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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    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
























      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













      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













      share|cite|improve this question











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      asked Nov 18 at 13:35









      PabloZ392

      1386




      1386






















          1 Answer
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          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























          • 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











          Your Answer





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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























          • 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















          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























          • 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













          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














          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














          share|cite|improve this answer



          share|cite|improve this answer








          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


















          • 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


















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