at 11:05
blue_noteblue_note
51948
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add a comment |
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What the title says. If
- I draw a random value $x_1 sim mathcal{N}(mu, sigma)$
- a minute later, I draw another $x_2 sim mathcal{N}(mu, sigma)$
they come from identical distributions. Is there any scenario in which that does not necessarily imply that they are independent?
random-variables independence
asked Dec 14 '18 at 11:05
blue_noteblue_note
51948
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In your scenario you implicitly assume idndependent and nowhere do you use the fact that the distributions are identical.
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– Somos
Dec 14 '18 at 11:17
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If you are asking in context of sampling, then drawing without replacement means $X_1,X_2$ are not independent. Drawing with replacement (usual random sample) means they are independent and identically distributed.
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– StubbornAtom
Dec 14 '18 at 11:17
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@StubbornAtom: yes, but if you are drawing without replacement, they are not identically distributed either.
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– blue_note
Dec 14 '18 at 11:19
add a comment |
$begingroup$
What the title says. If
- I draw a random value $x_1 sim mathcal{N}(mu, sigma)$
- a minute later, I draw another $x_2 sim mathcal{N}(mu, sigma)$
they come from identical distributions. Is there any scenario in which that does not necessarily imply that they are independent?
random-variables independence
asked Dec 14 '18 at 11:05
blue_noteblue_note
51948
$endgroup$
What the title says. If
- I draw a random value $x_1 sim mathcal{N}(mu, sigma)$
- a minute later, I draw another $x_2 sim mathcal{N}(mu, sigma)$
they come from identical distributions. Is there any scenario in which that does not necessarily imply that they are independent?
random-variables independence
random-variables independence
asked Dec 14 '18 at 11:05
blue_noteblue_note
51948
asked Dec 14 '18 at 11:05
blue_noteblue_note
51948
asked Dec 14 '18 at 11:05
blue_noteblue_note
51948
asked Dec 14 '18 at 11:05
blue_noteblue_note
51948
asked Dec 14 '18 at 11:05
blue_noteblue_note
51948
51948
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In your scenario you implicitly assume idndependent and nowhere do you use the fact that the distributions are identical.
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– Somos
Dec 14 '18 at 11:17
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If you are asking in context of sampling, then drawing without replacement means $X_1,X_2$ are not independent. Drawing with replacement (usual random sample) means they are independent and identically distributed.
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– StubbornAtom
Dec 14 '18 at 11:17
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@StubbornAtom: yes, but if you are drawing without replacement, they are not identically distributed either.
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– blue_note
Dec 14 '18 at 11:19
add a comment |
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In your scenario you implicitly assume idndependent and nowhere do you use the fact that the distributions are identical.
$endgroup$
– Somos
Dec 14 '18 at 11:17
$begingroup$
If you are asking in context of sampling, then drawing without replacement means $X_1,X_2$ are not independent. Drawing with replacement (usual random sample) means they are independent and identically distributed.
$endgroup$
– StubbornAtom
Dec 14 '18 at 11:17
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@StubbornAtom: yes, but if you are drawing without replacement, they are not identically distributed either.
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– blue_note
Dec 14 '18 at 11:19
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In your scenario you implicitly assume idndependent and nowhere do you use the fact that the distributions are identical.
$endgroup$
– Somos
Dec 14 '18 at 11:17
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In your scenario you implicitly assume idndependent and nowhere do you use the fact that the distributions are identical.
$endgroup$
– Somos
Dec 14 '18 at 11:17
$begingroup$
If you are asking in context of sampling, then drawing without replacement means $X_1,X_2$ are not independent. Drawing with replacement (usual random sample) means they are independent and identically distributed.
$endgroup$
– StubbornAtom
Dec 14 '18 at 11:17
$begingroup$
If you are asking in context of sampling, then drawing without replacement means $X_1,X_2$ are not independent. Drawing with replacement (usual random sample) means they are independent and identically distributed.
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– StubbornAtom
Dec 14 '18 at 11:17
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@StubbornAtom: yes, but if you are drawing without replacement, they are not identically distributed either.
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– blue_note
Dec 14 '18 at 11:19
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@StubbornAtom: yes, but if you are drawing without replacement, they are not identically distributed either.
$endgroup$
– blue_note
Dec 14 '18 at 11:19
add a comment |
2 Answers
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Here's a simple example. Roll a die, to give a random number $N$. Let $X$ be the variable $lfloor N/4rfloor$, and $Y$ be the random variable $Npmod 2$.
$X$ and $Y$ are identically distributed $mathrm{Bernoulli}(1/2)$ random variables, but they are not independent: $Pr(X=Y=1)=1/6$.
answered Dec 14 '18 at 11:15
Especially LimeEspecially Lime
22.3k22858
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If e.g. $mu=0$ in your example then $x_1$ and $-x_1$ are identically distributed but are not independent.
answered Dec 14 '18 at 11:18
drhabdrhab
102k545136
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
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$begingroup$
Here's a simple example. Roll a die, to give a random number $N$. Let $X$ be the variable $lfloor N/4rfloor$, and $Y$ be the random variable $Npmod 2$.
$X$ and $Y$ are identically distributed $mathrm{Bernoulli}(1/2)$ random variables, but they are not independent: $Pr(X=Y=1)=1/6$.
answered Dec 14 '18 at 11:15
Especially LimeEspecially Lime
22.3k22858
$endgroup$
add a comment |
$begingroup$
Here's a simple example. Roll a die, to give a random number $N$. Let $X$ be the variable $lfloor N/4rfloor$, and $Y$ be the random variable $Npmod 2$.
$X$ and $Y$ are identically distributed $mathrm{Bernoulli}(1/2)$ random variables, but they are not independent: $Pr(X=Y=1)=1/6$.
answered Dec 14 '18 at 11:15
Especially LimeEspecially Lime
22.3k22858
$endgroup$
add a comment |
$begingroup$
Here's a simple example. Roll a die, to give a random number $N$. Let $X$ be the variable $lfloor N/4rfloor$, and $Y$ be the random variable $Npmod 2$.
$X$ and $Y$ are identically distributed $mathrm{Bernoulli}(1/2)$ random variables, but they are not independent: $Pr(X=Y=1)=1/6$.
answered Dec 14 '18 at 11:15
Especially LimeEspecially Lime
22.3k22858
$endgroup$
Here's a simple example. Roll a die, to give a random number $N$. Let $X$ be the variable $lfloor N/4rfloor$, and $Y$ be the random variable $Npmod 2$.
$X$ and $Y$ are identically distributed $mathrm{Bernoulli}(1/2)$ random variables, but they are not independent: $Pr(X=Y=1)=1/6$.
answered Dec 14 '18 at 11:15
Especially LimeEspecially Lime
22.3k22858
answered Dec 14 '18 at 11:15
Especially LimeEspecially Lime
22.3k22858
answered Dec 14 '18 at 11:15
Especially LimeEspecially Lime
22.3k22858
answered Dec 14 '18 at 11:15
Especially LimeEspecially Lime
22.3k22858
22.3k22858
add a comment |
add a comment |
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If e.g. $mu=0$ in your example then $x_1$ and $-x_1$ are identically distributed but are not independent.
answered Dec 14 '18 at 11:18
drhabdrhab
102k545136
$endgroup$
add a comment |
$begingroup$
If e.g. $mu=0$ in your example then $x_1$ and $-x_1$ are identically distributed but are not independent.
answered Dec 14 '18 at 11:18
drhabdrhab
102k545136
$endgroup$
add a comment |
$begingroup$
If e.g. $mu=0$ in your example then $x_1$ and $-x_1$ are identically distributed but are not independent.
answered Dec 14 '18 at 11:18
drhabdrhab
102k545136
$endgroup$
If e.g. $mu=0$ in your example then $x_1$ and $-x_1$ are identically distributed but are not independent.
answered Dec 14 '18 at 11:18
drhabdrhab
102k545136
answered Dec 14 '18 at 11:18
drhabdrhab
102k545136
answered Dec 14 '18 at 11:18
drhabdrhab
102k545136
answered Dec 14 '18 at 11:18
drhabdrhab
102k545136
102k545136
add a comment |
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$begingroup$
In your scenario you implicitly assume idndependent and nowhere do you use the fact that the distributions are identical.
$endgroup$
– Somos
Dec 14 '18 at 11:17
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If you are asking in context of sampling, then drawing without replacement means $X_1,X_2$ are not independent. Drawing with replacement (usual random sample) means they are independent and identically distributed.
$endgroup$
– StubbornAtom
Dec 14 '18 at 11:17
$begingroup$
@StubbornAtom: yes, but if you are drawing without replacement, they are not identically distributed either.
$endgroup$
– blue_note
Dec 14 '18 at 11:19