Convergence in probability












4












$begingroup$


enter image description here



Can anyone tell me how they got the regions $0<epsilon<theta$ and $epsilon >0 $.



Also to clarify, is the last step where it says $lim_{n to infty} P(|Y_n-theta|>epsilon)=0$






probability sequences-and-series convergence random-variables uniform-distribution







share|cite|improve this question











$endgroup$

















    4












    $begingroup$


    enter image description here



    Can anyone tell me how they got the regions $0<epsilon<theta$ and $epsilon >0 $.



    Also to clarify, is the last step where it says $lim_{n to infty} P(|Y_n-theta|>epsilon)=0$






    probability sequences-and-series convergence random-variables uniform-distribution







    share|cite|improve this question











    $endgroup$















      4












      4








      4





      $begingroup$


      enter image description here



      Can anyone tell me how they got the regions $0<epsilon<theta$ and $epsilon >0 $.



      Also to clarify, is the last step where it says $lim_{n to infty} P(|Y_n-theta|>epsilon)=0$






      probability sequences-and-series convergence random-variables uniform-distribution







      share|cite|improve this question











      $endgroup$




      enter image description here



      Can anyone tell me how they got the regions $0<epsilon<theta$ and $epsilon >0 $.



      Also to clarify, is the last step where it says $lim_{n to infty} P(|Y_n-theta|>epsilon)=0$






      probability sequences-and-series convergence random-variables uniform-distribution




      probability sequences-and-series convergence random-variables uniform-distribution






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 17 '18 at 22:54









      Davide Giraudo

      127k17154268




      127k17154268










      asked Jun 5 '16 at 14:23









      HueHueHueHue

      12310




      12310






















          2 Answers
          2






          active

          oldest

          votes


















          3












          $begingroup$

          begin{align}
          mathbb{P}(|Y_n - theta | geq epsilon) &= 1- mathbb{P}(-epsilon leq Y_n - theta leq epsilon) \
          &= 1 - mathbb{P}(Y_n leq theta + epsilon) + mathbb{P}(Y_n leq theta - epsilon) \
          &= 1 - F_{Y_n}(theta + epsilon) + F_{Y_n}(theta-epsilon). \
          &= F_{Y_n}(theta- epsilon).
          end{align}
          Note that the last equality follows from the fact that $theta + epsilon geq theta$ for any $epsilon>0$ and consequently $F_{Y_n}(theta+ epsilon) = 1$. According to the definition of the distribution function of $Y_n$, there are two cases to be considered:



          (1) Assume $0<epsilon<theta$, then $0 < theta-epsilon < theta$ and consequently
          $$mathbb{P}(|Y_n - theta| geq epsilon) = left(frac{theta-epsilon}{theta}right)^n to 0,$$
          as $n to infty$.



          (2) Assume $epsilon > theta$, then $theta - epsilon < 0$ and consequently $F_{Y_n}(theta- epsilon) = 0, forall n geq 1$. Therefore, the result holds in the limit.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I'm a little confused with your statement, Shouldn't $F_{yn}(theta -epsilon)=(frac{theta -epsilon}{theta})^{n}$
            $endgroup$
            – HueHue
            Jun 5 '16 at 16:14








          • 1




            $begingroup$
            Yes ... and $mathbb{P}(|Y_n -theta | geq epsilon) = F_{Y_n}(theta-epsilon)$ as showed above.
            $endgroup$
            – Siron
            Jun 5 '16 at 16:26












          • $begingroup$
            Sorry I meant I was confused with the bit where you said "consider $epsilon geq theta$. In that case..."
            $endgroup$
            – HueHue
            Jun 5 '16 at 16:31










          • $begingroup$
            What exactly are you confused about?
            $endgroup$
            – Siron
            Jun 5 '16 at 16:45



















          -1












          $begingroup$

          Imagine we have $X$ representing the toss of a fair die.



          Then $text{Range}(X) = {1,2,cdots,6}$



          $$P(X = x) = frac16$$



          So:



          $$P(X le 1) = frac16$$
          $$P(X le 2) = frac26$$
          $$P(X le 3) = frac36$$
          $$P(X le 4) = frac46$$
          $$P(X le 5) = frac56$$
          $$P(X le 6) = frac66 = 1$$



          Also:



          $$P(X le 7) = frac66 = 1$$
          $$P(X le 0) = frac06$$
          $$P(X le 0.5) = frac06$$
          $$P(X le -2000) = frac06$$
          $$P(X le 4.5) = frac46$$
          $$P(X > 7) = frac06$$



          Thus:



          The '$varepsilon > theta$' (I'll assume '$varepsilon > 0$' is a typo) is like '$X > 7$'



          The '$0 < varepsilon < theta$' (I'll assume '$varepsilon > 0$' is a typo) is like '$1 < X < 6$'




          Also



          Show that $Y_n$ converges in probability






          share|cite|improve this answer











          $endgroup$













            Your Answer





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            2 Answers
            2






            active

            oldest

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            2 Answers
            2






            active

            oldest

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            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$

            begin{align}
            mathbb{P}(|Y_n - theta | geq epsilon) &= 1- mathbb{P}(-epsilon leq Y_n - theta leq epsilon) \
            &= 1 - mathbb{P}(Y_n leq theta + epsilon) + mathbb{P}(Y_n leq theta - epsilon) \
            &= 1 - F_{Y_n}(theta + epsilon) + F_{Y_n}(theta-epsilon). \
            &= F_{Y_n}(theta- epsilon).
            end{align}
            Note that the last equality follows from the fact that $theta + epsilon geq theta$ for any $epsilon>0$ and consequently $F_{Y_n}(theta+ epsilon) = 1$. According to the definition of the distribution function of $Y_n$, there are two cases to be considered:



            (1) Assume $0<epsilon<theta$, then $0 < theta-epsilon < theta$ and consequently
            $$mathbb{P}(|Y_n - theta| geq epsilon) = left(frac{theta-epsilon}{theta}right)^n to 0,$$
            as $n to infty$.



            (2) Assume $epsilon > theta$, then $theta - epsilon < 0$ and consequently $F_{Y_n}(theta- epsilon) = 0, forall n geq 1$. Therefore, the result holds in the limit.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I'm a little confused with your statement, Shouldn't $F_{yn}(theta -epsilon)=(frac{theta -epsilon}{theta})^{n}$
              $endgroup$
              – HueHue
              Jun 5 '16 at 16:14








            • 1




              $begingroup$
              Yes ... and $mathbb{P}(|Y_n -theta | geq epsilon) = F_{Y_n}(theta-epsilon)$ as showed above.
              $endgroup$
              – Siron
              Jun 5 '16 at 16:26












            • $begingroup$
              Sorry I meant I was confused with the bit where you said "consider $epsilon geq theta$. In that case..."
              $endgroup$
              – HueHue
              Jun 5 '16 at 16:31










            • $begingroup$
              What exactly are you confused about?
              $endgroup$
              – Siron
              Jun 5 '16 at 16:45
















            3












            $begingroup$

            begin{align}
            mathbb{P}(|Y_n - theta | geq epsilon) &= 1- mathbb{P}(-epsilon leq Y_n - theta leq epsilon) \
            &= 1 - mathbb{P}(Y_n leq theta + epsilon) + mathbb{P}(Y_n leq theta - epsilon) \
            &= 1 - F_{Y_n}(theta + epsilon) + F_{Y_n}(theta-epsilon). \
            &= F_{Y_n}(theta- epsilon).
            end{align}
            Note that the last equality follows from the fact that $theta + epsilon geq theta$ for any $epsilon>0$ and consequently $F_{Y_n}(theta+ epsilon) = 1$. According to the definition of the distribution function of $Y_n$, there are two cases to be considered:



            (1) Assume $0<epsilon<theta$, then $0 < theta-epsilon < theta$ and consequently
            $$mathbb{P}(|Y_n - theta| geq epsilon) = left(frac{theta-epsilon}{theta}right)^n to 0,$$
            as $n to infty$.



            (2) Assume $epsilon > theta$, then $theta - epsilon < 0$ and consequently $F_{Y_n}(theta- epsilon) = 0, forall n geq 1$. Therefore, the result holds in the limit.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I'm a little confused with your statement, Shouldn't $F_{yn}(theta -epsilon)=(frac{theta -epsilon}{theta})^{n}$
              $endgroup$
              – HueHue
              Jun 5 '16 at 16:14








            • 1




              $begingroup$
              Yes ... and $mathbb{P}(|Y_n -theta | geq epsilon) = F_{Y_n}(theta-epsilon)$ as showed above.
              $endgroup$
              – Siron
              Jun 5 '16 at 16:26












            • $begingroup$
              Sorry I meant I was confused with the bit where you said "consider $epsilon geq theta$. In that case..."
              $endgroup$
              – HueHue
              Jun 5 '16 at 16:31










            • $begingroup$
              What exactly are you confused about?
              $endgroup$
              – Siron
              Jun 5 '16 at 16:45














            3












            3








            3





            $begingroup$

            begin{align}
            mathbb{P}(|Y_n - theta | geq epsilon) &= 1- mathbb{P}(-epsilon leq Y_n - theta leq epsilon) \
            &= 1 - mathbb{P}(Y_n leq theta + epsilon) + mathbb{P}(Y_n leq theta - epsilon) \
            &= 1 - F_{Y_n}(theta + epsilon) + F_{Y_n}(theta-epsilon). \
            &= F_{Y_n}(theta- epsilon).
            end{align}
            Note that the last equality follows from the fact that $theta + epsilon geq theta$ for any $epsilon>0$ and consequently $F_{Y_n}(theta+ epsilon) = 1$. According to the definition of the distribution function of $Y_n$, there are two cases to be considered:



            (1) Assume $0<epsilon<theta$, then $0 < theta-epsilon < theta$ and consequently
            $$mathbb{P}(|Y_n - theta| geq epsilon) = left(frac{theta-epsilon}{theta}right)^n to 0,$$
            as $n to infty$.



            (2) Assume $epsilon > theta$, then $theta - epsilon < 0$ and consequently $F_{Y_n}(theta- epsilon) = 0, forall n geq 1$. Therefore, the result holds in the limit.






            share|cite|improve this answer











            $endgroup$



            begin{align}
            mathbb{P}(|Y_n - theta | geq epsilon) &= 1- mathbb{P}(-epsilon leq Y_n - theta leq epsilon) \
            &= 1 - mathbb{P}(Y_n leq theta + epsilon) + mathbb{P}(Y_n leq theta - epsilon) \
            &= 1 - F_{Y_n}(theta + epsilon) + F_{Y_n}(theta-epsilon). \
            &= F_{Y_n}(theta- epsilon).
            end{align}
            Note that the last equality follows from the fact that $theta + epsilon geq theta$ for any $epsilon>0$ and consequently $F_{Y_n}(theta+ epsilon) = 1$. According to the definition of the distribution function of $Y_n$, there are two cases to be considered:



            (1) Assume $0<epsilon<theta$, then $0 < theta-epsilon < theta$ and consequently
            $$mathbb{P}(|Y_n - theta| geq epsilon) = left(frac{theta-epsilon}{theta}right)^n to 0,$$
            as $n to infty$.



            (2) Assume $epsilon > theta$, then $theta - epsilon < 0$ and consequently $F_{Y_n}(theta- epsilon) = 0, forall n geq 1$. Therefore, the result holds in the limit.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 17 '18 at 22:18

























            answered Jun 5 '16 at 15:42









            SironSiron

            1,205614




            1,205614












            • $begingroup$
              I'm a little confused with your statement, Shouldn't $F_{yn}(theta -epsilon)=(frac{theta -epsilon}{theta})^{n}$
              $endgroup$
              – HueHue
              Jun 5 '16 at 16:14








            • 1




              $begingroup$
              Yes ... and $mathbb{P}(|Y_n -theta | geq epsilon) = F_{Y_n}(theta-epsilon)$ as showed above.
              $endgroup$
              – Siron
              Jun 5 '16 at 16:26












            • $begingroup$
              Sorry I meant I was confused with the bit where you said "consider $epsilon geq theta$. In that case..."
              $endgroup$
              – HueHue
              Jun 5 '16 at 16:31










            • $begingroup$
              What exactly are you confused about?
              $endgroup$
              – Siron
              Jun 5 '16 at 16:45


















            • $begingroup$
              I'm a little confused with your statement, Shouldn't $F_{yn}(theta -epsilon)=(frac{theta -epsilon}{theta})^{n}$
              $endgroup$
              – HueHue
              Jun 5 '16 at 16:14








            • 1




              $begingroup$
              Yes ... and $mathbb{P}(|Y_n -theta | geq epsilon) = F_{Y_n}(theta-epsilon)$ as showed above.
              $endgroup$
              – Siron
              Jun 5 '16 at 16:26












            • $begingroup$
              Sorry I meant I was confused with the bit where you said "consider $epsilon geq theta$. In that case..."
              $endgroup$
              – HueHue
              Jun 5 '16 at 16:31










            • $begingroup$
              What exactly are you confused about?
              $endgroup$
              – Siron
              Jun 5 '16 at 16:45
















            $begingroup$
            I'm a little confused with your statement, Shouldn't $F_{yn}(theta -epsilon)=(frac{theta -epsilon}{theta})^{n}$
            $endgroup$
            – HueHue
            Jun 5 '16 at 16:14






            $begingroup$
            I'm a little confused with your statement, Shouldn't $F_{yn}(theta -epsilon)=(frac{theta -epsilon}{theta})^{n}$
            $endgroup$
            – HueHue
            Jun 5 '16 at 16:14






            1




            1




            $begingroup$
            Yes ... and $mathbb{P}(|Y_n -theta | geq epsilon) = F_{Y_n}(theta-epsilon)$ as showed above.
            $endgroup$
            – Siron
            Jun 5 '16 at 16:26






            $begingroup$
            Yes ... and $mathbb{P}(|Y_n -theta | geq epsilon) = F_{Y_n}(theta-epsilon)$ as showed above.
            $endgroup$
            – Siron
            Jun 5 '16 at 16:26














            $begingroup$
            Sorry I meant I was confused with the bit where you said "consider $epsilon geq theta$. In that case..."
            $endgroup$
            – HueHue
            Jun 5 '16 at 16:31




            $begingroup$
            Sorry I meant I was confused with the bit where you said "consider $epsilon geq theta$. In that case..."
            $endgroup$
            – HueHue
            Jun 5 '16 at 16:31












            $begingroup$
            What exactly are you confused about?
            $endgroup$
            – Siron
            Jun 5 '16 at 16:45




            $begingroup$
            What exactly are you confused about?
            $endgroup$
            – Siron
            Jun 5 '16 at 16:45











            -1












            $begingroup$

            Imagine we have $X$ representing the toss of a fair die.



            Then $text{Range}(X) = {1,2,cdots,6}$



            $$P(X = x) = frac16$$



            So:



            $$P(X le 1) = frac16$$
            $$P(X le 2) = frac26$$
            $$P(X le 3) = frac36$$
            $$P(X le 4) = frac46$$
            $$P(X le 5) = frac56$$
            $$P(X le 6) = frac66 = 1$$



            Also:



            $$P(X le 7) = frac66 = 1$$
            $$P(X le 0) = frac06$$
            $$P(X le 0.5) = frac06$$
            $$P(X le -2000) = frac06$$
            $$P(X le 4.5) = frac46$$
            $$P(X > 7) = frac06$$



            Thus:



            The '$varepsilon > theta$' (I'll assume '$varepsilon > 0$' is a typo) is like '$X > 7$'



            The '$0 < varepsilon < theta$' (I'll assume '$varepsilon > 0$' is a typo) is like '$1 < X < 6$'




            Also



            Show that $Y_n$ converges in probability






            share|cite|improve this answer











            $endgroup$


















              -1












              $begingroup$

              Imagine we have $X$ representing the toss of a fair die.



              Then $text{Range}(X) = {1,2,cdots,6}$



              $$P(X = x) = frac16$$



              So:



              $$P(X le 1) = frac16$$
              $$P(X le 2) = frac26$$
              $$P(X le 3) = frac36$$
              $$P(X le 4) = frac46$$
              $$P(X le 5) = frac56$$
              $$P(X le 6) = frac66 = 1$$



              Also:



              $$P(X le 7) = frac66 = 1$$
              $$P(X le 0) = frac06$$
              $$P(X le 0.5) = frac06$$
              $$P(X le -2000) = frac06$$
              $$P(X le 4.5) = frac46$$
              $$P(X > 7) = frac06$$



              Thus:



              The '$varepsilon > theta$' (I'll assume '$varepsilon > 0$' is a typo) is like '$X > 7$'



              The '$0 < varepsilon < theta$' (I'll assume '$varepsilon > 0$' is a typo) is like '$1 < X < 6$'




              Also



              Show that $Y_n$ converges in probability






              share|cite|improve this answer











              $endgroup$
















                -1












                -1








                -1





                $begingroup$

                Imagine we have $X$ representing the toss of a fair die.



                Then $text{Range}(X) = {1,2,cdots,6}$



                $$P(X = x) = frac16$$



                So:



                $$P(X le 1) = frac16$$
                $$P(X le 2) = frac26$$
                $$P(X le 3) = frac36$$
                $$P(X le 4) = frac46$$
                $$P(X le 5) = frac56$$
                $$P(X le 6) = frac66 = 1$$



                Also:



                $$P(X le 7) = frac66 = 1$$
                $$P(X le 0) = frac06$$
                $$P(X le 0.5) = frac06$$
                $$P(X le -2000) = frac06$$
                $$P(X le 4.5) = frac46$$
                $$P(X > 7) = frac06$$



                Thus:



                The '$varepsilon > theta$' (I'll assume '$varepsilon > 0$' is a typo) is like '$X > 7$'



                The '$0 < varepsilon < theta$' (I'll assume '$varepsilon > 0$' is a typo) is like '$1 < X < 6$'




                Also



                Show that $Y_n$ converges in probability






                share|cite|improve this answer











                $endgroup$



                Imagine we have $X$ representing the toss of a fair die.



                Then $text{Range}(X) = {1,2,cdots,6}$



                $$P(X = x) = frac16$$



                So:



                $$P(X le 1) = frac16$$
                $$P(X le 2) = frac26$$
                $$P(X le 3) = frac36$$
                $$P(X le 4) = frac46$$
                $$P(X le 5) = frac56$$
                $$P(X le 6) = frac66 = 1$$



                Also:



                $$P(X le 7) = frac66 = 1$$
                $$P(X le 0) = frac06$$
                $$P(X le 0.5) = frac06$$
                $$P(X le -2000) = frac06$$
                $$P(X le 4.5) = frac46$$
                $$P(X > 7) = frac06$$



                Thus:



                The '$varepsilon > theta$' (I'll assume '$varepsilon > 0$' is a typo) is like '$X > 7$'



                The '$0 < varepsilon < theta$' (I'll assume '$varepsilon > 0$' is a typo) is like '$1 < X < 6$'




                Also



                Show that $Y_n$ converges in probability







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Apr 13 '17 at 12:19









                Community♦

                1




                1










                answered Nov 2 '16 at 15:04









                BCLCBCLC

                1




                1






























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                    Musa

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                    Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
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