Mixing
Let $mathcal F=${$emptyset, Omega$} prove that $X : Omega to Bbb R$ is a random variable if $X$ is constant.
$begingroup$
I have to prove this:
Let $(Omega, mathcal F)$ be a measurable space, $mathcal F=${$emptyset, Omega$} prove that $X : Omega to Bbb R$ is a random variable if and only if $X$ is constant.
I've tried using that $X$ is a random variable iff $X^{-1}B in mathcal F$ if $B$ is a Borel set but I cannot conclude anything, also I´ve tried with the other definition: $X$ is a random variable iff $(X le x) in mathcal F$ $ forall x in R$.
Any hint or idea about what definition of random variable should I use?
probability probability-theory measure-theory random-variables
edited Jan 11 at 18:57
6005
36.2k751125
asked Jan 11 at 18:45
Alex TurnerAlex Turner
336210
$endgroup$
add a comment |
$begingroup$
I have to prove this:
Let $(Omega, mathcal F)$ be a measurable space, $mathcal F=${$emptyset, Omega$} prove that $X : Omega to Bbb R$ is a random variable if and only if $X$ is constant.
I've tried using that $X$ is a random variable iff $X^{-1}B in mathcal F$ if $B$ is a Borel set but I cannot conclude anything, also I´ve tried with the other definition: $X$ is a random variable iff $(X le x) in mathcal F$ $ forall x in R$.
Any hint or idea about what definition of random variable should I use?
probability probability-theory measure-theory random-variables
edited Jan 11 at 18:57
6005
36.2k751125
asked Jan 11 at 18:45
Alex TurnerAlex Turner
336210
$endgroup$
1
$begingroup$
Using the inequality version is the approach I would use.
$endgroup$
– Joe
Jan 11 at 18:54
add a comment |
$begingroup$
I have to prove this:
Let $(Omega, mathcal F)$ be a measurable space, $mathcal F=${$emptyset, Omega$} prove that $X : Omega to Bbb R$ is a random variable if and only if $X$ is constant.
I've tried using that $X$ is a random variable iff $X^{-1}B in mathcal F$ if $B$ is a Borel set but I cannot conclude anything, also I´ve tried with the other definition: $X$ is a random variable iff $(X le x) in mathcal F$ $ forall x in R$.
Any hint or idea about what definition of random variable should I use?
probability probability-theory measure-theory random-variables
edited Jan 11 at 18:57
6005
36.2k751125
asked Jan 11 at 18:45
Alex TurnerAlex Turner
336210
$endgroup$
I have to prove this:
Let $(Omega, mathcal F)$ be a measurable space, $mathcal F=${$emptyset, Omega$} prove that $X : Omega to Bbb R$ is a random variable if and only if $X$ is constant.
I've tried using that $X$ is a random variable iff $X^{-1}B in mathcal F$ if $B$ is a Borel set but I cannot conclude anything, also I´ve tried with the other definition: $X$ is a random variable iff $(X le x) in mathcal F$ $ forall x in R$.
Any hint or idea about what definition of random variable should I use?
probability probability-theory measure-theory random-variables
probability probability-theory measure-theory random-variables
edited Jan 11 at 18:57
6005
36.2k751125
asked Jan 11 at 18:45
Alex TurnerAlex Turner
336210
edited Jan 11 at 18:57
6005
36.2k751125
asked Jan 11 at 18:45
Alex TurnerAlex Turner
336210
edited Jan 11 at 18:57
6005
36.2k751125
edited Jan 11 at 18:57
6005
36.2k751125
edited Jan 11 at 18:57
6005
36.2k751125
36.2k751125
asked Jan 11 at 18:45
Alex TurnerAlex Turner
336210
asked Jan 11 at 18:45
Alex TurnerAlex Turner
336210
asked Jan 11 at 18:45
Alex TurnerAlex Turner
336210
336210
1
$begingroup$
Using the inequality version is the approach I would use.
$endgroup$
– Joe
Jan 11 at 18:54
add a comment |
1
$begingroup$
Using the inequality version is the approach I would use.
$endgroup$
– Joe
Jan 11 at 18:54
1
1
$begingroup$
Using the inequality version is the approach I would use.
$endgroup$
– Joe
Jan 11 at 18:54
$begingroup$
Using the inequality version is the approach I would use.
$endgroup$
– Joe
Jan 11 at 18:54
add a comment |
2 Answers
2
active
oldest
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$begingroup$
I've tried using that $X$ is a random variable iff $X^{-1}B in mathcal F$ if $B$ is a Borel set but I cannot conclude anything, also I've tried with the other definition: $X$ is a random variable iff $(X le x) in mathcal F$ $ forall x in R$
You are on the right track. Either idea will work, but let's go with your second idea. Since $mathcal{F} = {varnothing, Omega}$ (there are only two measurable sets), the statement "${X le x} in mathcal{F}$" is equivalent to
$$
{X le x} = varnothing qquadtext{OR}qquad {X le x } = Omega
$$
And that has to be true for all $x$. Now (assuming $Omega$ nonempty), let $a in Omega$, and consider $f(a)$. What can you say about
$$
{X le f(a)}?
$$
Also, what can you say about
$$
{X le y}
$$
for any $y < f(a)$?
answered Jan 11 at 18:55
60056005
36.2k751125
$endgroup$
add a comment |
$begingroup$
Use the second definition. Let $c=sup {x:{Xleq x} neq Omega}$. Verify that $0<c<infty$. Note that ${Xleq x}=Omega$ for $x>c$ and conclude that $Xleq c$. Next,note that ${Xleq x}=emptyset$ for all $x<c$. Conclude that $X=c$.
answered Jan 11 at 23:57
Kavi Rama MurthyKavi Rama Murthy
59k42161
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
I've tried using that $X$ is a random variable iff $X^{-1}B in mathcal F$ if $B$ is a Borel set but I cannot conclude anything, also I've tried with the other definition: $X$ is a random variable iff $(X le x) in mathcal F$ $ forall x in R$
You are on the right track. Either idea will work, but let's go with your second idea. Since $mathcal{F} = {varnothing, Omega}$ (there are only two measurable sets), the statement "${X le x} in mathcal{F}$" is equivalent to
$$
{X le x} = varnothing qquadtext{OR}qquad {X le x } = Omega
$$
And that has to be true for all $x$. Now (assuming $Omega$ nonempty), let $a in Omega$, and consider $f(a)$. What can you say about
$$
{X le f(a)}?
$$
Also, what can you say about
$$
{X le y}
$$
for any $y < f(a)$?
answered Jan 11 at 18:55
60056005
36.2k751125
$endgroup$
add a comment |
$begingroup$
I've tried using that $X$ is a random variable iff $X^{-1}B in mathcal F$ if $B$ is a Borel set but I cannot conclude anything, also I've tried with the other definition: $X$ is a random variable iff $(X le x) in mathcal F$ $ forall x in R$
You are on the right track. Either idea will work, but let's go with your second idea. Since $mathcal{F} = {varnothing, Omega}$ (there are only two measurable sets), the statement "${X le x} in mathcal{F}$" is equivalent to
$$
{X le x} = varnothing qquadtext{OR}qquad {X le x } = Omega
$$
And that has to be true for all $x$. Now (assuming $Omega$ nonempty), let $a in Omega$, and consider $f(a)$. What can you say about
$$
{X le f(a)}?
$$
Also, what can you say about
$$
{X le y}
$$
for any $y < f(a)$?
answered Jan 11 at 18:55
60056005
36.2k751125
$endgroup$
add a comment |
$begingroup$
I've tried using that $X$ is a random variable iff $X^{-1}B in mathcal F$ if $B$ is a Borel set but I cannot conclude anything, also I've tried with the other definition: $X$ is a random variable iff $(X le x) in mathcal F$ $ forall x in R$
You are on the right track. Either idea will work, but let's go with your second idea. Since $mathcal{F} = {varnothing, Omega}$ (there are only two measurable sets), the statement "${X le x} in mathcal{F}$" is equivalent to
$$
{X le x} = varnothing qquadtext{OR}qquad {X le x } = Omega
$$
And that has to be true for all $x$. Now (assuming $Omega$ nonempty), let $a in Omega$, and consider $f(a)$. What can you say about
$$
{X le f(a)}?
$$
Also, what can you say about
$$
{X le y}
$$
for any $y < f(a)$?
answered Jan 11 at 18:55
60056005
36.2k751125
$endgroup$
I've tried using that $X$ is a random variable iff $X^{-1}B in mathcal F$ if $B$ is a Borel set but I cannot conclude anything, also I've tried with the other definition: $X$ is a random variable iff $(X le x) in mathcal F$ $ forall x in R$
You are on the right track. Either idea will work, but let's go with your second idea. Since $mathcal{F} = {varnothing, Omega}$ (there are only two measurable sets), the statement "${X le x} in mathcal{F}$" is equivalent to
$$
{X le x} = varnothing qquadtext{OR}qquad {X le x } = Omega
$$
And that has to be true for all $x$. Now (assuming $Omega$ nonempty), let $a in Omega$, and consider $f(a)$. What can you say about
$$
{X le f(a)}?
$$
Also, what can you say about
$$
{X le y}
$$
for any $y < f(a)$?
answered Jan 11 at 18:55
60056005
36.2k751125
answered Jan 11 at 18:55
60056005
36.2k751125
answered Jan 11 at 18:55
60056005
36.2k751125
answered Jan 11 at 18:55
60056005
36.2k751125
36.2k751125
add a comment |
add a comment |
$begingroup$
Use the second definition. Let $c=sup {x:{Xleq x} neq Omega}$. Verify that $0<c<infty$. Note that ${Xleq x}=Omega$ for $x>c$ and conclude that $Xleq c$. Next,note that ${Xleq x}=emptyset$ for all $x<c$. Conclude that $X=c$.
answered Jan 11 at 23:57
Kavi Rama MurthyKavi Rama Murthy
59k42161
$endgroup$
add a comment |
$begingroup$
Use the second definition. Let $c=sup {x:{Xleq x} neq Omega}$. Verify that $0<c<infty$. Note that ${Xleq x}=Omega$ for $x>c$ and conclude that $Xleq c$. Next,note that ${Xleq x}=emptyset$ for all $x<c$. Conclude that $X=c$.
answered Jan 11 at 23:57
Kavi Rama MurthyKavi Rama Murthy
59k42161
$endgroup$
add a comment |
$begingroup$
Use the second definition. Let $c=sup {x:{Xleq x} neq Omega}$. Verify that $0<c<infty$. Note that ${Xleq x}=Omega$ for $x>c$ and conclude that $Xleq c$. Next,note that ${Xleq x}=emptyset$ for all $x<c$. Conclude that $X=c$.
answered Jan 11 at 23:57
Kavi Rama MurthyKavi Rama Murthy
59k42161
$endgroup$
Use the second definition. Let $c=sup {x:{Xleq x} neq Omega}$. Verify that $0<c<infty$. Note that ${Xleq x}=Omega$ for $x>c$ and conclude that $Xleq c$. Next,note that ${Xleq x}=emptyset$ for all $x<c$. Conclude that $X=c$.
answered Jan 11 at 23:57
Kavi Rama MurthyKavi Rama Murthy
59k42161
answered Jan 11 at 23:57
Kavi Rama MurthyKavi Rama Murthy
59k42161
answered Jan 11 at 23:57
Kavi Rama MurthyKavi Rama Murthy
59k42161
answered Jan 11 at 23:57
Kavi Rama MurthyKavi Rama Murthy
59k42161
59k42161
add a comment |
add a comment |
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1
$begingroup$
Using the inequality version is the approach I would use.
$endgroup$
– Joe
Jan 11 at 18:54