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Proving if $X$ is a random variable and $f$ is a continuous function, then $f(X)$ is a random variable
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I'm reading "Fundamental of Mathematical Statistics" by Gupta and Kapoor and the authors make the claim that "if $X$ is a random variable and $f$ is a continuous function, then $f(X)$ is a random variable". The proof has been skipped as it was beyond the scope of the textbook.
However, here's my question: There's not much information about given about $f$, so I assume $f$ is a real valued function on $mathbbR$. So clearly the composition $f(X)$ is also a random variable. The hypothesis that $f$ is continuous is never used. So, does it mean that $f$ does not have be continuous? Or am I wrong somewhere?
statistics
asked Jul 15 at 4:27
Ashish K
440312
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up vote
1
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I'm reading "Fundamental of Mathematical Statistics" by Gupta and Kapoor and the authors make the claim that "if $X$ is a random variable and $f$ is a continuous function, then $f(X)$ is a random variable". The proof has been skipped as it was beyond the scope of the textbook.
However, here's my question: There's not much information about given about $f$, so I assume $f$ is a real valued function on $mathbbR$. So clearly the composition $f(X)$ is also a random variable. The hypothesis that $f$ is continuous is never used. So, does it mean that $f$ does not have be continuous? Or am I wrong somewhere?
statistics
asked Jul 15 at 4:27
Ashish K
440312
Could you provide the text in which this statement was found?
– Kevin Li
Jul 15 at 4:30
@KevinLi I've edited my question details.
– Ashish K
Jul 15 at 4:33
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm reading "Fundamental of Mathematical Statistics" by Gupta and Kapoor and the authors make the claim that "if $X$ is a random variable and $f$ is a continuous function, then $f(X)$ is a random variable". The proof has been skipped as it was beyond the scope of the textbook.
However, here's my question: There's not much information about given about $f$, so I assume $f$ is a real valued function on $mathbbR$. So clearly the composition $f(X)$ is also a random variable. The hypothesis that $f$ is continuous is never used. So, does it mean that $f$ does not have be continuous? Or am I wrong somewhere?
statistics
asked Jul 15 at 4:27
Ashish K
440312
I'm reading "Fundamental of Mathematical Statistics" by Gupta and Kapoor and the authors make the claim that "if $X$ is a random variable and $f$ is a continuous function, then $f(X)$ is a random variable". The proof has been skipped as it was beyond the scope of the textbook.
However, here's my question: There's not much information about given about $f$, so I assume $f$ is a real valued function on $mathbbR$. So clearly the composition $f(X)$ is also a random variable. The hypothesis that $f$ is continuous is never used. So, does it mean that $f$ does not have be continuous? Or am I wrong somewhere?
statistics
asked Jul 15 at 4:27
Ashish K
440312
edited Jul 15 at 4:33
asked Jul 15 at 4:27
Ashish K
440312
asked Jul 15 at 4:27
Ashish K
440312
asked Jul 15 at 4:27
Ashish K
440312
440312
Could you provide the text in which this statement was found?
– Kevin Li
Jul 15 at 4:30
@KevinLi I've edited my question details.
– Ashish K
Jul 15 at 4:33
add a comment |Â
Could you provide the text in which this statement was found?
– Kevin Li
Jul 15 at 4:30
@KevinLi I've edited my question details.
– Ashish K
Jul 15 at 4:33
Could you provide the text in which this statement was found?
– Kevin Li
Jul 15 at 4:30
Could you provide the text in which this statement was found?
– Kevin Li
Jul 15 at 4:30
@KevinLi I've edited my question details.
– Ashish K
Jul 15 at 4:33
@KevinLi I've edited my question details.
– Ashish K
Jul 15 at 4:33
add a comment |Â
3 Answers
3
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3
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The proof is most likely out of the scope of most undergraduate mathematical statistics/probability courses.
Formally, a random variable is a measurable function from $ Omega $ to $ E $ where
$ (Omega, mathcalF, P) $ is a probability space and $ (E, mathcalE) $
is a measurable space.
A well-known result in measure-theoretic probability is that if $ X : Omega to E $, and $ f : E to E $ is
continuous, then $ f circ X $ is also a measurable function with respect to the previous probability space. This is the result to which the authors are referring.
answered Jul 15 at 4:37
Kevin Li
22829
2
"A well-known result in measure-theoretic probability is..." Assuming $mathcal E$ is the Borel sigma-algebra on $E$, I guess.
– Did
Jul 15 at 5:00
add a comment |Â
up vote
1
down vote
Composition of measurable functions results in a measurable function.
So it is enough for $f$ to be measurable as a function $f:(mathbb R,mathcal B)to(mathbb R,mathcal B)$ where $mathcal B$ denotes the Borel $sigma$-algebra.
So it boils down to the statement that every continuous function is Borel-measurable.
If $tau$ denotes the topology on $mathbb R$ then the continuity of $f$ assures that $f^-1(tau)subseteqtau$.
From this it follows directly that: $$sigma(f^-1(tau))subseteqsigma(tau)=mathcal B$$
It is true in general that $$sigma(f^-1(mathcal A))=f^-1(sigma(mathcal A))$$
For a proof of that see here.
Applying this we find: $$f^-1(mathcal B)=f^-1(sigma(tau))subseteqsigma(tau)=mathcal B$$stating exactly that $f$ is measurable.
answered Jul 15 at 5:24
drhab
86.7k541118
add a comment |Â
up vote
0
down vote
Yes. You're right that $f$ does not have to be continuous. It just merely needs to be measurable. You can think of measurable functions as functions that can be approximated by continuous functions (Lusin's theorem). Measurable functions arise in the study of Lebesgue integration.
This case might be familiar to you as an undergraduate student (like me):
The Lebesgue Theorem for Riemann integration says that a function $f$ is Riemann integrable on $[a,b]$ if and only if the set of the discontinuities of $f$ is a null-set. So, this gives you a class of functions that are not necessarily continuous, but they're measurable.
answered Jul 15 at 4:43
stressed out
3,1081431
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
The proof is most likely out of the scope of most undergraduate mathematical statistics/probability courses.
Formally, a random variable is a measurable function from $ Omega $ to $ E $ where
$ (Omega, mathcalF, P) $ is a probability space and $ (E, mathcalE) $
is a measurable space.
A well-known result in measure-theoretic probability is that if $ X : Omega to E $, and $ f : E to E $ is
continuous, then $ f circ X $ is also a measurable function with respect to the previous probability space. This is the result to which the authors are referring.
answered Jul 15 at 4:37
Kevin Li
22829
2
"A well-known result in measure-theoretic probability is..." Assuming $mathcal E$ is the Borel sigma-algebra on $E$, I guess.
– Did
Jul 15 at 5:00
add a comment |Â
up vote
3
down vote
The proof is most likely out of the scope of most undergraduate mathematical statistics/probability courses.
Formally, a random variable is a measurable function from $ Omega $ to $ E $ where
$ (Omega, mathcalF, P) $ is a probability space and $ (E, mathcalE) $
is a measurable space.
A well-known result in measure-theoretic probability is that if $ X : Omega to E $, and $ f : E to E $ is
continuous, then $ f circ X $ is also a measurable function with respect to the previous probability space. This is the result to which the authors are referring.
answered Jul 15 at 4:37
Kevin Li
22829
2
"A well-known result in measure-theoretic probability is..." Assuming $mathcal E$ is the Borel sigma-algebra on $E$, I guess.
– Did
Jul 15 at 5:00
add a comment |Â
up vote
3
down vote
up vote
3
down vote
The proof is most likely out of the scope of most undergraduate mathematical statistics/probability courses.
Formally, a random variable is a measurable function from $ Omega $ to $ E $ where
$ (Omega, mathcalF, P) $ is a probability space and $ (E, mathcalE) $
is a measurable space.
A well-known result in measure-theoretic probability is that if $ X : Omega to E $, and $ f : E to E $ is
continuous, then $ f circ X $ is also a measurable function with respect to the previous probability space. This is the result to which the authors are referring.
answered Jul 15 at 4:37
Kevin Li
22829
The proof is most likely out of the scope of most undergraduate mathematical statistics/probability courses.
Formally, a random variable is a measurable function from $ Omega $ to $ E $ where
$ (Omega, mathcalF, P) $ is a probability space and $ (E, mathcalE) $
is a measurable space.
A well-known result in measure-theoretic probability is that if $ X : Omega to E $, and $ f : E to E $ is
continuous, then $ f circ X $ is also a measurable function with respect to the previous probability space. This is the result to which the authors are referring.
answered Jul 15 at 4:37
Kevin Li
22829
answered Jul 15 at 4:37
Kevin Li
22829
answered Jul 15 at 4:37
Kevin Li
22829
answered Jul 15 at 4:37
Kevin Li
22829
22829
2
"A well-known result in measure-theoretic probability is..." Assuming $mathcal E$ is the Borel sigma-algebra on $E$, I guess.
– Did
Jul 15 at 5:00
add a comment |Â
2
"A well-known result in measure-theoretic probability is..." Assuming $mathcal E$ is the Borel sigma-algebra on $E$, I guess.
– Did
Jul 15 at 5:00
2
2
"A well-known result in measure-theoretic probability is..." Assuming $mathcal E$ is the Borel sigma-algebra on $E$, I guess.
– Did
Jul 15 at 5:00
"A well-known result in measure-theoretic probability is..." Assuming $mathcal E$ is the Borel sigma-algebra on $E$, I guess.
– Did
Jul 15 at 5:00
add a comment |Â
up vote
1
down vote
Composition of measurable functions results in a measurable function.
So it is enough for $f$ to be measurable as a function $f:(mathbb R,mathcal B)to(mathbb R,mathcal B)$ where $mathcal B$ denotes the Borel $sigma$-algebra.
So it boils down to the statement that every continuous function is Borel-measurable.
If $tau$ denotes the topology on $mathbb R$ then the continuity of $f$ assures that $f^-1(tau)subseteqtau$.
From this it follows directly that: $$sigma(f^-1(tau))subseteqsigma(tau)=mathcal B$$
It is true in general that $$sigma(f^-1(mathcal A))=f^-1(sigma(mathcal A))$$
For a proof of that see here.
Applying this we find: $$f^-1(mathcal B)=f^-1(sigma(tau))subseteqsigma(tau)=mathcal B$$stating exactly that $f$ is measurable.
answered Jul 15 at 5:24
drhab
86.7k541118
add a comment |Â
up vote
1
down vote
Composition of measurable functions results in a measurable function.
So it is enough for $f$ to be measurable as a function $f:(mathbb R,mathcal B)to(mathbb R,mathcal B)$ where $mathcal B$ denotes the Borel $sigma$-algebra.
So it boils down to the statement that every continuous function is Borel-measurable.
If $tau$ denotes the topology on $mathbb R$ then the continuity of $f$ assures that $f^-1(tau)subseteqtau$.
From this it follows directly that: $$sigma(f^-1(tau))subseteqsigma(tau)=mathcal B$$
It is true in general that $$sigma(f^-1(mathcal A))=f^-1(sigma(mathcal A))$$
For a proof of that see here.
Applying this we find: $$f^-1(mathcal B)=f^-1(sigma(tau))subseteqsigma(tau)=mathcal B$$stating exactly that $f$ is measurable.
answered Jul 15 at 5:24
drhab
86.7k541118
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Composition of measurable functions results in a measurable function.
So it is enough for $f$ to be measurable as a function $f:(mathbb R,mathcal B)to(mathbb R,mathcal B)$ where $mathcal B$ denotes the Borel $sigma$-algebra.
So it boils down to the statement that every continuous function is Borel-measurable.
If $tau$ denotes the topology on $mathbb R$ then the continuity of $f$ assures that $f^-1(tau)subseteqtau$.
From this it follows directly that: $$sigma(f^-1(tau))subseteqsigma(tau)=mathcal B$$
It is true in general that $$sigma(f^-1(mathcal A))=f^-1(sigma(mathcal A))$$
For a proof of that see here.
Applying this we find: $$f^-1(mathcal B)=f^-1(sigma(tau))subseteqsigma(tau)=mathcal B$$stating exactly that $f$ is measurable.
answered Jul 15 at 5:24
drhab
86.7k541118
Composition of measurable functions results in a measurable function.
So it is enough for $f$ to be measurable as a function $f:(mathbb R,mathcal B)to(mathbb R,mathcal B)$ where $mathcal B$ denotes the Borel $sigma$-algebra.
So it boils down to the statement that every continuous function is Borel-measurable.
If $tau$ denotes the topology on $mathbb R$ then the continuity of $f$ assures that $f^-1(tau)subseteqtau$.
From this it follows directly that: $$sigma(f^-1(tau))subseteqsigma(tau)=mathcal B$$
It is true in general that $$sigma(f^-1(mathcal A))=f^-1(sigma(mathcal A))$$
For a proof of that see here.
Applying this we find: $$f^-1(mathcal B)=f^-1(sigma(tau))subseteqsigma(tau)=mathcal B$$stating exactly that $f$ is measurable.
answered Jul 15 at 5:24
drhab
86.7k541118
edited Jul 15 at 5:39
answered Jul 15 at 5:24
drhab
86.7k541118
answered Jul 15 at 5:24
drhab
86.7k541118
answered Jul 15 at 5:24
drhab
86.7k541118
86.7k541118
add a comment |Â
add a comment |Â
up vote
0
down vote
Yes. You're right that $f$ does not have to be continuous. It just merely needs to be measurable. You can think of measurable functions as functions that can be approximated by continuous functions (Lusin's theorem). Measurable functions arise in the study of Lebesgue integration.
This case might be familiar to you as an undergraduate student (like me):
The Lebesgue Theorem for Riemann integration says that a function $f$ is Riemann integrable on $[a,b]$ if and only if the set of the discontinuities of $f$ is a null-set. So, this gives you a class of functions that are not necessarily continuous, but they're measurable.
answered Jul 15 at 4:43
stressed out
3,1081431
add a comment |Â
up vote
0
down vote
Yes. You're right that $f$ does not have to be continuous. It just merely needs to be measurable. You can think of measurable functions as functions that can be approximated by continuous functions (Lusin's theorem). Measurable functions arise in the study of Lebesgue integration.
This case might be familiar to you as an undergraduate student (like me):
The Lebesgue Theorem for Riemann integration says that a function $f$ is Riemann integrable on $[a,b]$ if and only if the set of the discontinuities of $f$ is a null-set. So, this gives you a class of functions that are not necessarily continuous, but they're measurable.
answered Jul 15 at 4:43
stressed out
3,1081431
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Yes. You're right that $f$ does not have to be continuous. It just merely needs to be measurable. You can think of measurable functions as functions that can be approximated by continuous functions (Lusin's theorem). Measurable functions arise in the study of Lebesgue integration.
This case might be familiar to you as an undergraduate student (like me):
The Lebesgue Theorem for Riemann integration says that a function $f$ is Riemann integrable on $[a,b]$ if and only if the set of the discontinuities of $f$ is a null-set. So, this gives you a class of functions that are not necessarily continuous, but they're measurable.
answered Jul 15 at 4:43
stressed out
3,1081431
Yes. You're right that $f$ does not have to be continuous. It just merely needs to be measurable. You can think of measurable functions as functions that can be approximated by continuous functions (Lusin's theorem). Measurable functions arise in the study of Lebesgue integration.
This case might be familiar to you as an undergraduate student (like me):
The Lebesgue Theorem for Riemann integration says that a function $f$ is Riemann integrable on $[a,b]$ if and only if the set of the discontinuities of $f$ is a null-set. So, this gives you a class of functions that are not necessarily continuous, but they're measurable.
answered Jul 15 at 4:43
stressed out
3,1081431
edited Jul 15 at 4:49
answered Jul 15 at 4:43
stressed out
3,1081431
answered Jul 15 at 4:43
stressed out
3,1081431
answered Jul 15 at 4:43
stressed out
3,1081431
3,1081431
add a comment |Â
add a comment |Â
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Could you provide the text in which this statement was found?
– Kevin Li
Jul 15 at 4:30
@KevinLi I've edited my question details.
– Ashish K
Jul 15 at 4:33