Random variable and maximum metric.
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Let $Omega := [0,1] times [0,1]$. Consider on $sigma$-algebra Borel sets with Lebesgue measure .Let $X(w)$ describe the distance (in maximum metric) between the point and the nearest corner of the square. Is $X(w)$ a random variable? If so, find probability distribution and CDF.
I need help,beacuse it is quite difficult.
probability probability-distributions
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up vote
1
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Let $Omega := [0,1] times [0,1]$. Consider on $sigma$-algebra Borel sets with Lebesgue measure .Let $X(w)$ describe the distance (in maximum metric) between the point and the nearest corner of the square. Is $X(w)$ a random variable? If so, find probability distribution and CDF.
I need help,beacuse it is quite difficult.
probability probability-distributions
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $Omega := [0,1] times [0,1]$. Consider on $sigma$-algebra Borel sets with Lebesgue measure .Let $X(w)$ describe the distance (in maximum metric) between the point and the nearest corner of the square. Is $X(w)$ a random variable? If so, find probability distribution and CDF.
I need help,beacuse it is quite difficult.
probability probability-distributions
Let $Omega := [0,1] times [0,1]$. Consider on $sigma$-algebra Borel sets with Lebesgue measure .Let $X(w)$ describe the distance (in maximum metric) between the point and the nearest corner of the square. Is $X(w)$ a random variable? If so, find probability distribution and CDF.
I need help,beacuse it is quite difficult.
probability probability-distributions
probability probability-distributions
asked Nov 18 at 13:35
PabloZ392
1386
1386
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1 Answer
1
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oldest
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1
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accepted
Hint
Let $omega =(omega _1,omega _2)in Omega $. Then,$$X(omega )=infBig{sup{omega _1,omega _2}, sup{omega _1,1-omega _2 }, sup{1-omega _1,1-omega _2}, sup{1-omega _2,omega_2}Big}.$$
Added
Set
begin{align*}
X_1(omega )&=sup{omega _1,omega _2}\
X_2(omega )&=sup{omega _1,1-omega _2}\
X_3(omega )&=sup{1-omega _1,omega _2}\
X_4(omega )&=sup{1-omega _1,1-omega _2}.
end{align*}
$$mathbb P{Xleq x}=mathbb P{Xboldsymbol 1_{[0,1/2]^2}leq x}+mathbb P{Xboldsymbol 1_{[0,1/2]times [1/2,1]}leq x}+mathbb P{Xboldsymbol 1_{[1/2,1]times [0,1/2]}leq x}+mathbb P{Xboldsymbol 1_{[1/2,1]^2}leq x}.$$
Now $$mathbb P{Xboldsymbol 1_{[0,1/2]^2}leq x}=mathbb P{omega in [0,1/2]^2mid X_1leq x}=mathbb P{omegain [0,1/2]^2mid omega _1leq x, omega _2leq x }=mathbb P{omega _1in [0,1/2]mid omega _1leq x}mathbb P{omega _2in [0,1/2]mid omega _2leq x},$$
the last inequality come from independence of $omega _1mapsto omega _1$ and $omega _2mapsto omega _2$. Notice that that r.v. follow uniform law on $[0,1/2]$.
The rest goes the same.
We know, that $X(w) in[0,1/2]$. Now I have to "take" any Borel set $B in[0,1/2]. Is $X^{-1}(B)$ measurable? I think so
– PabloZ392
Nov 18 at 19:52
I have problem with probability distribution and CDF.
– PabloZ392
Nov 18 at 23:12
@PabloZ392: I edited my answer.
– idm
Nov 19 at 14:31
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Hint
Let $omega =(omega _1,omega _2)in Omega $. Then,$$X(omega )=infBig{sup{omega _1,omega _2}, sup{omega _1,1-omega _2 }, sup{1-omega _1,1-omega _2}, sup{1-omega _2,omega_2}Big}.$$
Added
Set
begin{align*}
X_1(omega )&=sup{omega _1,omega _2}\
X_2(omega )&=sup{omega _1,1-omega _2}\
X_3(omega )&=sup{1-omega _1,omega _2}\
X_4(omega )&=sup{1-omega _1,1-omega _2}.
end{align*}
$$mathbb P{Xleq x}=mathbb P{Xboldsymbol 1_{[0,1/2]^2}leq x}+mathbb P{Xboldsymbol 1_{[0,1/2]times [1/2,1]}leq x}+mathbb P{Xboldsymbol 1_{[1/2,1]times [0,1/2]}leq x}+mathbb P{Xboldsymbol 1_{[1/2,1]^2}leq x}.$$
Now $$mathbb P{Xboldsymbol 1_{[0,1/2]^2}leq x}=mathbb P{omega in [0,1/2]^2mid X_1leq x}=mathbb P{omegain [0,1/2]^2mid omega _1leq x, omega _2leq x }=mathbb P{omega _1in [0,1/2]mid omega _1leq x}mathbb P{omega _2in [0,1/2]mid omega _2leq x},$$
the last inequality come from independence of $omega _1mapsto omega _1$ and $omega _2mapsto omega _2$. Notice that that r.v. follow uniform law on $[0,1/2]$.
The rest goes the same.
We know, that $X(w) in[0,1/2]$. Now I have to "take" any Borel set $B in[0,1/2]. Is $X^{-1}(B)$ measurable? I think so
– PabloZ392
Nov 18 at 19:52
I have problem with probability distribution and CDF.
– PabloZ392
Nov 18 at 23:12
@PabloZ392: I edited my answer.
– idm
Nov 19 at 14:31
add a comment |
up vote
1
down vote
accepted
Hint
Let $omega =(omega _1,omega _2)in Omega $. Then,$$X(omega )=infBig{sup{omega _1,omega _2}, sup{omega _1,1-omega _2 }, sup{1-omega _1,1-omega _2}, sup{1-omega _2,omega_2}Big}.$$
Added
Set
begin{align*}
X_1(omega )&=sup{omega _1,omega _2}\
X_2(omega )&=sup{omega _1,1-omega _2}\
X_3(omega )&=sup{1-omega _1,omega _2}\
X_4(omega )&=sup{1-omega _1,1-omega _2}.
end{align*}
$$mathbb P{Xleq x}=mathbb P{Xboldsymbol 1_{[0,1/2]^2}leq x}+mathbb P{Xboldsymbol 1_{[0,1/2]times [1/2,1]}leq x}+mathbb P{Xboldsymbol 1_{[1/2,1]times [0,1/2]}leq x}+mathbb P{Xboldsymbol 1_{[1/2,1]^2}leq x}.$$
Now $$mathbb P{Xboldsymbol 1_{[0,1/2]^2}leq x}=mathbb P{omega in [0,1/2]^2mid X_1leq x}=mathbb P{omegain [0,1/2]^2mid omega _1leq x, omega _2leq x }=mathbb P{omega _1in [0,1/2]mid omega _1leq x}mathbb P{omega _2in [0,1/2]mid omega _2leq x},$$
the last inequality come from independence of $omega _1mapsto omega _1$ and $omega _2mapsto omega _2$. Notice that that r.v. follow uniform law on $[0,1/2]$.
The rest goes the same.
We know, that $X(w) in[0,1/2]$. Now I have to "take" any Borel set $B in[0,1/2]. Is $X^{-1}(B)$ measurable? I think so
– PabloZ392
Nov 18 at 19:52
I have problem with probability distribution and CDF.
– PabloZ392
Nov 18 at 23:12
@PabloZ392: I edited my answer.
– idm
Nov 19 at 14:31
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Hint
Let $omega =(omega _1,omega _2)in Omega $. Then,$$X(omega )=infBig{sup{omega _1,omega _2}, sup{omega _1,1-omega _2 }, sup{1-omega _1,1-omega _2}, sup{1-omega _2,omega_2}Big}.$$
Added
Set
begin{align*}
X_1(omega )&=sup{omega _1,omega _2}\
X_2(omega )&=sup{omega _1,1-omega _2}\
X_3(omega )&=sup{1-omega _1,omega _2}\
X_4(omega )&=sup{1-omega _1,1-omega _2}.
end{align*}
$$mathbb P{Xleq x}=mathbb P{Xboldsymbol 1_{[0,1/2]^2}leq x}+mathbb P{Xboldsymbol 1_{[0,1/2]times [1/2,1]}leq x}+mathbb P{Xboldsymbol 1_{[1/2,1]times [0,1/2]}leq x}+mathbb P{Xboldsymbol 1_{[1/2,1]^2}leq x}.$$
Now $$mathbb P{Xboldsymbol 1_{[0,1/2]^2}leq x}=mathbb P{omega in [0,1/2]^2mid X_1leq x}=mathbb P{omegain [0,1/2]^2mid omega _1leq x, omega _2leq x }=mathbb P{omega _1in [0,1/2]mid omega _1leq x}mathbb P{omega _2in [0,1/2]mid omega _2leq x},$$
the last inequality come from independence of $omega _1mapsto omega _1$ and $omega _2mapsto omega _2$. Notice that that r.v. follow uniform law on $[0,1/2]$.
The rest goes the same.
Hint
Let $omega =(omega _1,omega _2)in Omega $. Then,$$X(omega )=infBig{sup{omega _1,omega _2}, sup{omega _1,1-omega _2 }, sup{1-omega _1,1-omega _2}, sup{1-omega _2,omega_2}Big}.$$
Added
Set
begin{align*}
X_1(omega )&=sup{omega _1,omega _2}\
X_2(omega )&=sup{omega _1,1-omega _2}\
X_3(omega )&=sup{1-omega _1,omega _2}\
X_4(omega )&=sup{1-omega _1,1-omega _2}.
end{align*}
$$mathbb P{Xleq x}=mathbb P{Xboldsymbol 1_{[0,1/2]^2}leq x}+mathbb P{Xboldsymbol 1_{[0,1/2]times [1/2,1]}leq x}+mathbb P{Xboldsymbol 1_{[1/2,1]times [0,1/2]}leq x}+mathbb P{Xboldsymbol 1_{[1/2,1]^2}leq x}.$$
Now $$mathbb P{Xboldsymbol 1_{[0,1/2]^2}leq x}=mathbb P{omega in [0,1/2]^2mid X_1leq x}=mathbb P{omegain [0,1/2]^2mid omega _1leq x, omega _2leq x }=mathbb P{omega _1in [0,1/2]mid omega _1leq x}mathbb P{omega _2in [0,1/2]mid omega _2leq x},$$
the last inequality come from independence of $omega _1mapsto omega _1$ and $omega _2mapsto omega _2$. Notice that that r.v. follow uniform law on $[0,1/2]$.
The rest goes the same.
edited Nov 19 at 14:31
answered Nov 18 at 13:41
idm
8,54621345
8,54621345
We know, that $X(w) in[0,1/2]$. Now I have to "take" any Borel set $B in[0,1/2]. Is $X^{-1}(B)$ measurable? I think so
– PabloZ392
Nov 18 at 19:52
I have problem with probability distribution and CDF.
– PabloZ392
Nov 18 at 23:12
@PabloZ392: I edited my answer.
– idm
Nov 19 at 14:31
add a comment |
We know, that $X(w) in[0,1/2]$. Now I have to "take" any Borel set $B in[0,1/2]. Is $X^{-1}(B)$ measurable? I think so
– PabloZ392
Nov 18 at 19:52
I have problem with probability distribution and CDF.
– PabloZ392
Nov 18 at 23:12
@PabloZ392: I edited my answer.
– idm
Nov 19 at 14:31
We know, that $X(w) in[0,1/2]$. Now I have to "take" any Borel set $B in[0,1/2]. Is $X^{-1}(B)$ measurable? I think so
– PabloZ392
Nov 18 at 19:52
We know, that $X(w) in[0,1/2]$. Now I have to "take" any Borel set $B in[0,1/2]. Is $X^{-1}(B)$ measurable? I think so
– PabloZ392
Nov 18 at 19:52
I have problem with probability distribution and CDF.
– PabloZ392
Nov 18 at 23:12
I have problem with probability distribution and CDF.
– PabloZ392
Nov 18 at 23:12
@PabloZ392: I edited my answer.
– idm
Nov 19 at 14:31
@PabloZ392: I edited my answer.
– idm
Nov 19 at 14:31
add a comment |
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