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Do Random Variables have to be bijective? [closed]
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This question has been asked before.
Let me clarify what I don't get about each of the responses.
The first response (Michael Hardy): To me, it seems like this answer is comparing apples to oranges(no offense to the answerer: I am sure I am wrong). Michael defines a new random variable (X1,X2,X3), which looks bijective to me. However, he goes on to say that X1 is not bijective. This makes little sense to me.
The second response (Qiaochu Yuan):
$X^−1$ is not the inverse. It is the inverse image.
I believe the answer's saying that the Real numbers is infinite but the event space could be defined to have finite size, thus bijectivity doesn't apply because the size of codomain is larger than the size of the domain. While this is true, I can always change the set getting mapped to from real numbers to the numbers I map to. Ok, I am cheating a bit. Can anyone elaborate on this?
The third response (M Turgeon):
No, it doesn't have to be bijective - take a constant function.
Ok, this one logically makes sense. Unfortunately, overly constructed contradictions sometimes fail to change my instinct, and this is one of those cases. An example of a use case for a constant random variable would be helpful.
Please note that I am not familiar with Groups/Algebra, Topology or Analysis/Measure Theory.
What I want: I would like answers that expound on the answers to the question I have linked. Addressing the concerns I have mentioned in the question would be nice, but isn't necessary in interest of the response being useful to the community.
probability random-variables
asked Jul 27 at 3:41
Yolo Voe
134
closed as unclear what you're asking by copper.hat, Somos, Jyrki Lahtonen, José Carlos Santos, Mostafa Ayaz Jul 27 at 17:05
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
 |Â
show 2 more comments
up vote
2
down vote
favorite
This question has been asked before.
Let me clarify what I don't get about each of the responses.
The first response (Michael Hardy): To me, it seems like this answer is comparing apples to oranges(no offense to the answerer: I am sure I am wrong). Michael defines a new random variable (X1,X2,X3), which looks bijective to me. However, he goes on to say that X1 is not bijective. This makes little sense to me.
The second response (Qiaochu Yuan):
$X^−1$ is not the inverse. It is the inverse image.
I believe the answer's saying that the Real numbers is infinite but the event space could be defined to have finite size, thus bijectivity doesn't apply because the size of codomain is larger than the size of the domain. While this is true, I can always change the set getting mapped to from real numbers to the numbers I map to. Ok, I am cheating a bit. Can anyone elaborate on this?
The third response (M Turgeon):
No, it doesn't have to be bijective - take a constant function.
Ok, this one logically makes sense. Unfortunately, overly constructed contradictions sometimes fail to change my instinct, and this is one of those cases. An example of a use case for a constant random variable would be helpful.
Please note that I am not familiar with Groups/Algebra, Topology or Analysis/Measure Theory.
What I want: I would like answers that expound on the answers to the question I have linked. Addressing the concerns I have mentioned in the question would be nice, but isn't necessary in interest of the response being useful to the community.
probability random-variables
asked Jul 27 at 3:41
Yolo Voe
134
closed as unclear what you're asking by copper.hat, Somos, Jyrki Lahtonen, José Carlos Santos, Mostafa Ayaz Jul 27 at 17:05
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
What is your question? The function $X_1$ mentioned above is a simple example of a measurable function that it not bijective. Qiaochu's answer is clarifying what appears to be some confusion on your part, which is that $X^-1$ refers to the inverse image, and not an inverse function. A random variable is just a measurable function. You asked if all random variables are bijective and a constant function is a reasonable example. You should add comments to ask for clarification on the other question rather than asking a question about a question.
– copper.hat
Jul 27 at 3:54
Its okay to not know the topics you list in your last line, but do you at the very least understand elementary set theory and the definition of what it means to be bijective? MichaelHardy's answer gives an example of a sample space and a random variable on that sample space such that there are many elementary events corresponding to the same outcome, something which should not happen for a bijective function.
– JMoravitz
Jul 27 at 3:55
In very simple terms, the random variable is a map for sample space to probability. Meaning different elements in the probability space might have the same probability. Take the experiment of tossing two coins. P(0 head)=P(2 heads)=0.25. But if you take the cumulative distribution function of random variable that is in fact bijective on $Sto [0,1]$. where $S$ is the sample space.
– Piyush Divyanakar
Jul 27 at 3:59
2
Also, side note: writing "no offense to the answerer" is more or less incompatible with following it by "The guy [...]."
– Clement C.
Jul 27 at 3:59
@ClementC. Thanks for letting me know. I was debating that when writing the question. In retrospect, I should have called him by his name.
– Yolo Voe
Jul 27 at 4:06
 |Â
show 2 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
This question has been asked before.
Let me clarify what I don't get about each of the responses.
The first response (Michael Hardy): To me, it seems like this answer is comparing apples to oranges(no offense to the answerer: I am sure I am wrong). Michael defines a new random variable (X1,X2,X3), which looks bijective to me. However, he goes on to say that X1 is not bijective. This makes little sense to me.
The second response (Qiaochu Yuan):
$X^−1$ is not the inverse. It is the inverse image.
I believe the answer's saying that the Real numbers is infinite but the event space could be defined to have finite size, thus bijectivity doesn't apply because the size of codomain is larger than the size of the domain. While this is true, I can always change the set getting mapped to from real numbers to the numbers I map to. Ok, I am cheating a bit. Can anyone elaborate on this?
The third response (M Turgeon):
No, it doesn't have to be bijective - take a constant function.
Ok, this one logically makes sense. Unfortunately, overly constructed contradictions sometimes fail to change my instinct, and this is one of those cases. An example of a use case for a constant random variable would be helpful.
Please note that I am not familiar with Groups/Algebra, Topology or Analysis/Measure Theory.
What I want: I would like answers that expound on the answers to the question I have linked. Addressing the concerns I have mentioned in the question would be nice, but isn't necessary in interest of the response being useful to the community.
probability random-variables
asked Jul 27 at 3:41
Yolo Voe
134
This question has been asked before.
Let me clarify what I don't get about each of the responses.
The first response (Michael Hardy): To me, it seems like this answer is comparing apples to oranges(no offense to the answerer: I am sure I am wrong). Michael defines a new random variable (X1,X2,X3), which looks bijective to me. However, he goes on to say that X1 is not bijective. This makes little sense to me.
The second response (Qiaochu Yuan):
$X^−1$ is not the inverse. It is the inverse image.
I believe the answer's saying that the Real numbers is infinite but the event space could be defined to have finite size, thus bijectivity doesn't apply because the size of codomain is larger than the size of the domain. While this is true, I can always change the set getting mapped to from real numbers to the numbers I map to. Ok, I am cheating a bit. Can anyone elaborate on this?
The third response (M Turgeon):
No, it doesn't have to be bijective - take a constant function.
Ok, this one logically makes sense. Unfortunately, overly constructed contradictions sometimes fail to change my instinct, and this is one of those cases. An example of a use case for a constant random variable would be helpful.
Please note that I am not familiar with Groups/Algebra, Topology or Analysis/Measure Theory.
What I want: I would like answers that expound on the answers to the question I have linked. Addressing the concerns I have mentioned in the question would be nice, but isn't necessary in interest of the response being useful to the community.
probability random-variables
asked Jul 27 at 3:41
Yolo Voe
134
edited Jul 27 at 18:31
asked Jul 27 at 3:41
Yolo Voe
134
asked Jul 27 at 3:41
Yolo Voe
134
asked Jul 27 at 3:41
Yolo Voe
134
134
closed as unclear what you're asking by copper.hat, Somos, Jyrki Lahtonen, José Carlos Santos, Mostafa Ayaz Jul 27 at 17:05
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by copper.hat, Somos, Jyrki Lahtonen, José Carlos Santos, Mostafa Ayaz Jul 27 at 17:05
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
What is your question? The function $X_1$ mentioned above is a simple example of a measurable function that it not bijective. Qiaochu's answer is clarifying what appears to be some confusion on your part, which is that $X^-1$ refers to the inverse image, and not an inverse function. A random variable is just a measurable function. You asked if all random variables are bijective and a constant function is a reasonable example. You should add comments to ask for clarification on the other question rather than asking a question about a question.
– copper.hat
Jul 27 at 3:54
Its okay to not know the topics you list in your last line, but do you at the very least understand elementary set theory and the definition of what it means to be bijective? MichaelHardy's answer gives an example of a sample space and a random variable on that sample space such that there are many elementary events corresponding to the same outcome, something which should not happen for a bijective function.
– JMoravitz
Jul 27 at 3:55
In very simple terms, the random variable is a map for sample space to probability. Meaning different elements in the probability space might have the same probability. Take the experiment of tossing two coins. P(0 head)=P(2 heads)=0.25. But if you take the cumulative distribution function of random variable that is in fact bijective on $Sto [0,1]$. where $S$ is the sample space.
– Piyush Divyanakar
Jul 27 at 3:59
2
Also, side note: writing "no offense to the answerer" is more or less incompatible with following it by "The guy [...]."
– Clement C.
Jul 27 at 3:59
@ClementC. Thanks for letting me know. I was debating that when writing the question. In retrospect, I should have called him by his name.
– Yolo Voe
Jul 27 at 4:06
 |Â
show 2 more comments
1
What is your question? The function $X_1$ mentioned above is a simple example of a measurable function that it not bijective. Qiaochu's answer is clarifying what appears to be some confusion on your part, which is that $X^-1$ refers to the inverse image, and not an inverse function. A random variable is just a measurable function. You asked if all random variables are bijective and a constant function is a reasonable example. You should add comments to ask for clarification on the other question rather than asking a question about a question.
– copper.hat
Jul 27 at 3:54
Its okay to not know the topics you list in your last line, but do you at the very least understand elementary set theory and the definition of what it means to be bijective? MichaelHardy's answer gives an example of a sample space and a random variable on that sample space such that there are many elementary events corresponding to the same outcome, something which should not happen for a bijective function.
– JMoravitz
Jul 27 at 3:55
In very simple terms, the random variable is a map for sample space to probability. Meaning different elements in the probability space might have the same probability. Take the experiment of tossing two coins. P(0 head)=P(2 heads)=0.25. But if you take the cumulative distribution function of random variable that is in fact bijective on $Sto [0,1]$. where $S$ is the sample space.
– Piyush Divyanakar
Jul 27 at 3:59
2
Also, side note: writing "no offense to the answerer" is more or less incompatible with following it by "The guy [...]."
– Clement C.
Jul 27 at 3:59
@ClementC. Thanks for letting me know. I was debating that when writing the question. In retrospect, I should have called him by his name.
– Yolo Voe
Jul 27 at 4:06
1
1
What is your question? The function $X_1$ mentioned above is a simple example of a measurable function that it not bijective. Qiaochu's answer is clarifying what appears to be some confusion on your part, which is that $X^-1$ refers to the inverse image, and not an inverse function. A random variable is just a measurable function. You asked if all random variables are bijective and a constant function is a reasonable example. You should add comments to ask for clarification on the other question rather than asking a question about a question.
– copper.hat
Jul 27 at 3:54
What is your question? The function $X_1$ mentioned above is a simple example of a measurable function that it not bijective. Qiaochu's answer is clarifying what appears to be some confusion on your part, which is that $X^-1$ refers to the inverse image, and not an inverse function. A random variable is just a measurable function. You asked if all random variables are bijective and a constant function is a reasonable example. You should add comments to ask for clarification on the other question rather than asking a question about a question.
– copper.hat
Jul 27 at 3:54
Its okay to not know the topics you list in your last line, but do you at the very least understand elementary set theory and the definition of what it means to be bijective? MichaelHardy's answer gives an example of a sample space and a random variable on that sample space such that there are many elementary events corresponding to the same outcome, something which should not happen for a bijective function.
– JMoravitz
Jul 27 at 3:55
Its okay to not know the topics you list in your last line, but do you at the very least understand elementary set theory and the definition of what it means to be bijective? MichaelHardy's answer gives an example of a sample space and a random variable on that sample space such that there are many elementary events corresponding to the same outcome, something which should not happen for a bijective function.
– JMoravitz
Jul 27 at 3:55
In very simple terms, the random variable is a map for sample space to probability. Meaning different elements in the probability space might have the same probability. Take the experiment of tossing two coins. P(0 head)=P(2 heads)=0.25. But if you take the cumulative distribution function of random variable that is in fact bijective on $Sto [0,1]$. where $S$ is the sample space.
– Piyush Divyanakar
Jul 27 at 3:59
In very simple terms, the random variable is a map for sample space to probability. Meaning different elements in the probability space might have the same probability. Take the experiment of tossing two coins. P(0 head)=P(2 heads)=0.25. But if you take the cumulative distribution function of random variable that is in fact bijective on $Sto [0,1]$. where $S$ is the sample space.
– Piyush Divyanakar
Jul 27 at 3:59
2
2
Also, side note: writing "no offense to the answerer" is more or less incompatible with following it by "The guy [...]."
– Clement C.
Jul 27 at 3:59
Also, side note: writing "no offense to the answerer" is more or less incompatible with following it by "The guy [...]."
– Clement C.
Jul 27 at 3:59
@ClementC. Thanks for letting me know. I was debating that when writing the question. In retrospect, I should have called him by his name.
– Yolo Voe
Jul 27 at 4:06
@ClementC. Thanks for letting me know. I was debating that when writing the question. In retrospect, I should have called him by his name.
– Yolo Voe
Jul 27 at 4:06
 |Â
show 2 more comments
1 Answer
1
active
oldest
votes
up vote
6
down vote
accepted
When you have a random vector $(X_1,X_2,X_3)$, each coordinate $X_i$ is a random variable. Michael Hardy is saying that the random variable $X_1$ is not injective.
You are right; you can always change the codomain to make the random variable surjective.
Consider the opposite problem; suppose that the probability space $Omega$ is infinite, while the range of the random variable is finite. Something like, flip a coin over and over until the result is heads, and let $X=1$ if the number of flips is odd and $X=0$ otherwise. The sample space is $H,TH,TTH,TTTH,dots$, but the range is $0,1$. This random variable is essentially not bijective.Imagine an urn with $100$ white balls and $101$ black balls. Over and over, remove two random balls from the urn. If they are the same color, throw in a white ball. If they are different colors, throw in a black ball. Continue (for 200 draws total) until there is only one ball left.
You would agree that the color of the last ball is a random variable, yes? It is the result of a series of random decisions. However, you can prove that no matter what sequence of random draws you make, the last ball will always be black! This is a constant random variable which arises "in the real world."
edited Jul 27 at 8:07
Niu
31
answered Jul 27 at 3:59
Mike Earnest
14.9k11644
I understand the second point; I need to think more about the third point, but I think I can understand it after some more thinking. I would like to ask some clarification for the first point: don't we care about the random vector being injective (instead of the random variables that comprise the random vector)?
– Yolo Voe
Jul 27 at 4:12
@YoloVoe Michael Hardy's goal was to find a counterexample, and $X_1$ proved to be a counter example. In light of this, it does matter whether the vector $(X_1,X_2,X_3)$ is injective. Just forget about the other variables $X_2,X_3$, and consider $X_1$ on the same probability space, and there is your non-injective counterexample.
– Mike Earnest
Jul 27 at 4:16
@YoloVoe Also, there are less silly examples of important constant random variables. For example, the Strong Law of Large numbers implies that when $X_1,X_2,dots,$ is a sequence of independent coin flips, then $lim_ntoinfty (text# heads in X_1,X_2,dots,X_n)/n$ is (almost surely) constant and equal to $frac12$. The limit depends on a sequence of random variables, so is itself random, but it is almost surely constant; most sequences result in 1/2. The point is that constant random variables are not an arbitrary counterexample, they show up in important and unexpected places.
– Mike Earnest
Jul 27 at 4:20
1
Thanks! Your explanation and comments were really helpful and got to the crux of the matter and clarified my confusion! :)
– Yolo Voe
Jul 27 at 4:25
1
@YoloVoe Happy to talk about probability theory with an interested audience!
– Mike Earnest
Jul 27 at 4:35
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
When you have a random vector $(X_1,X_2,X_3)$, each coordinate $X_i$ is a random variable. Michael Hardy is saying that the random variable $X_1$ is not injective.
You are right; you can always change the codomain to make the random variable surjective.
Consider the opposite problem; suppose that the probability space $Omega$ is infinite, while the range of the random variable is finite. Something like, flip a coin over and over until the result is heads, and let $X=1$ if the number of flips is odd and $X=0$ otherwise. The sample space is $H,TH,TTH,TTTH,dots$, but the range is $0,1$. This random variable is essentially not bijective.Imagine an urn with $100$ white balls and $101$ black balls. Over and over, remove two random balls from the urn. If they are the same color, throw in a white ball. If they are different colors, throw in a black ball. Continue (for 200 draws total) until there is only one ball left.
You would agree that the color of the last ball is a random variable, yes? It is the result of a series of random decisions. However, you can prove that no matter what sequence of random draws you make, the last ball will always be black! This is a constant random variable which arises "in the real world."
edited Jul 27 at 8:07
Niu
31
answered Jul 27 at 3:59
Mike Earnest
14.9k11644
I understand the second point; I need to think more about the third point, but I think I can understand it after some more thinking. I would like to ask some clarification for the first point: don't we care about the random vector being injective (instead of the random variables that comprise the random vector)?
– Yolo Voe
Jul 27 at 4:12
@YoloVoe Michael Hardy's goal was to find a counterexample, and $X_1$ proved to be a counter example. In light of this, it does matter whether the vector $(X_1,X_2,X_3)$ is injective. Just forget about the other variables $X_2,X_3$, and consider $X_1$ on the same probability space, and there is your non-injective counterexample.
– Mike Earnest
Jul 27 at 4:16
@YoloVoe Also, there are less silly examples of important constant random variables. For example, the Strong Law of Large numbers implies that when $X_1,X_2,dots,$ is a sequence of independent coin flips, then $lim_ntoinfty (text# heads in X_1,X_2,dots,X_n)/n$ is (almost surely) constant and equal to $frac12$. The limit depends on a sequence of random variables, so is itself random, but it is almost surely constant; most sequences result in 1/2. The point is that constant random variables are not an arbitrary counterexample, they show up in important and unexpected places.
– Mike Earnest
Jul 27 at 4:20
1
Thanks! Your explanation and comments were really helpful and got to the crux of the matter and clarified my confusion! :)
– Yolo Voe
Jul 27 at 4:25
1
@YoloVoe Happy to talk about probability theory with an interested audience!
– Mike Earnest
Jul 27 at 4:35
add a comment |Â
up vote
6
down vote
accepted
When you have a random vector $(X_1,X_2,X_3)$, each coordinate $X_i$ is a random variable. Michael Hardy is saying that the random variable $X_1$ is not injective.
You are right; you can always change the codomain to make the random variable surjective.
Consider the opposite problem; suppose that the probability space $Omega$ is infinite, while the range of the random variable is finite. Something like, flip a coin over and over until the result is heads, and let $X=1$ if the number of flips is odd and $X=0$ otherwise. The sample space is $H,TH,TTH,TTTH,dots$, but the range is $0,1$. This random variable is essentially not bijective.Imagine an urn with $100$ white balls and $101$ black balls. Over and over, remove two random balls from the urn. If they are the same color, throw in a white ball. If they are different colors, throw in a black ball. Continue (for 200 draws total) until there is only one ball left.
You would agree that the color of the last ball is a random variable, yes? It is the result of a series of random decisions. However, you can prove that no matter what sequence of random draws you make, the last ball will always be black! This is a constant random variable which arises "in the real world."
edited Jul 27 at 8:07
Niu
31
answered Jul 27 at 3:59
Mike Earnest
14.9k11644
I understand the second point; I need to think more about the third point, but I think I can understand it after some more thinking. I would like to ask some clarification for the first point: don't we care about the random vector being injective (instead of the random variables that comprise the random vector)?
– Yolo Voe
Jul 27 at 4:12
@YoloVoe Michael Hardy's goal was to find a counterexample, and $X_1$ proved to be a counter example. In light of this, it does matter whether the vector $(X_1,X_2,X_3)$ is injective. Just forget about the other variables $X_2,X_3$, and consider $X_1$ on the same probability space, and there is your non-injective counterexample.
– Mike Earnest
Jul 27 at 4:16
@YoloVoe Also, there are less silly examples of important constant random variables. For example, the Strong Law of Large numbers implies that when $X_1,X_2,dots,$ is a sequence of independent coin flips, then $lim_ntoinfty (text# heads in X_1,X_2,dots,X_n)/n$ is (almost surely) constant and equal to $frac12$. The limit depends on a sequence of random variables, so is itself random, but it is almost surely constant; most sequences result in 1/2. The point is that constant random variables are not an arbitrary counterexample, they show up in important and unexpected places.
– Mike Earnest
Jul 27 at 4:20
1
Thanks! Your explanation and comments were really helpful and got to the crux of the matter and clarified my confusion! :)
– Yolo Voe
Jul 27 at 4:25
1
@YoloVoe Happy to talk about probability theory with an interested audience!
– Mike Earnest
Jul 27 at 4:35
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
When you have a random vector $(X_1,X_2,X_3)$, each coordinate $X_i$ is a random variable. Michael Hardy is saying that the random variable $X_1$ is not injective.
You are right; you can always change the codomain to make the random variable surjective.
Consider the opposite problem; suppose that the probability space $Omega$ is infinite, while the range of the random variable is finite. Something like, flip a coin over and over until the result is heads, and let $X=1$ if the number of flips is odd and $X=0$ otherwise. The sample space is $H,TH,TTH,TTTH,dots$, but the range is $0,1$. This random variable is essentially not bijective.Imagine an urn with $100$ white balls and $101$ black balls. Over and over, remove two random balls from the urn. If they are the same color, throw in a white ball. If they are different colors, throw in a black ball. Continue (for 200 draws total) until there is only one ball left.
You would agree that the color of the last ball is a random variable, yes? It is the result of a series of random decisions. However, you can prove that no matter what sequence of random draws you make, the last ball will always be black! This is a constant random variable which arises "in the real world."
edited Jul 27 at 8:07
Niu
31
answered Jul 27 at 3:59
Mike Earnest
14.9k11644
When you have a random vector $(X_1,X_2,X_3)$, each coordinate $X_i$ is a random variable. Michael Hardy is saying that the random variable $X_1$ is not injective.
You are right; you can always change the codomain to make the random variable surjective.
Consider the opposite problem; suppose that the probability space $Omega$ is infinite, while the range of the random variable is finite. Something like, flip a coin over and over until the result is heads, and let $X=1$ if the number of flips is odd and $X=0$ otherwise. The sample space is $H,TH,TTH,TTTH,dots$, but the range is $0,1$. This random variable is essentially not bijective.Imagine an urn with $100$ white balls and $101$ black balls. Over and over, remove two random balls from the urn. If they are the same color, throw in a white ball. If they are different colors, throw in a black ball. Continue (for 200 draws total) until there is only one ball left.
You would agree that the color of the last ball is a random variable, yes? It is the result of a series of random decisions. However, you can prove that no matter what sequence of random draws you make, the last ball will always be black! This is a constant random variable which arises "in the real world."
edited Jul 27 at 8:07
Niu
31
answered Jul 27 at 3:59
Mike Earnest
14.9k11644
edited Jul 27 at 8:07
Niu
31
edited Jul 27 at 8:07
Niu
31
edited Jul 27 at 8:07
Niu
31
31
answered Jul 27 at 3:59
Mike Earnest
14.9k11644
answered Jul 27 at 3:59
Mike Earnest
14.9k11644
answered Jul 27 at 3:59
Mike Earnest
14.9k11644
14.9k11644
I understand the second point; I need to think more about the third point, but I think I can understand it after some more thinking. I would like to ask some clarification for the first point: don't we care about the random vector being injective (instead of the random variables that comprise the random vector)?
– Yolo Voe
Jul 27 at 4:12
@YoloVoe Michael Hardy's goal was to find a counterexample, and $X_1$ proved to be a counter example. In light of this, it does matter whether the vector $(X_1,X_2,X_3)$ is injective. Just forget about the other variables $X_2,X_3$, and consider $X_1$ on the same probability space, and there is your non-injective counterexample.
– Mike Earnest
Jul 27 at 4:16
@YoloVoe Also, there are less silly examples of important constant random variables. For example, the Strong Law of Large numbers implies that when $X_1,X_2,dots,$ is a sequence of independent coin flips, then $lim_ntoinfty (text# heads in X_1,X_2,dots,X_n)/n$ is (almost surely) constant and equal to $frac12$. The limit depends on a sequence of random variables, so is itself random, but it is almost surely constant; most sequences result in 1/2. The point is that constant random variables are not an arbitrary counterexample, they show up in important and unexpected places.
– Mike Earnest
Jul 27 at 4:20
1
Thanks! Your explanation and comments were really helpful and got to the crux of the matter and clarified my confusion! :)
– Yolo Voe
Jul 27 at 4:25
1
@YoloVoe Happy to talk about probability theory with an interested audience!
– Mike Earnest
Jul 27 at 4:35
add a comment |Â
I understand the second point; I need to think more about the third point, but I think I can understand it after some more thinking. I would like to ask some clarification for the first point: don't we care about the random vector being injective (instead of the random variables that comprise the random vector)?
– Yolo Voe
Jul 27 at 4:12
@YoloVoe Michael Hardy's goal was to find a counterexample, and $X_1$ proved to be a counter example. In light of this, it does matter whether the vector $(X_1,X_2,X_3)$ is injective. Just forget about the other variables $X_2,X_3$, and consider $X_1$ on the same probability space, and there is your non-injective counterexample.
– Mike Earnest
Jul 27 at 4:16
@YoloVoe Also, there are less silly examples of important constant random variables. For example, the Strong Law of Large numbers implies that when $X_1,X_2,dots,$ is a sequence of independent coin flips, then $lim_ntoinfty (text# heads in X_1,X_2,dots,X_n)/n$ is (almost surely) constant and equal to $frac12$. The limit depends on a sequence of random variables, so is itself random, but it is almost surely constant; most sequences result in 1/2. The point is that constant random variables are not an arbitrary counterexample, they show up in important and unexpected places.
– Mike Earnest
Jul 27 at 4:20
1
Thanks! Your explanation and comments were really helpful and got to the crux of the matter and clarified my confusion! :)
– Yolo Voe
Jul 27 at 4:25
1
@YoloVoe Happy to talk about probability theory with an interested audience!
– Mike Earnest
Jul 27 at 4:35
I understand the second point; I need to think more about the third point, but I think I can understand it after some more thinking. I would like to ask some clarification for the first point: don't we care about the random vector being injective (instead of the random variables that comprise the random vector)?
– Yolo Voe
Jul 27 at 4:12
I understand the second point; I need to think more about the third point, but I think I can understand it after some more thinking. I would like to ask some clarification for the first point: don't we care about the random vector being injective (instead of the random variables that comprise the random vector)?
– Yolo Voe
Jul 27 at 4:12
@YoloVoe Michael Hardy's goal was to find a counterexample, and $X_1$ proved to be a counter example. In light of this, it does matter whether the vector $(X_1,X_2,X_3)$ is injective. Just forget about the other variables $X_2,X_3$, and consider $X_1$ on the same probability space, and there is your non-injective counterexample.
– Mike Earnest
Jul 27 at 4:16
@YoloVoe Michael Hardy's goal was to find a counterexample, and $X_1$ proved to be a counter example. In light of this, it does matter whether the vector $(X_1,X_2,X_3)$ is injective. Just forget about the other variables $X_2,X_3$, and consider $X_1$ on the same probability space, and there is your non-injective counterexample.
– Mike Earnest
Jul 27 at 4:16
@YoloVoe Also, there are less silly examples of important constant random variables. For example, the Strong Law of Large numbers implies that when $X_1,X_2,dots,$ is a sequence of independent coin flips, then $lim_ntoinfty (text# heads in X_1,X_2,dots,X_n)/n$ is (almost surely) constant and equal to $frac12$. The limit depends on a sequence of random variables, so is itself random, but it is almost surely constant; most sequences result in 1/2. The point is that constant random variables are not an arbitrary counterexample, they show up in important and unexpected places.
– Mike Earnest
Jul 27 at 4:20
@YoloVoe Also, there are less silly examples of important constant random variables. For example, the Strong Law of Large numbers implies that when $X_1,X_2,dots,$ is a sequence of independent coin flips, then $lim_ntoinfty (text# heads in X_1,X_2,dots,X_n)/n$ is (almost surely) constant and equal to $frac12$. The limit depends on a sequence of random variables, so is itself random, but it is almost surely constant; most sequences result in 1/2. The point is that constant random variables are not an arbitrary counterexample, they show up in important and unexpected places.
– Mike Earnest
Jul 27 at 4:20
1
1
Thanks! Your explanation and comments were really helpful and got to the crux of the matter and clarified my confusion! :)
– Yolo Voe
Jul 27 at 4:25
Thanks! Your explanation and comments were really helpful and got to the crux of the matter and clarified my confusion! :)
– Yolo Voe
Jul 27 at 4:25
1
1
@YoloVoe Happy to talk about probability theory with an interested audience!
– Mike Earnest
Jul 27 at 4:35
@YoloVoe Happy to talk about probability theory with an interested audience!
– Mike Earnest
Jul 27 at 4:35
add a comment |Â
1
What is your question? The function $X_1$ mentioned above is a simple example of a measurable function that it not bijective. Qiaochu's answer is clarifying what appears to be some confusion on your part, which is that $X^-1$ refers to the inverse image, and not an inverse function. A random variable is just a measurable function. You asked if all random variables are bijective and a constant function is a reasonable example. You should add comments to ask for clarification on the other question rather than asking a question about a question.
– copper.hat
Jul 27 at 3:54
Its okay to not know the topics you list in your last line, but do you at the very least understand elementary set theory and the definition of what it means to be bijective? MichaelHardy's answer gives an example of a sample space and a random variable on that sample space such that there are many elementary events corresponding to the same outcome, something which should not happen for a bijective function.
– JMoravitz
Jul 27 at 3:55
In very simple terms, the random variable is a map for sample space to probability. Meaning different elements in the probability space might have the same probability. Take the experiment of tossing two coins. P(0 head)=P(2 heads)=0.25. But if you take the cumulative distribution function of random variable that is in fact bijective on $Sto [0,1]$. where $S$ is the sample space.
– Piyush Divyanakar
Jul 27 at 3:59
2
Also, side note: writing "no offense to the answerer" is more or less incompatible with following it by "The guy [...]."
– Clement C.
Jul 27 at 3:59
@ClementC. Thanks for letting me know. I was debating that when writing the question. In retrospect, I should have called him by his name.
– Yolo Voe
Jul 27 at 4:06