Mixing
Find $P(X_1lt X_2lt X_3)$ and $P(X_1gt X_2|X_2lt X_3)$ for $X_1, X_2, X_3$ i.i.d random variable.
$begingroup$
Find $P(X_1lt X_2lt X_3)$ and $P(X_1gt X_2|X_2lt X_3)$ for $X_1, X_2, X_3$ i.i.d random variable.
For $P(X_1lt X_2lt X_3)$, since there exists six possible orders, $P(X_1lt X_2lt X_3)={1over 6}$. Is this true for any distribution? Can I get $P(X_1lt X_2lt X_3lt...lt X_n)={1over n!}$?
For $P(X_1gt X_2|X_2lt X_3)$, since $X_1gt X_2$ and $X_2lt X_3$ are independent event (am I right?), I just simply get $P(X_1gt X_2|X_2lt X_3)={1over2}$
probability
asked Jan 6 at 3:43
Yibei HeYibei He
2229
$endgroup$
add a comment |
$begingroup$
Find $P(X_1lt X_2lt X_3)$ and $P(X_1gt X_2|X_2lt X_3)$ for $X_1, X_2, X_3$ i.i.d random variable.
For $P(X_1lt X_2lt X_3)$, since there exists six possible orders, $P(X_1lt X_2lt X_3)={1over 6}$. Is this true for any distribution? Can I get $P(X_1lt X_2lt X_3lt...lt X_n)={1over n!}$?
For $P(X_1gt X_2|X_2lt X_3)$, since $X_1gt X_2$ and $X_2lt X_3$ are independent event (am I right?), I just simply get $P(X_1gt X_2|X_2lt X_3)={1over2}$
probability
asked Jan 6 at 3:43
Yibei HeYibei He
2229
$endgroup$
add a comment |
$begingroup$
Find $P(X_1lt X_2lt X_3)$ and $P(X_1gt X_2|X_2lt X_3)$ for $X_1, X_2, X_3$ i.i.d random variable.
For $P(X_1lt X_2lt X_3)$, since there exists six possible orders, $P(X_1lt X_2lt X_3)={1over 6}$. Is this true for any distribution? Can I get $P(X_1lt X_2lt X_3lt...lt X_n)={1over n!}$?
For $P(X_1gt X_2|X_2lt X_3)$, since $X_1gt X_2$ and $X_2lt X_3$ are independent event (am I right?), I just simply get $P(X_1gt X_2|X_2lt X_3)={1over2}$
probability
asked Jan 6 at 3:43
Yibei HeYibei He
2229
$endgroup$
Find $P(X_1lt X_2lt X_3)$ and $P(X_1gt X_2|X_2lt X_3)$ for $X_1, X_2, X_3$ i.i.d random variable.
For $P(X_1lt X_2lt X_3)$, since there exists six possible orders, $P(X_1lt X_2lt X_3)={1over 6}$. Is this true for any distribution? Can I get $P(X_1lt X_2lt X_3lt...lt X_n)={1over n!}$?
For $P(X_1gt X_2|X_2lt X_3)$, since $X_1gt X_2$ and $X_2lt X_3$ are independent event (am I right?), I just simply get $P(X_1gt X_2|X_2lt X_3)={1over2}$
probability
probability
asked Jan 6 at 3:43
Yibei HeYibei He
2229
asked Jan 6 at 3:43
Yibei HeYibei He
2229
asked Jan 6 at 3:43
Yibei HeYibei He
2229
asked Jan 6 at 3:43
Yibei HeYibei He
2229
asked Jan 6 at 3:43
Yibei HeYibei He
2229
2229
add a comment |
add a comment |
1 Answer
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$begingroup$
For the first part, this is true for any i.i.d random variables as long as the probabilities of any two being equal is 0. So if they are continuous random variables it is true, if they are discrete then no.
For the second part, you need to consider your $6$ equally likely orderings
$X_1 < X_2 < X_3$, $X_2 < X_1 < X_3$, $X_1 < X_3 < X_2$, $X_3 < X_1 < X_2$, $X_2 < X_3 < X_1$, and $X_3 < X_2 < X_1$. If $X_2 < X_3$ then there are only $3$ of these orderings possible, $X_1 < X_2 < X_3$, $X_2 < X_1 < X_3$, and $X_2 < X_3 < X_1$. Of these $X_1 > X_2$ in $2$ of them, so the probability is actually $frac23$.
Intuitively this is because $X_2 < X_3$ and $X_1 > X_2$ are not independent. If you think about it, $X_2 < X_3$ means that $X_2$ is more likely to be small, and so it is more likely that $X_1$ is greater than it.
answered Jan 6 at 3:52
Erik ParkinsonErik Parkinson
8549
$endgroup$
$begingroup$
would it be $2over 3$ instead of $1over 3$?
$endgroup$
– Yibei He
Jan 6 at 4:03
$begingroup$
@YibeiHe Good catch! Just fixed it.
$endgroup$
– Erik Parkinson
Jan 6 at 4:05
add a comment |
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1 Answer
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1 Answer
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$begingroup$
For the first part, this is true for any i.i.d random variables as long as the probabilities of any two being equal is 0. So if they are continuous random variables it is true, if they are discrete then no.
For the second part, you need to consider your $6$ equally likely orderings
$X_1 < X_2 < X_3$, $X_2 < X_1 < X_3$, $X_1 < X_3 < X_2$, $X_3 < X_1 < X_2$, $X_2 < X_3 < X_1$, and $X_3 < X_2 < X_1$. If $X_2 < X_3$ then there are only $3$ of these orderings possible, $X_1 < X_2 < X_3$, $X_2 < X_1 < X_3$, and $X_2 < X_3 < X_1$. Of these $X_1 > X_2$ in $2$ of them, so the probability is actually $frac23$.
Intuitively this is because $X_2 < X_3$ and $X_1 > X_2$ are not independent. If you think about it, $X_2 < X_3$ means that $X_2$ is more likely to be small, and so it is more likely that $X_1$ is greater than it.
answered Jan 6 at 3:52
Erik ParkinsonErik Parkinson
8549
$endgroup$
$begingroup$
would it be $2over 3$ instead of $1over 3$?
$endgroup$
– Yibei He
Jan 6 at 4:03
$begingroup$
@YibeiHe Good catch! Just fixed it.
$endgroup$
– Erik Parkinson
Jan 6 at 4:05
add a comment |
$begingroup$
For the first part, this is true for any i.i.d random variables as long as the probabilities of any two being equal is 0. So if they are continuous random variables it is true, if they are discrete then no.
For the second part, you need to consider your $6$ equally likely orderings
$X_1 < X_2 < X_3$, $X_2 < X_1 < X_3$, $X_1 < X_3 < X_2$, $X_3 < X_1 < X_2$, $X_2 < X_3 < X_1$, and $X_3 < X_2 < X_1$. If $X_2 < X_3$ then there are only $3$ of these orderings possible, $X_1 < X_2 < X_3$, $X_2 < X_1 < X_3$, and $X_2 < X_3 < X_1$. Of these $X_1 > X_2$ in $2$ of them, so the probability is actually $frac23$.
Intuitively this is because $X_2 < X_3$ and $X_1 > X_2$ are not independent. If you think about it, $X_2 < X_3$ means that $X_2$ is more likely to be small, and so it is more likely that $X_1$ is greater than it.
answered Jan 6 at 3:52
Erik ParkinsonErik Parkinson
8549
$endgroup$
$begingroup$
would it be $2over 3$ instead of $1over 3$?
$endgroup$
– Yibei He
Jan 6 at 4:03
$begingroup$
@YibeiHe Good catch! Just fixed it.
$endgroup$
– Erik Parkinson
Jan 6 at 4:05
add a comment |
$begingroup$
For the first part, this is true for any i.i.d random variables as long as the probabilities of any two being equal is 0. So if they are continuous random variables it is true, if they are discrete then no.
For the second part, you need to consider your $6$ equally likely orderings
$X_1 < X_2 < X_3$, $X_2 < X_1 < X_3$, $X_1 < X_3 < X_2$, $X_3 < X_1 < X_2$, $X_2 < X_3 < X_1$, and $X_3 < X_2 < X_1$. If $X_2 < X_3$ then there are only $3$ of these orderings possible, $X_1 < X_2 < X_3$, $X_2 < X_1 < X_3$, and $X_2 < X_3 < X_1$. Of these $X_1 > X_2$ in $2$ of them, so the probability is actually $frac23$.
Intuitively this is because $X_2 < X_3$ and $X_1 > X_2$ are not independent. If you think about it, $X_2 < X_3$ means that $X_2$ is more likely to be small, and so it is more likely that $X_1$ is greater than it.
answered Jan 6 at 3:52
Erik ParkinsonErik Parkinson
8549
$endgroup$
For the first part, this is true for any i.i.d random variables as long as the probabilities of any two being equal is 0. So if they are continuous random variables it is true, if they are discrete then no.
For the second part, you need to consider your $6$ equally likely orderings
$X_1 < X_2 < X_3$, $X_2 < X_1 < X_3$, $X_1 < X_3 < X_2$, $X_3 < X_1 < X_2$, $X_2 < X_3 < X_1$, and $X_3 < X_2 < X_1$. If $X_2 < X_3$ then there are only $3$ of these orderings possible, $X_1 < X_2 < X_3$, $X_2 < X_1 < X_3$, and $X_2 < X_3 < X_1$. Of these $X_1 > X_2$ in $2$ of them, so the probability is actually $frac23$.
Intuitively this is because $X_2 < X_3$ and $X_1 > X_2$ are not independent. If you think about it, $X_2 < X_3$ means that $X_2$ is more likely to be small, and so it is more likely that $X_1$ is greater than it.
answered Jan 6 at 3:52
Erik ParkinsonErik Parkinson
8549
edited Jan 6 at 4:04
answered Jan 6 at 3:52
Erik ParkinsonErik Parkinson
8549
answered Jan 6 at 3:52
Erik ParkinsonErik Parkinson
8549
answered Jan 6 at 3:52
Erik ParkinsonErik Parkinson
8549
8549
$begingroup$
would it be $2over 3$ instead of $1over 3$?
$endgroup$
– Yibei He
Jan 6 at 4:03
$begingroup$
@YibeiHe Good catch! Just fixed it.
$endgroup$
– Erik Parkinson
Jan 6 at 4:05
add a comment |
$begingroup$
would it be $2over 3$ instead of $1over 3$?
$endgroup$
– Yibei He
Jan 6 at 4:03
$begingroup$
@YibeiHe Good catch! Just fixed it.
$endgroup$
– Erik Parkinson
Jan 6 at 4:05
$begingroup$
would it be $2over 3$ instead of $1over 3$?
$endgroup$
– Yibei He
Jan 6 at 4:03
$begingroup$
would it be $2over 3$ instead of $1over 3$?
$endgroup$
– Yibei He
Jan 6 at 4:03
$begingroup$
@YibeiHe Good catch! Just fixed it.
$endgroup$
– Erik Parkinson
Jan 6 at 4:05
$begingroup$
@YibeiHe Good catch! Just fixed it.
$endgroup$
– Erik Parkinson
Jan 6 at 4:05
add a comment |
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