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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asked 2 days ago

PabloZ392

1356

add a comment | 

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

share|cite|improve this question

asked 2 days ago

PabloZ392

1356

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

share|cite|improve this question

asked 2 days ago

PabloZ392

1356

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

share|cite|improve this question

asked 2 days ago

PabloZ392

1356

share|cite|improve this question

asked 2 days ago

PabloZ392

1356

share|cite|improve this question

share|cite|improve this question

asked 2 days ago

PabloZ392

1356

asked 2 days ago

PabloZ392

1356

asked 2 days ago

PabloZ392

1356

1356

add a comment | 

add a comment | 

1 Answer
1

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.

share|cite|improve this answer

edited yesterday

answered 2 days ago

idm

8,53021245

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I have problem with probability distribution and CDF.
  – PabloZ392
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @PabloZ392: I edited my answer.
  – idm
  yesterday
  

  
  
  
  

add a comment | 

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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.

share|cite|improve this answer

edited yesterday

answered 2 days ago

idm

8,53021245

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I have problem with probability distribution and CDF.
  – PabloZ392
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @PabloZ392: I edited my answer.
  – idm
  yesterday
  

  
  
  
  

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.

share|cite|improve this answer

edited yesterday

answered 2 days ago

idm

8,53021245

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I have problem with probability distribution and CDF.
  – PabloZ392
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @PabloZ392: I edited my answer.
  – idm
  yesterday
  

  
  
  
  

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.

share|cite|improve this answer

edited yesterday

answered 2 days ago

idm

8,53021245

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.

share|cite|improve this answer

edited yesterday

answered 2 days ago

idm

8,53021245

share|cite|improve this answer

share|cite|improve this answer

edited yesterday

edited yesterday

edited yesterday

answered 2 days ago

idm

8,53021245

answered 2 days ago

idm

8,53021245

answered 2 days ago

idm

8,53021245

8,53021245

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I have problem with probability distribution and CDF.
  – PabloZ392
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @PabloZ392: I edited my answer.
  – idm
  yesterday
  

  
  
  
  

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
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I have problem with probability distribution and CDF.
  – PabloZ392
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @PabloZ392: I edited my answer.
  – idm
  yesterday
  

  
  
  
  

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
2 days ago

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
2 days ago

I have problem with probability distribution and CDF.
– PabloZ392
2 days ago

I have problem with probability distribution and CDF.
– PabloZ392
2 days ago

@PabloZ392: I edited my answer.
– idm
yesterday

@PabloZ392: I edited my answer.
– idm
yesterday

add a comment | 

 

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