Mixing
If X and Y are equal almost surely, then they have the same distribution, but the reverse direction is not...
Show that if two random variables X and Y are equal almost surely, then they
have the same distribution. Show that the reverse direction is not correct.
If $2$ r.v are equal a.s. can we write $mathbb P((Xin B)triangle (Yin B))=0$ (How to write this better ?)
then
$mathbb P(Xin B)-mathbb P(Yin B)=mathbb P(Xin B setminus Yin B)le mathbb P((Xin B)triangle (Yin B))=0$
$Longrightarrow P(Xin B)=mathbb P(Yin B)$
but the other direction makes no sense for me, i don't know how this can be true.
probability-theory measure-theory probability-distributions random-variables almost-everywhere
edited May 15 '18 at 10:55
BCLC
1
asked Apr 3 '14 at 12:14
derivative
1,130923
add a comment |
Show that if two random variables X and Y are equal almost surely, then they
have the same distribution. Show that the reverse direction is not correct.
If $2$ r.v are equal a.s. can we write $mathbb P((Xin B)triangle (Yin B))=0$ (How to write this better ?)
then
$mathbb P(Xin B)-mathbb P(Yin B)=mathbb P(Xin B setminus Yin B)le mathbb P((Xin B)triangle (Yin B))=0$
$Longrightarrow P(Xin B)=mathbb P(Yin B)$
but the other direction makes no sense for me, i don't know how this can be true.
probability-theory measure-theory probability-distributions random-variables almost-everywhere
edited May 15 '18 at 10:55
BCLC
1
asked Apr 3 '14 at 12:14
derivative
1,130923
1
FYI, the identity $mathbb P(Xin B)-mathbb P(Yin B)=mathbb P(Xin B setminus Yin B)$ in the question, is wrong in general (and wrong here).
– Did
Sep 14 '15 at 20:21
1
For the first direction you can use: If $Acap R=Acap S$ and $P(A)=1$ then: $P(R)=P(Acap R)=P(Acap S)=P(S)$. Now take $A={X=Y}$, $R={Xin B}$ and $S={Yin B}$. If you throw exactly one fair coin and $X$ is the number of heads that fall and $Y$ the number of tails, then $Xneq Y$ but they have the same distribution.
– drhab
Sep 14 '15 at 20:35
add a comment |
Show that if two random variables X and Y are equal almost surely, then they
have the same distribution. Show that the reverse direction is not correct.
If $2$ r.v are equal a.s. can we write $mathbb P((Xin B)triangle (Yin B))=0$ (How to write this better ?)
then
$mathbb P(Xin B)-mathbb P(Yin B)=mathbb P(Xin B setminus Yin B)le mathbb P((Xin B)triangle (Yin B))=0$
$Longrightarrow P(Xin B)=mathbb P(Yin B)$
but the other direction makes no sense for me, i don't know how this can be true.
probability-theory measure-theory probability-distributions random-variables almost-everywhere
edited May 15 '18 at 10:55
BCLC
1
asked Apr 3 '14 at 12:14
derivative
1,130923
Show that if two random variables X and Y are equal almost surely, then they
have the same distribution. Show that the reverse direction is not correct.
If $2$ r.v are equal a.s. can we write $mathbb P((Xin B)triangle (Yin B))=0$ (How to write this better ?)
then
$mathbb P(Xin B)-mathbb P(Yin B)=mathbb P(Xin B setminus Yin B)le mathbb P((Xin B)triangle (Yin B))=0$
$Longrightarrow P(Xin B)=mathbb P(Yin B)$
but the other direction makes no sense for me, i don't know how this can be true.
probability-theory measure-theory probability-distributions random-variables almost-everywhere
probability-theory measure-theory probability-distributions random-variables almost-everywhere
edited May 15 '18 at 10:55
BCLC
1
asked Apr 3 '14 at 12:14
derivative
1,130923
edited May 15 '18 at 10:55
BCLC
1
asked Apr 3 '14 at 12:14
derivative
1,130923
edited May 15 '18 at 10:55
BCLC
1
edited May 15 '18 at 10:55
BCLC
1
edited May 15 '18 at 10:55
BCLC
1
1
asked Apr 3 '14 at 12:14
derivative
1,130923
asked Apr 3 '14 at 12:14
derivative
1,130923
asked Apr 3 '14 at 12:14
derivative
1,130923
1,130923
1
FYI, the identity $mathbb P(Xin B)-mathbb P(Yin B)=mathbb P(Xin B setminus Yin B)$ in the question, is wrong in general (and wrong here).
– Did
Sep 14 '15 at 20:21
1
For the first direction you can use: If $Acap R=Acap S$ and $P(A)=1$ then: $P(R)=P(Acap R)=P(Acap S)=P(S)$. Now take $A={X=Y}$, $R={Xin B}$ and $S={Yin B}$. If you throw exactly one fair coin and $X$ is the number of heads that fall and $Y$ the number of tails, then $Xneq Y$ but they have the same distribution.
– drhab
Sep 14 '15 at 20:35
add a comment |
1
FYI, the identity $mathbb P(Xin B)-mathbb P(Yin B)=mathbb P(Xin B setminus Yin B)$ in the question, is wrong in general (and wrong here).
– Did
Sep 14 '15 at 20:21
1
For the first direction you can use: If $Acap R=Acap S$ and $P(A)=1$ then: $P(R)=P(Acap R)=P(Acap S)=P(S)$. Now take $A={X=Y}$, $R={Xin B}$ and $S={Yin B}$. If you throw exactly one fair coin and $X$ is the number of heads that fall and $Y$ the number of tails, then $Xneq Y$ but they have the same distribution.
– drhab
Sep 14 '15 at 20:35
1
1
FYI, the identity $mathbb P(Xin B)-mathbb P(Yin B)=mathbb P(Xin B setminus Yin B)$ in the question, is wrong in general (and wrong here).
– Did
Sep 14 '15 at 20:21
FYI, the identity $mathbb P(Xin B)-mathbb P(Yin B)=mathbb P(Xin B setminus Yin B)$ in the question, is wrong in general (and wrong here).
– Did
Sep 14 '15 at 20:21
1
1
For the first direction you can use: If $Acap R=Acap S$ and $P(A)=1$ then: $P(R)=P(Acap R)=P(Acap S)=P(S)$. Now take $A={X=Y}$, $R={Xin B}$ and $S={Yin B}$. If you throw exactly one fair coin and $X$ is the number of heads that fall and $Y$ the number of tails, then $Xneq Y$ but they have the same distribution.
– drhab
Sep 14 '15 at 20:35
For the first direction you can use: If $Acap R=Acap S$ and $P(A)=1$ then: $P(R)=P(Acap R)=P(Acap S)=P(S)$. Now take $A={X=Y}$, $R={Xin B}$ and $S={Yin B}$. If you throw exactly one fair coin and $X$ is the number of heads that fall and $Y$ the number of tails, then $Xneq Y$ but they have the same distribution.
– drhab
Sep 14 '15 at 20:35
add a comment |
3 Answers
3
active
oldest
votes
Take $X$ and $Y$ with probabilities $P(X=1)=P(X=2)=P(Y=1)=P(Y=2)=0.5$ and which are independent. Then $$P(X=Y) = P(X=1, Y=1) + P(X=2,Y=2) =\= P(X=1)P(Y=1)+P(Y=2)P(Y=2)=0.5,$$ meaning that $X=Y$ holds with probability $0.5$, not $1$.
answered Apr 3 '14 at 12:20
5xum
89.6k393161
add a comment |
Consider flipping a fair coin. Then $1_H$ and $1_T$ have the same distribution.
However, they are more than just not equal almost surely ($P(X ne Y)>0$): they are almost surely not equal ($P(X ne Y)=1$)!
answered May 15 '18 at 10:54
BCLC
1
add a comment |
If $X$ is a random variable following uniform $mathcal U(-1,1)$ distribution, then $X$ and $-X$ are identically distributed, but obviously $X$ and $-X$ are not almost surely equal, in fact $P(X=-X)=0$
answered Nov 20 '18 at 23:06
user521337
9781315
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Take $X$ and $Y$ with probabilities $P(X=1)=P(X=2)=P(Y=1)=P(Y=2)=0.5$ and which are independent. Then $$P(X=Y) = P(X=1, Y=1) + P(X=2,Y=2) =\= P(X=1)P(Y=1)+P(Y=2)P(Y=2)=0.5,$$ meaning that $X=Y$ holds with probability $0.5$, not $1$.
answered Apr 3 '14 at 12:20
5xum
89.6k393161
add a comment |
Take $X$ and $Y$ with probabilities $P(X=1)=P(X=2)=P(Y=1)=P(Y=2)=0.5$ and which are independent. Then $$P(X=Y) = P(X=1, Y=1) + P(X=2,Y=2) =\= P(X=1)P(Y=1)+P(Y=2)P(Y=2)=0.5,$$ meaning that $X=Y$ holds with probability $0.5$, not $1$.
answered Apr 3 '14 at 12:20
5xum
89.6k393161
add a comment |
Take $X$ and $Y$ with probabilities $P(X=1)=P(X=2)=P(Y=1)=P(Y=2)=0.5$ and which are independent. Then $$P(X=Y) = P(X=1, Y=1) + P(X=2,Y=2) =\= P(X=1)P(Y=1)+P(Y=2)P(Y=2)=0.5,$$ meaning that $X=Y$ holds with probability $0.5$, not $1$.
answered Apr 3 '14 at 12:20
5xum
89.6k393161
Take $X$ and $Y$ with probabilities $P(X=1)=P(X=2)=P(Y=1)=P(Y=2)=0.5$ and which are independent. Then $$P(X=Y) = P(X=1, Y=1) + P(X=2,Y=2) =\= P(X=1)P(Y=1)+P(Y=2)P(Y=2)=0.5,$$ meaning that $X=Y$ holds with probability $0.5$, not $1$.
answered Apr 3 '14 at 12:20
5xum
89.6k393161
answered Apr 3 '14 at 12:20
5xum
89.6k393161
answered Apr 3 '14 at 12:20
5xum
89.6k393161
answered Apr 3 '14 at 12:20
5xum
89.6k393161
89.6k393161
add a comment |
add a comment |
Consider flipping a fair coin. Then $1_H$ and $1_T$ have the same distribution.
However, they are more than just not equal almost surely ($P(X ne Y)>0$): they are almost surely not equal ($P(X ne Y)=1$)!
answered May 15 '18 at 10:54
BCLC
1
add a comment |
Consider flipping a fair coin. Then $1_H$ and $1_T$ have the same distribution.
However, they are more than just not equal almost surely ($P(X ne Y)>0$): they are almost surely not equal ($P(X ne Y)=1$)!
answered May 15 '18 at 10:54
BCLC
1
add a comment |
Consider flipping a fair coin. Then $1_H$ and $1_T$ have the same distribution.
However, they are more than just not equal almost surely ($P(X ne Y)>0$): they are almost surely not equal ($P(X ne Y)=1$)!
answered May 15 '18 at 10:54
BCLC
1
Consider flipping a fair coin. Then $1_H$ and $1_T$ have the same distribution.
However, they are more than just not equal almost surely ($P(X ne Y)>0$): they are almost surely not equal ($P(X ne Y)=1$)!
answered May 15 '18 at 10:54
BCLC
1
answered May 15 '18 at 10:54
BCLC
1
answered May 15 '18 at 10:54
BCLC
1
answered May 15 '18 at 10:54
BCLC
1
1
add a comment |
add a comment |
If $X$ is a random variable following uniform $mathcal U(-1,1)$ distribution, then $X$ and $-X$ are identically distributed, but obviously $X$ and $-X$ are not almost surely equal, in fact $P(X=-X)=0$
answered Nov 20 '18 at 23:06
user521337
9781315
add a comment |
If $X$ is a random variable following uniform $mathcal U(-1,1)$ distribution, then $X$ and $-X$ are identically distributed, but obviously $X$ and $-X$ are not almost surely equal, in fact $P(X=-X)=0$
answered Nov 20 '18 at 23:06
user521337
9781315
add a comment |
If $X$ is a random variable following uniform $mathcal U(-1,1)$ distribution, then $X$ and $-X$ are identically distributed, but obviously $X$ and $-X$ are not almost surely equal, in fact $P(X=-X)=0$
answered Nov 20 '18 at 23:06
user521337
9781315
If $X$ is a random variable following uniform $mathcal U(-1,1)$ distribution, then $X$ and $-X$ are identically distributed, but obviously $X$ and $-X$ are not almost surely equal, in fact $P(X=-X)=0$
answered Nov 20 '18 at 23:06
user521337
9781315
answered Nov 20 '18 at 23:06
user521337
9781315
answered Nov 20 '18 at 23:06
user521337
9781315
answered Nov 20 '18 at 23:06
user521337
9781315
9781315
add a comment |
add a comment |
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1
FYI, the identity $mathbb P(Xin B)-mathbb P(Yin B)=mathbb P(Xin B setminus Yin B)$ in the question, is wrong in general (and wrong here).
– Did
Sep 14 '15 at 20:21
1
For the first direction you can use: If $Acap R=Acap S$ and $P(A)=1$ then: $P(R)=P(Acap R)=P(Acap S)=P(S)$. Now take $A={X=Y}$, $R={Xin B}$ and $S={Yin B}$. If you throw exactly one fair coin and $X$ is the number of heads that fall and $Y$ the number of tails, then $Xneq Y$ but they have the same distribution.
– drhab
Sep 14 '15 at 20:35